Dr Yukawa talked about Rotation and Spin (or physical concepts of Moment of Inertia, Angular Momentum) at several occasions in his 3-day lecture.

Day One

9. 質点と剛体 Point Mass and Rigid Body
10. 角運動量の問題 Problems of Angular Momentum
11.ひずみと応力について About Strain and Stress
　　
Day Two

3. ニュートン力学における空間 Space of Newton Mechanics
6.見かけの力と本当の力 Fictitious Force and Real Force
7.マッハの解釈 Mach's interpretation

Day Three

12. 入れ物（時空）と中身（物質） Container (Space-Time) and Contents (Matters)

He raises a question or doubt every time he talks about rotation. To understand why he raises a question or doubt it is worthwhile to review the commonly accepted concepts of Moment of Inertia like in the following review (I have added Torque) and if possible with questions and doubt like Dr Yukawa.

1. What is Moment of Inertia ?

From wiki <Moment of inertia> (12-July-2014)

"
Moment of inertia is the mass property of a rigid body that determines the torque needed for a desired angular acceleration about an axis of rotation.

"
To understand this brief explanation we need to have at least some idea about Moment, Inertia, Torque and Angular Acceleration. Moment of inertia determines the torque, which means in turn that the torque equally determines Moment of inertia as we shall see in the equations later. (Dr Yukawa refers to the relation of the left hand side and the right hand side of an equation when he speaks about Mach.) This relation is analogous to m (mass) F=ma as we can say "Mass is the mass property of a rigid body that determines the force needed for a desired rectilinear acceleration ." Although analogous Moment of inertia seems quite different from mass.

First, 1) "What is Moment ? " and 2) "What is Inertia ? "

1) What is Moment ?

From wiki <Moment (physics)> (03-July-2014)

"
Moment is a tendency to produce motion, especially, a rotation about a point or axis.

"
2) What is Inertia ?

From wiki <Inertia> (03-July-2014)

"
Classical inertia

The law of inertia states that it is the tendency of an object to resist a change in motion.

"

or Newton's original explanation (also from the same wiki)

"

Unless acted upon by a net unbalanced force, an object will maintain a constant velocity.

Note that "velocity" in this context is defined as a vector, thus Newton's "constant velocity" implies both constant speed and constant direction (and also includes the case of zero speed, or no motion).

"

Please note that vector concept and vector calculation did not exist in Newton's time.

Moment of Inertia is only appears in rotation and spin and supposed to be related with inertia. So for memory of a concept one of the old names, rotational inertia sounds better. As <moment is a tendency to produce motion, especially, a rotation about a point or axis> then Moment of Inertia means <a tendency to produce motion in the form of inertia, especially, a rotation about a point or axis.>. But what does it mean ? Does this make sense ? As < Inertia is the tendency of an object to resist a change in motion.>. It seems Moment = - (minus) Inertia. or Moment = 1 / Inertia (or Inertia = 1/ Moment) ?

This reminds me of the following Newton's physics law (not mathematical calculation).

F = G \frac{m_1 m_2}{r^2}\

F (Gravitation) relates with "r" (distance between mass m1 and mass m2) but in 1/

r 2

.

Moment of Inertia ( I ) is expressed as

I=mr^2.

This equation is usually <is given> or <is expressed> . Who or what gives or expresses this ?

Where does it come from? and What does it mean, especially

r 2

?

An analogy is often used (I is analogous to m, or Moment of Inertia is analogous to Mass) but analogy is an analogy and not a proof. Analogy gives you some good hint for understanding but sometimes leads you to misunderstanding or insufficient understanding or deviates you from the truth or the correct meaning.

In this case, the difference between Moment of Inertia (I) and Mass is evident. Mass is mass (m) only while Moment of Inertia (I) = m

r 2

.

What does

< r 2

> of

I = mr 2

mean ?

Does < r 2

> mean an area - square ?

I = m

r 2

does not come directly from <m x r x r > but comes form Torque or Angular Momentum or Angular Acceleration or Angular Velocity or the definitions of these, and from these, some analogies of Moment of Inertia to Rectilinear Motion are made up. Anyway I = m

r 2

is supposed to relate with force or Torque. Torque is defined as ＜F x r＞ in a vector form and also called Moment of Force and indicates rotational force. Anyway "r" is strongly connected with rotation as it is a radial of the rotation circle.

One of the typical things about rotation or circle is that the point or position on a circle moves back to the starting point (position) after one round or after traveling 2

π

r. "r" works as a radius (straight line) as well as a part of traveling distance (a curve, a part of circle) which gives velocity and acceleration when time is involved.

Very simple calculations

2

π

r/r = 2

π (no unit, just a number, in radian)

2 π r /

2 π =

r (length, unit in m (meter))

Please also note the following basic relations.

θ s
----- = -----
2

π

πr

where θ = an angle in radian, s = the corresponding arc length on the circle of the radius r. Or

θ = angular displacement, s = displacement on the circle of the radius r.

Then it becomes simply

s
θ = -----

(bot sides are in radian, the right hand side s and r are in m, so no unit and just a

r number)

or

θ r = s

When considering velocity

that the angular velocity

ω

where v is the linear velocity along the circle of the radius r. This is crucially important.

The left hand side. The unit of ω is radian / sec

The right had side the unit of v in meter / sec and r is in meter. Then v/r is in a number (divided by r) / sec. Please recall that <

radian> is not a physics unit but a number.

This can be applied to the above

s
θ = -----

r

θ = an angle in radian.

and similarly or analogously the angular acceleration ⍺ is

where at is a liner Tangential Acceleration corresponding to the linear velocity v. ⍺ is attain by at
< divided by

r

> so the unit in radian / sec square.

------

Back to

Moment of Inertia ( I ) is expressed as

I=mr^2.

Can we consider "r" as an inertia like "m" of F = ma ? If not, the word Moment of Inertia sounds strange.

From wiki <Moment of Inertia>

"
Simple pendulum
A simple pendulum is a point mass suspended by a string so that its movement is constrained to a circle around a pivot point. The mass of a simple pendulum supported by a light string accelerates due to the force of gravity.
The moment of inertia of the pendulum about the pivot point is its resistance to movement due to the torque due to gravity. Mathematically, it is the ratio of the torque due to gravity about the pivot of a pendulum to its angular acceleration about that pivot point. For a simple pendulum this is found to be the product of the mass of the particle with the square of its distance to the pivot. This is shown as follows:
The force of gravity on the mass of a simple pendulum generates a torque

\boldsymbol \tau = \mathbf{r}\times \mathbf{F}

around the axis perpendicular to the plane of the pendulum movement. Here F is the tangential component of the net force on the mass. Associated with this torque is an angular acceleration,

\boldsymbol\alpha

, of the string and mass around this axis. Since the mass is constrained to a circle the tangential acceleration of the mass is (underlined by sptt)

$\bold a = \boldsymbol\alpha \times \bold r$

and since

F = ma

the torque equation becomes:

\boldsymbol\tau = \mathbf{r}\times \mathbf{F} = \mathbf{r}\times (m \boldsymbol\alpha \times \bold r) = (mr^2)\boldsymbol\alpha = I\alpha \bold e,

where e is a unit vector perpendicular to the plane of the pendulum. (The second to the last step occurs because of the BAC-CAB rule (*) using the fact that

\boldsymbol\alpha

is always perpendicular to r.) The quantity

I = mr 2

is the moment of inertia of this single mass around the pivot point. (underline by sptt and this is a definition or naming)

The quantity

I = mr 2

also appears in the angular momentum (L) of a simple pendulum, which is calculated from the velocity

v = ω \times r

of the pendulum mass around the pivot, where

ω

is the angular velocity of the mass about the pivot point. This angular momentum is given by (underline by sptt and this is also a definition or naming)

\mathbf{L} = \mathbf{r}\times(m\mathbf{v}) = (mr^2)\boldsymbol\omega = I\omega\bold e,

using the math similar to that used to derive the previous equation.

"

<mv> is a linear momentum.

Angular momentum (L) = mv x r

a = Tangential acceleration (time derivative of Tangential velocity) of the mass, which direction continuously changes going along the circle of the radius. Tangential velocity (velocity is a vector) has the magnitude (speed, which is scalar) and direction. Even the speed remains the same Tangential acceleration (time derivative) changes as the direction changes.
Angular acceleration, ⍺ (time derivative of Angular velocity,

ω

), on the other hand, is considered to be zero (0) (?) when Angular velocity is constant.

$\bold a = \boldsymbol\alpha \times \bold r$

Angular acceleration ⍺ = a (Tangential acceleration) / r

Torque = F x r = ma x r (as F = ma) = m (⍺ x r) x r = m $r 2$ ⍺

Analogously

F = ma

Torque = m $r 2$ ⍺

Therefore again analogously

m $r 2$ (I or Moment of Inertia) is analogous to m.

Please recall

⍺ is an angular <acceleration>, the unit of which is supposed to be which is (<meter / sec

2

) / r

2

>.

Therefore in the angular world, m $r 2$ can be regarded as m or Inertia. Then m $r 2$ (I) can be called <Moment of Inertia> or <Rotational Inertia> or <Angular Inertia>, corresponding to <m/r

2

Angular acceleration (wiki - 02-Aug-2016)

Angular acceleration

From Wikipedia, the free encyclopedia

Radians per second squared
Unit system	SI derived unit
Unit of	Angular acceleration
Symbol	rad/s² or rad⋅s⁻²

Angular acceleration is the rate of change of angular velocity. In SI units, it is measured in radians per second squared (rad/s²), and is usually denoted by the Greek letter alpha (α).^[1]

Mathematical definition

The angular acceleration can be defined as either:

{\alpha} = \frac{{d\omega}}{dt} = \frac{d^2{\theta}}{dt^2}

, or

{\alpha} = \frac{a_T}{r}

where

{\omega }

is the angular velocity,

a_T

is the linear tangential acceleration, and

r

, (usually defined as the radius of the circular path of which a point moving along), is the distance from the origin of the coordinate system that defines

\theta

and

\omega

to the point of interest.

Equations of motion

For two-dimensional rotational motion (constant

\hat L

), Newton's second law can be adapted to describe the relation between torque and angular acceleration:

{\tau} = I\ {\alpha}

where

{\tau }

is the total torque exerted on the body, and

I

is the mass moment of inertia of the body.

Constant acceleration

For all constant values of the torque,

{\tau }

, of an object, the angular acceleration will also be constant. For this special case of constant angular acceleration, the above equation will produce a definitive, constant value for the angular acceleration:

{\alpha} = \frac{\tau}{I}.

Non-constant acceleration

For any non-constant torque, the angular acceleration of an object will change with time. The equation becomes a differential equation instead of a constant value. This differential equation is known as the equation of motion of the system and can completely describe the motion of the object. It is also the best way to calculate the angular velocity.

"

(*) BAC-CAB rule - vector calculation

Please note that the bold letters are all vectors, having the magnitude and direction. And

\times

is

vector cross product, which is rather a mathematical rule (invention) than what we can see and can be proved.

wiki <Cross Product> 02-Aug-2016)

[ The cross product a × b is defined as a vector c that is perpendicular to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.
The cross product is defined by the formula^[3]^[4]

\mathbf {a} \times \mathbf {b} =\left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\sin(\theta )\ \mathbf {n}

where θ is the angle between a and b in the plane containing them (hence, it is between 0° and 180°), ‖a‖ and ‖b‖ are the magnitudes of vectors a and b, and n is a unit vector perpendicular to the plane containing a and b in the direction given by the right-hand rule (illustrated). If the vectors a and b are parallel (i.e., the angle θ between them is either 0° or 180°), by the above formula, the cross product of a and b is the zero vector 0. ]

"

Somehow repeating,

Relation between Moment of Inertia ( I ), Angular Momentum ( L ) and Torque

\mathbf{L} = \mathbf{r}\times(m\mathbf{v}) = (mr^2)\boldsymbol\omega = I\omega\bold e,

\tau = I \alpha.

\boldsymbol\tau = \mathbf{r}\times \mathbf{F} = \mathbf{r}\times (m \boldsymbol\alpha \times \bold r) = (mr^2)\boldsymbol\alpha = I\alpha \bold e,

where

r is Position (location) vector.
ω is Angular Velocity vector.
e is the unit vector. ⍺ (not bold type) is a scalar.

⍺ is Angular Acceleration vector.

Please note that these are vectors.

τ = I⍺

This relation (simply) shows that either m (mass) or r (radius) or ⍺ increases then τ increases. As

I=mr^2.

Please recall the above underlined wiki explanation.

Mathematically, it is the ratio of the torque due to gravity about the pivot of a pendulum to its angular acceleration about that pivot point. For a simple pendulum this is found to be the product of the mass of the particle with the square of its distance to the pivot.

I want to emphasize <Mathematically>.

Meanwhile,

Angular Acceleration	$\boldsymbol{\alpha} =\frac{d\boldsymbol{\omega}}{dt}$ , where

Angular Velocity	$\boldsymbol{\omega} =\frac{d\boldsymbol{\theta}}{dt}$

Angular velocity is not a normal velocity related with distance but related with only angle (which has no relation with distance). Angular acceleration is not a normal acceleration related with distance, or rectilinear acceleration but related with only angle, too.

Angular velocity is complicated if you look into it even in two dimension (plane, plate) and especially in three dimension (3D rotation or curl). But we can see it superficially or in a simplified manner under a certain limited condition.

From wiki <Angular velocity> (14-July-2014)
Particle in two dimensions

The angular velocity of the particle at P with respect to the origin O is determined by the perpendicular component of the velocity vector v.

In two dimensions the angular velocity ω is given by

\omega = \frac{d\phi}{dt}

This is related to the cross-radial (tangential) velocity by:^[1]

\mathrm{v}_\perp=r\,\frac{d\phi}{dt}

An explicit formula for v_⊥ in terms of v and θ is:

\mathrm{v}_\perp=|\mathrm{\mathbf{v}}|\,\sin(\theta)

Combining the above equations gives a formula for ω:

\omega=\frac{|\mathrm{\mathbf{v}}|\sin(\theta)}{|\mathrm{\mathbf{r}}|}

In two dimensions the angular velocity is a single number that has no direction, but it does have a sense or orientation. In two dimensions the angular velocity is a pseudoscalar, a quantity that changes its sign under a parity inversion (for example if one of the axes is inverted or if they are swapped). The positive direction of rotation is taken, by convention, to be in the direction towards the y axis from the x axis. If parity is inverted, but the sense of a rotation does not, then the sign of the angular velocity changes.
There are three types of angular velocity involved in the movement on an ellipse corresponding to the three anomalies (true, eccentric and mean).

------

The above wiki explanation is somehow redundant not very straight so it is difficult to understand (also due to lack of some further knowledge about like - an ellipse corresponding to the three anomalies (true, eccentric and mean)) - why explained this way.

Among the above equations, first look at

"
In two dimensions the angular velocity ω is given by

\mathrm{v}_\perp=r\,\frac{d\phi}{dt}

"

This not a definition but an equation. What does this mean ? This means

1) The tangential velocity is different from the angular velocity.
2) "r" is a multiplying factor to the angular velocity against the tangential velocity. The same thing but we can rearrange this equation to see the relation between the tangential velocity, angular velocity and "r" (radius)

v (Tangential velocity)

--------------------- = ω (Angular velocity)

r

Even when the angular velocity remains the same the tangential velocity varies according to the radius. The shorter the radius, the slower the v (t.v.) while the longer the radius, the faster the v (t.v.). Very simple relation but important and this is the fact, not a definition. But this is regarded as the definition of Angular Velocity (ω) by rearranging.

v (velocity)
ω = -------------
r (radius)

However, if we regard Angular velocity

\boldsymbol{\omega} =\frac{d\boldsymbol{\theta}}{dt}

and do not regard this equation as vectors here. The unit of

\boldsymbol{\omega} =\frac{d\boldsymbol{\theta}}{dt}

is m/sec, the unit of "r" is m, so the unit of ω is 1/sec, which is frequency.

In the last equation of the above

\omega=\frac{|\mathrm{\mathbf{v}}|\sin(\theta)}{|\mathrm{\mathbf{r}}|}

" 1/ $|r|$ " makes the meaning of rotation as simply an angle.

	Back to I = m $r 2$ . $\boldsymbol\tau = \mathbf{r}\times \mathbf{F} = \mathbf{r}\times (m \boldsymbol\alpha \times \bold r) = (mr^2)\boldsymbol\alpha = I\alpha \bold e,$

When " r " increases with Angular Acceleration (

\boldsymbol\alpha

) being unchanged I and therefore τ increases by

r 2

, very rapidly. On the other hand, when " r " decreases I and τ decreases by $r 2$ , very rapidly. How about when " r " is approaching 0 (zero) ? to keep the same τ (m being the same as well) extremely high Angular Acceleration is required. Or τ becomes o (zero) as well.

Although

mr 2

appears in the equations these are confusing and did not directly explain why and how.

Moment of Inertia ( I ) is

I=mr^2.

Again repeating

a = Tangential Acceleration of the mass, which direction changes (at every infinitesimal point) going along the circle of the radius r is defined (definition) as

$\bold a = \boldsymbol\alpha \times \bold r$

This is a definition as the vector product itself is a definition. This equation is explained as follows:

The direction changes at every infinitesimal point and goes along the circle of the radius r. The length of the circle is 2 $π$ r. ⍺ is Angular Acceleration vector, which is

	$\boldsymbol{\alpha} =\frac{d\boldsymbol{\omega}}{dt}$

ω is Angular Velocity vector, which is not the velocity of <distance / time> but the velocity of <angle / time> (how fast or how slow a point on the circle rotates) and no relation with r. Meanwhile a (Tangential Acceleration) relates with r. The bigger r the bigger a when the Angular Acceleration remains the same.

\boldsymbol\tau = \mathbf{r}\times \mathbf{F} = \mathbf{r}\times (m \boldsymbol\alpha \times \bold r) = (mr^2)\boldsymbol\alpha = I\alpha \bold e,

In this equation,

m (⍺ x r) is regarded as "m⍺ x r" and "r x (m⍺ x r)" is "(m $r 2$ )⍺".

The underlined part <the mass is constrained to a circle the tangential acceleration of the mass> implies some force. But torque itself is very confusing especially its direction - perpendicular to the plane of the pendulum.

We may be able to consider I = m

r 2

as follows:

Torque = F x r (vector cross product)

F = Torque / r = (m $r 2$ )⍺ / r = mr⍺

a = ⍺ x r (vector cross product) ---> ⍺ = a / r

Therefore

F = mr⍺ ---> mra / r = ma

Actually this calculation process is almost the same as

\boldsymbol\tau = \mathbf{r}\times \mathbf{F} = \mathbf{r}\times (m \boldsymbol\alpha \times \bold r) = (mr^2)\boldsymbol\alpha = I\alpha \bold e,

and the problem is that the process of <vector cross product> is neglected. We must be more familiar with <vector cross product> (refer to wiki <cross product> at the end of sources).

2. Why the direction of torque is the direction of the cross product of "r" and "F" ?

Answer One

From: Hyperphysics <Rotation Vectors>

"

Rotation Vectors

Angular motion has direction associated with it and is inherently a vector process. But a point on a rotating wheel is continuously changing direction and it is inconvenient to track that direction. The only fixed, unique direction for a rotating wheel is the axis of rotation, so it is logical to choose this axis direction as the direction of the angular velocity. Left with two choices about direction, it is customary to use the right hand rule to specify the direction of angular quantities.

"

This explanation is not very convincing and " logical ".

Answer Two
Also From: Hyperphysics <Right Hand Rule for Torque>

Right Hand Rule for Torque

Torque is inherently a vector quantity. Part of the torque calculation is the determination of direction. The direction is perpendicular to both the radius from the axis and to the force. It is conventional to choose it in the right hand rule direction along the axis of rotation. The torque is in the direction of the angular velocity which would be produced by it in the absence of other influences. In general, the change in angular velocity is in the direction of the torque.

This explanation is a rule, which is a definition, so actually not a explanation. Besides it siays <The torque is in the direction of the angular velocity which would be produced by it in the absence of other influences>.

The direction of the angular velocity is perpendicular to the plane of rotation is also a definition.

Yohoo Answers

Why is torque perpendicular to force?

please explain why along with a mathematical proof is possible....thank you!

Best Answer

It might be be easiest to think of torque in terms of the time change of angular momentum. That makes it pretty obvious that the torque is perpendicular to the applied force since angular momentum is represented as a vector thru the axis of rotation of an object. Whenever you exert a force that causes an object to rotate or changes its rotational speed, the torque is along the axis of rotation. Obviously the direction of the force that causes an object to rotate is perpendicular to the axis of rotation. So, from the equation below you can see that the torque vector has to be in the same direction as the angular momentum vector (axis of rotation)

T = dL/dt where T = torque vector, dL = increment of change in angular momentum vector (along axis of rotation) and dt is the increment of time over which the change in angular momentum dL occurred.

From wiki: <angular momentum>

The time derivative of angular momentum is called torque:

\mathbf{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \times \mathbf{p} + \mathbf{r} \times \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = 0 + \mathbf{r} \times \mathbf{F} = \mathbf{r} \times \mathbf{F}

-----

Angular Momentum is defined (analogous to momentum p) as

\mathbf{L} = \mathbf{r}\times(m\mathbf{v}) = (mr^2)\boldsymbol\omega = I\omega\bold e,

But the same question to Angular Momentum.

Then why is Angular Momentum perpendicular to velocity?

Other Answers (2)

It is a "pseudo force" something that acts like a vector though its source is not apparent.

The source is as follows: consider a wheel (bicycle wheel) spinning with its axle vertical. Now, imagine a little block of mass at the rim on your right and left side. The mass on your right is going away from you (horizontal) tangential to the wheel and the mass on the left is coming toward you. You are holding the axle with your right hand on top. You try to tip (to tilt) the axle toward you. The masses are now moving in a different direction, the mass on the right now has an upward component and the mass on the left has a downward component. By Newtons's 1st Law you had to apply a force to the axle to tip the when and change the velocity vectors of the masses. That force FEELS like there was a vector vertically out of the axle that was pointed UP. It really was the forces applied to the masses, but it does not feel like that.

TORQUE is the CROSS PRODUCT of 2 vectors and the result of that mathematical operation is a Psudo Vector at right angle to the plane the 2 vectors lie in.

CwCc answered 9 months ago

It's not always perpendicular. The vector equation for torque is
T = r (x) F
Where T is the torque vector, r is the lever arm (a vector that points from the pivot to the site where force is applied), and F is the force. I'm using (x) for the vector cross product. If we use magnitudes instead of vectors, the equation can be written as
T = rF sin(a)
Where a is the angle between the lever arm and the force. Note that the sine is maximized when r and F are perpendicular. Note also that the torque goes to zero when r and F are parallel.
To make sense of all of this, consider trying to open a door. To cause the door to swing open, you need to apply a certain amount of torque with respect to the hinge. You know from experience that it's much harder to open the door if you push near the hinge than if you push near the far end -- the reason is that the torque is proportional to the lever arm, which is the distance from the hinge at which you're pushing. You also know that it's easier to open the door if you push straight on instead of trying to pull the door sideways. This corresponds to the fact that torque is maximized when you push/pull perpendicular to the lever arm.

Besides the direction of Torque we may ask what ＜x r＞of (cross product) ＜Torque = F x r＞ mean. If we neglect the direction defined by cross product it is simply a multiplying factor or Torque is proportional to the value of "r" while F is the same.

----

All of the three including the best answer explain why because it is so defined or by using vector calculation of cross product. But Vector Cross Product itself seems a mathematical definition, a definition like Imaginary Numbers, which explains some physical phenomena very well but not real numbers. Confusion of the questioner and answers come from this - reality (physical world) and mathematical reality, both of which are unfortunately or confusingly regarded as true as they are proved. When using vectors the direction is needed since a vector needs a direction because so defined. Dr Yukawa talked about History of Vector in Day Two with his own opinion. Vector is a relatively new concept or mathematical invention in late 19th century.

The door example of the last answer is easy to understand the relation between torque, force and distance (radius).

Anyway, all these answers are "Why ? because mathematically (not physically) so defined ".

Let's consider Centripetal Force. The point is <the mass is constrained to a circle>. Some physical force is implied.

Also Wiki says in <Moment of Inertia>

"
Simple pendulum

A simple pendulum is a point mass suspended by a string so that its movement is constrained to a circle around a pivot point. The mass of a simple pendulum supported by a light string accelerates due to the force of gravity. (underlined by sptt)

"

<its movement is constrained to a circle around a pivot point.> means some physical meaning, or something like force.

3. What is Centripetal Force ?

From Wiki

"

Formula
The magnitude of the centripetal force on an object of mass m moving at tangential speed v along a path with radius of curvature r is:^[6]

F = ma_c = \frac{m v^2}{r}

where

a_c

is the centripetal acceleration. The direction of the force is toward the center of the circle in which the object is moving, or the osculating circle, the circle that best fits the local path of the object, if the path is not circular.The speed in the formula is squared, so twice the speed needs four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force. This force is also sometimes written in terms of the angular velocity ω of the object about the center of the circle:

F = m r \omega^2. \,

"

Now consider ω, Angular Velocity vector, especially the direction of it. If the direction of
Angular Velocity vector is perpendicular to the velocity, the direction of Angular Momentum vector is also perpendicular to the velocity due to the following equation.

\mathbf{L} = \mathbf{r}\times(m\mathbf{v}) = (mr^2)\boldsymbol\omega = I\omega\bold e,

Then the direction of torque vector is as well as due to the following equation.

T = dL/dt where T = torque vector, dL = increment of change in angular momentum vector (along axis of rotation) and dt is the increment of time over which the change in angular momentum dL occurred.

From wiki: <angular momentum>

The time derivative of angular momentum is called torque:

\mathbf{\tau} = \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \times \mathbf{p} + \mathbf{r} \times \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = 0 + \mathbf{r} \times \mathbf{F} = \mathbf{r} \times \mathbf{F}

-----

The second terns comes from

Differentiation of L ( = r x mv, nv = p (momentum))

Another method to get the centripetal acceleration.

From: Hyperphysics

"

"

The above explanation seems tricky by using equations. We must focus at the point when θ approaches to zero (0 angle) to get this conclusion (a = v $2$ /r), but not at zero (0 angle), which seems also tricky.

s/r can be regarded as an angle in radian - the direction is not sure as an angle has no direction unless it is defined as a vector, which must have a direction as well as magnitude (*). This explanation does not say but s and r may be vectors so the directions should be mentioned. Or they are not treated as vectors.

The above explanation does not say but the direction of Δv is to the center - centripetal. You can see this by moving the tail of V2 to the tail to V1, which will become the drawing in the middle.

Consider a very simple equation y = 1/x, the reciprocal or inverse of y = x in terms of x.

From wiki

y becomes extremely large when x approaches to zero (0) and no value when x = 0. A dramatic change at or around zero (0). Rotation and spin are seen or described around a center point or an axis.

Next consider an equation y = x $2$ .

Graph of y = x ²

y increases exponentially as x increases.

Then for a = v $2$ /r

a goes to where when ? both v increases and r approaches zero (0).

(*) Another story is found in "circular motion" in wiki (05-Aug-2014)

Uniform

Figure 1: Velocity v and acceleration a in uniform circular motion at angular rate ω; the speed is constant, but the velocity is always tangent to the orbit; the acceleration has constant magnitude, but always points toward the center of rotation

In physics, uniform circular motion describes the motion of a body traversing a circular path at constant speed. The distance of the body from the axis of rotation remains constant at all times. Though the body's speed is constant, its velocity is not constant: velocity, a vector quantity, depends on both the body's speed and its direction of travel. This changing velocity indicates the presence of an acceleration; this centripetal acceleration is of constant magnitude and directed at all times towards the axis of rotation. This acceleration is, in turn, produced by a centripetal force which is also constant in magnitude and directed towards the axis of rotation.
In the case of rotation around a fixed axis of a rigid body that is not negligibly small compared to the radius of the path, each particle of the body describes a uniform circular motion with the same angular velocity, but with velocity and acceleration varying with the position with respect to the axis.
Formulas
For motion in a circle of radius r, the circumference of the circle is C = 2π r. If the period for one rotation is T, the angular rate of rotation, also known as angular velocity, ω is:

\omega = \frac {2 \pi}{T} \

and the units are radians/sec

The speed of the object traveling the circle is:

v\, = \frac {2 \pi r } {T} = \omega r

The angle θ swept out in a time t is:

\theta = 2 \pi \frac{t}{T} = \omega t\,

The acceleration due to change in the direction is:

a\, = \frac {v^2} {r} \, = {\omega^2} {r}

Figure 2: The velocity vectors at time t and time t + dt are moved from the orbit on the left to new positions where their tails coincide, on the right. Because the velocity is fixed in magnitude at v = r ω, the velocity vectors also sweep out a circular path at angular rate ω. As dt → 0, the acceleration vector a becomes perpendicular to v, which means it points toward the center of the orbit in the circle on the left. Angle ω dt is the very small angle between the two velocities and tends to zero as dt→ 0 (underlined by sptt).

Please consider the underlined part: limit. Does this make sense ?

-------
And all these vectors related with rotation around an axis pseudovectors.
From: mathworld.wolfram <Pseudovector>
A typical vector (i.e., a vector such as the radius vector

) is transformed to its negative under inversion of its coordinate axes. Such "proper" vectors are known as polar vectors. A vector-like object which is invariant under inversion is called a pseudovector, also called an axial vector (as a result of such vectors frequently arising as vectors describing rotation; Arfken 1985, p. 128; Morse and Feshbach 1953). The cross product

(1)

is a pseudovector, whereas the vector triple product

(2)

is a polar vector. (Polar) vectors and pseudovectors are interrelated in the following ways under application of the cross product,

(3)

(4)

Examples of pseudovectors therefore include the angular velocity vector

, angular momentum

, torque

, auxiliary magnetic field

, and magnetic dipole moment

.

-------

Unfortunately we are still not able to find the meaning of Moment of Inertia or have come back to the starting point after one circle. We have, however, some idea of the relations between "r" and Torque, Angular Velocity, Angular Moment, Angular Acceleration.

I have found a more direct answer to this question. The answer is simple and also found in wiki <Centripetal force>(07-Aug-2014). Anyway "r" strongly related with force.

Calculus derivation
In two dimensions the position vector

\textbf{r}

which has magnitude (length)

r

and directed at an angle

\theta

above the x-axis can be expressed in Cartesian coordinates using the unit vectors

\hat{x}

and

y-hat

:^[11]

\textbf{r} = r \cos(\theta) \hat{x} + r \sin(\theta) \hat{y}.

Assume uniform circular motion, which requires three things.

The object moves only on a circle.

The radius of the circle $r$ does not change in time.

The object moves with constant angular velocity $\omega$ around the circle. Therefore $\theta = \omega t$ where $t$ is time.

Now find the velocity

\textbf{v}

and acceleration

\textbf{a}

of the motion by taking derivatives of position with respect to time.

\textbf{r} = r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y}

\dot{\textbf{r}} = \textbf{v} = - r \omega \sin(\omega t) \hat{x} + r \omega \cos(\omega t) \hat{y}

\ddot{\textbf{r}} = \textbf{a} = - r \omega^2 \cos(\omega t) \hat{x} - r \omega^2 \sin(\omega t) \hat{y}

\textbf{a} = - \omega^2 (r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y})

Notice that the term in parenthesis is the original expression of

\textbf{r}

in Cartesian coordinates. Consequently,

\textbf{a} = - \omega^2 \textbf{r}.

The negative shows that the acceleration is pointed towards the center of the circle (opposite the radius), hence it is called "centripetal" (i.e. "center-seeking"). While objects naturally follow a straight path (due to inertia), this centripetal acceleration describes the circular motion path caused by a centripetal force.

--------
Sources of the argument

From wiki <Moment of Inertia> (03-July-2014)

Moment of inertia

Moment of inertia is the mass property of a rigid body that determines the torque needed for a desired angular acceleration about an axis of rotation. Moment of inertia depends on the shape of the body and may be different around different axes of rotation. A larger moment of inertia around a given axis requires more torque to increase the rotation, or to stop the rotation, of a body about that axis. Moment of inertia depends on the amount and distribution of its mass, and can be found through the sum of moments of inertia of the masses making up the whole object, under the same conditions. For example, if m_a + m_b = m_c, then I_a + I_b = I_c. In classical mechanics, moment of inertia may also be called mass moment of inertia, rotational inertia, polar moment of inertia, or the angular mass.

For planar movement of a body, the trajectories of all of its points lie in parallel planes, and the rotation occurs only about an axis perpendicular to this plane. In this case, the body has a single moment of inertia, which is measured around this axis.

For spatial movement of a body, the moment of inertia is defined by its symmetric 3 × 3 inertia matrix. The inertia matrix is often described as a symmetric rank two tensor, having six independent components. The inertia matrix includes off-diagonal terms called products of inertia that couple torque around one axis to acceleration about another axis. Each body has a set of mutually perpendicular axes, called principal axes, for which the off-diagonal terms of the inertia matrix are zero, and a torque around a principal axis only affects the acceleration about that axis.
Introduction

When a body is rotating around an axis, a torque must be applied to change its angular momentum. The amount of torque needed for any given change in angular momentum is proportional to the size of that change. Moment of inertia may be expressed in terms of kilogram-square metres (kg·m²) in SI units and pound-square feet (lb_m·ft²) in imperial or US units.

In 1673 Christiaan Huygens introduced this parameter in his study of the oscillation of a body hanging from a pivot, known as a compound pendulum.^[1] The term moment of inertia was introduced by Leonhard Euler in his book Theoria motus corporum solidorum seu rigidorum in 1765,^[1]^[2] and it is incorporated into Euler's second law.

The natural frequency of oscillation of a compound pendulum is obtained from the ratio of the torque imposed by gravity on the mass of the pendulum to the resistance to acceleration defined by the moment of inertia. Comparison of this natural frequency to that of a simple pendulum consisting of a single point of mass provides a mathematical formulation for moment of inertia of an extended body.^[3]^[4]

Moment of inertia also appears in momentum, kinetic energy, and in Newton's laws of motion for a rigid body as a physical parameter that combines its shape and mass. There is an interesting difference in the way moment of inertia appears in planar and spatial movement. Planar movement has a single scalar that defines the moment of inertia, while for spatial movement the same calculations yield a 3 × 3 matrix of moments of inertia, called the inertia matrix or inertia tensor.^[5]^[6]

The moment of inertia of a rotating flywheel is used in a machine to resist variations in applied torque to smooth its rotational output. The moment of inertia of an airplane about its longitudinal, horizontal and vertical axes determines how steering forces on the control surfaces of its wings, elevators and tail affect the plane in roll, pitch and yaw.
Simple pendulum
A simple pendulum is a point mass suspended by a string so that its movement is constrained to a circle around a pivot point. The mass of a simple pendulum supported by a light string accelerates due to the force of gravity.
The moment of inertia of the pendulum about the pivot point is its resistance to movement due to the torque due to gravity. Mathematically, it is the ratio of the torque due to gravity about the pivot of a pendulum to its angular acceleration about that pivot point. For a simple pendulum this is found to be the product of the mass of the particle with the square of its distance to the pivot. This is shown as follows:
The force of gravity on the mass of a simple pendulum generates a torque

\boldsymbol \tau = \mathbf{r}\times \mathbf{F}

around the axis perpendicular to the plane of the pendulum movement. Here F is the tangential component of the net force on the mass. Associated with this torque is an angular acceleration,

\boldsymbol\alpha

, of the string and mass around this axis. Since the mass is constrained to a circle the tangential acceleration of the mass is

\bold a = \boldsymbol\alpha \times \bold r

. Since

F = ma

the torque equation becomes:

\boldsymbol\tau = \mathbf{r}\times \mathbf{F} = \mathbf{r}\times (m \boldsymbol\alpha \times \bold r) = (mr^2)\boldsymbol\alpha = I\alpha \bold e,

where e is a unit vector perpendicular to the plane of the pendulum. (The second to the last step occurs because of the BAC-CAB rule (*) using the fact that

\boldsymbol\alpha

is always perpendicular to r.) The quantity

I = mr 2

is the moment of inertia of this single mass around the pivot point.

The quantity

I = mr 2

also appears in the angular momentum of a simple pendulum, which is calculated from the velocity

v = ω \times r

of the pendulum mass around the pivot, where

ω

is the angular velocity of the mass about the pivot point. This angular momentum is given by

\mathbf{L} = \mathbf{r}\times(m\mathbf{v}) = (mr^2)\boldsymbol\omega = I\omega\bold e,

using math similar to that used to derive the previous equation.

Similarly, the kinetic energy of the pendulum mass is defined by the velocity of the pendulum around the pivot to yield

E_\text{K} = \frac{1}{2}m\mathbf{v}\cdot\mathbf{v} = \frac{1}{2}(mr^2)\omega^2 = \frac{1}{2}I\omega^2.

This shows that the quantity

I = mr 2

is how mass combines with the shape of a body to define rotational inertia. The moment of inertia of an arbitrarily shaped body is the sum of the values

mr

² for all of the elements of mass in the body.

(*) Vector calculation
Definition
The moment of inertia, I, is defined as the ratio of an applied torque to the angular acceleration along a principal axis of the object, where then τ,

\alpha

and I are scalars, that is

\tau = I \alpha.

An equivalent definition of I uses the angular momentum L as follows,

L = I \omega,

where

\omega

is the angular velocity of the object.

Apply this definition to a simple pendulum to see that the moment of inertia of the mass m about the pivot point at a distance r is

I=mr^2.

This generalizes to define the moment of inertia of a body about an axis S as the sum of all elemental point masses dm each multiplied by the square of its perpendicular distance r to the axis.

Hyperphysics <Basic Rotational Quantities>

Basic Rotational Quantities

In addition to any tangential acceleration, there is always the centripetal acceleration:

The angular displacement is defined by:

For a circular path it follows that the angular velocity is

and the angular acceleration is

where the acceleration here is the tangential acceleration.

The standard angle of a directed quantity is taken to be counterclockwise from the positive x axis.
------

From wiki <Torque> (03-July-2014)

Torque, moment or moment of force (see the terminology below), is the tendency of a force to rotate an object about an axis,^[1] fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Mathematically, torque is defined as the cross product of the lever-arm distance vector and the force vector, which tends to produce rotation.
Loosely speaking, torque is a measure of the turning force on an object such as a bolt or a flywheel. For example, pushing or pulling the handle of a wrench connected to a nut or bolt produces a torque (turning force) that loosens or tightens the nut or bolt.
The symbol for torque is typically τ, the Greek letter tau. When it is called moment, it is commonly denoted M.
The magnitude of torque depends on three quantities: the force applied, the length of the lever arm^[2] connecting the axis to the point of force application, and the angle between the force vector and the lever arm. In symbols:

\boldsymbol \tau = \mathbf{r}\times \mathbf{F}\,\!

\tau = \|\mathbf{r}\|\,\|\mathbf{F}\|\sin \theta\,\!

where

τ is the torque vector and τ is the magnitude of the torque,

r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied),

F is the force vector,

× denotes the cross product,

θ is the angle between the force vector and the lever arm vector.

From wiki <Moment of Inertia, Japanese versoin> (03-July-2014)

Rotational Motion and Linear Motion


Quantity	Rotational Motion		Linear Motion
Mechanical variable	Angle	$\boldsymbol{\theta}$	Location	$\boldsymbol{r}$
1st order derivative	Angular Velocity	$\boldsymbol{\omega} =\frac{d\boldsymbol{\theta}}{dt}$	Velocity	$\boldsymbol{v} =\frac{d\boldsymbol{r}}{dt}$
2nd order derivative	Angular Acceleration	$\boldsymbol{\alpha} =\frac{d\boldsymbol{\omega}}{dt}$	Acceleration	$\boldsymbol{a} =\frac{d\boldsymbol{v}}{dt}$
Inertia	Moment of Inetia	$I$	Mass	$m$
Momentum	Angular Momentum	$\boldsymbol{L} =\boldsymbol{r}\times\boldsymbol{p}$	Momentum	$\boldsymbol{p}=m\boldsymbol{v}$
Force	Moment of Force	$\boldsymbol{N} =\boldsymbol{r}\times\boldsymbol{F}$	Force	$\boldsymbol{F}$
Equation of Motion	$\frac{d\boldsymbol{L}}{dt} =\boldsymbol{N}$		$\frac{d\boldsymbol{p}}{dt} =\boldsymbol{F}$
Mechanical Enegery	$\frac{1}{2}I\omega^2$		$\frac{1}{2}mv^2$
Work	$\boldsymbol{N}\cdot\Delta\boldsymbol{\theta}$		$\boldsymbol{F}\cdot\Delta\boldsymbol{r}$
Power (*)	$\boldsymbol{N}\cdot\boldsymbol{\omega}$		$\boldsymbol{F}\cdot\boldsymbol{v}$

Power (*) is obtained first by integrating F (Force) in terms of location and second differentiating the result of integration in terns of time.

From Hyperphysics

Rotational-Linear Parallels

From Hyperphysics

Moment of Inertia

Moment of inertia is the name given to rotational inertia, the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation. For a point mass the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr². That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses.

From wiki <Torque> (03-July-2014)

What is Moment ?

Moment is a tendency to produce motion, especially, a rotation about a point or axis. Moment is about a physically measurable quantity in relation to a reference point. It is the product of a physical quantity, such as mass or force, and the distance of the body or point, for which such physical quantity pertains, from a fixed reference point.

What is Inertia ?

From wiki <Torque> (03-July-2014)

Classical inertia
The law of inertia states that it is the tendency of an object to resist a change in motion. According to Newton, an object will stay at rest or stay in motion (i.e. 'maintain its velocity' in modern terms) unless acted on by a net external force, whether it results from gravity, friction, contact, or some other source. The Aristotelian division of motion into mundane and celestial became increasingly problematic in the face of the conclusions of Nicolaus Copernicus in the 16th century, who argued that the earth (and everything on it) was in fact never "at rest", but was actually in constant motion around the sun.^[11] Galileo, in his further development of the Copernican model, recognized these problems with the then-accepted nature of motion and, at least partially as a result, included a restatement of Aristotle's description of motion in a void as a basic physical principle:

A body moving on a level surface will continue in the same direction at a constant speed unless disturbed.^[12]

Galileo writes that 'all external impediments removed, a heavy body on a spherical surface concentric with the earth will maintain itself in that state in which it has been; if placed in movement towards the west (for example), it will maintain itself in that movement'.^[13] This notion which is termed 'circular inertia' or 'horizontal circular inertia' by historians of science, is a precursor to, but distinct from, Newton's notion of rectilinear inertia.^[14]^[15] For Galileo, a motion is 'horizontal' if it does not carry the moving body towards or away from the centre of the earth, and for him 'a ship, for instance, having once received some impetus through the tranquil sea, would move continually around our globe without ever stopping'.^[16]^[17]
It is also worth noting that Galileo later went on to conclude that based on this initial premise of inertia, it is impossible to tell the difference between a moving object and a stationary one without some outside reference to compare it against.^[18] This observation ultimately came to be the basis for Einstein to develop the theory of Special Relativity.
Concepts of inertia in Galileo's writings would later come to be refined, modified and codified by Isaac Newton as the first of his Laws of Motion (first published in Newton's work, Philosophiae Naturalis Principia Mathematica, in 1687):

Unless acted upon by a net unbalanced force, an object will maintain a constant velocity.

Note that "velocity" in this context is defined as a vector, thus Newton's "constant velocity" implies both constant speed and constant direction (and also includes the case of zero speed, or no motion). Since initial publication, Newton's Laws of Motion (and by extension this first law) have come to form the basis for the branch of physics known as classical mechanics.^{[citation needed]}
The actual term "inertia" was first introduced by Johannes Kepler in his Epitome Astronomiae Copernicanae (published in three parts from 1618–1621); however, the meaning of Kepler's term (which he derived from the Latin word for "idleness" or "laziness") was not quite the same as its modern interpretation. Kepler defined inertia only in terms of a resistance to movement, once again based on the presumption that rest was a natural state which did not need explanation. It was not until the later work of Galileo and Newton unified rest and motion in one principle that the term "inertia" could be applied to these concepts as it is today.^{[citation needed]}
Nevertheless, despite defining the concept so elegantly in his laws of motion, even Newton did not actually use the term "inertia" to refer to his First Law. In fact, Newton originally viewed the phenomenon he described in his First Law of Motion as being caused by "innate forces" inherent in matter, which resisted any acceleration. Given this perspective, and borrowing from Kepler, Newton actually attributed the term "inertia" to mean "the innate force possessed by an object which resists changes in motion"; thus Newton defined "inertia" to mean the cause of the phenomenon, rather than the phenomenon itself. However, Newton's original ideas of "innate resistive force" were ultimately problematic for a variety of reasons, and thus most physicists no longer think in these terms. As no alternate mechanism has been readily accepted, and it is now generally accepted that there may not be one which we can know, the term "inertia" has come to mean simply the phenomenon itself, rather than any inherent mechanism. Thus, ultimately, "inertia" in modern classical physics has come to be a name for the same phenomenon described by Newton's First Law of Motion, and the two concepts are now considered to be equivalent.

----

Wiki <Angular velocity> (14-July-2014)
Particle in two dimensions

The angular velocity of the particle at P with respect to the origin O is determined by the perpendicular component of the velocity vector v.

The angular velocity describes the speed of rotation and the orientation of the instantaneous axis about which the rotation occurs. The direction of the angular velocity pseudovector will be along the axis of rotation; in this case (counter-clockwise rotation) the vector points up.

The angular velocity of a particle is measured around or relative to a point, called the origin. As shown in the diagram (with angles ɸ and θ in radians), if a line is drawn from the origin (O) to the particle (P), then the velocity (v) of the particle has a component along the radius (radial component, v_‖) and a component perpendicular to the radius (cross-radial component, v_⊥). If there is no radial component, then the particle moves in a circle. On the other hand, if there is no cross-radial component, then the particle moves along a straight line from the origin.
A radial motion produces no change in the direction of the particle relative to the origin, so for purposes of finding the angular velocity the radial component can be ignored. Therefore, the rotation is completely produced by the perpendicular motion around the origin, and the angular velocity is completely determined by this component.
In two dimensions the angular velocity ω is given by

\omega = \frac{d\phi}{dt}

This is related to the cross-radial (tangential) velocity by:^[1]

\mathrm{v}_\perp=r\,\frac{d\phi}{dt}

An explicit formula for v_⊥ in terms of v and θ is:

\mathrm{v}_\perp=|\mathrm{\mathbf{v}}|\,\sin(\theta)

Combining the above equations gives a formula for ω:

\omega=\frac{|\mathrm{\mathbf{v}}|\sin(\theta)}{|\mathrm{\mathbf{r}}|}

Angular acceleration (wiki - 14-Aug-2014)

Mathematical definition
The angular acceleration can be defined as either:

{\alpha} = \frac{{d\omega}}{dt} = \frac{d^2{\theta}}{dt^2}

, or

{\alpha} = \frac{a_T}{r}

where

{\omega}

is the angular velocity,

a_T

is the linear tangential acceleration, and

r

, (usually defined as the radius of the circular path of which a point moving along), is the distance from the origin of the coordinate system that defines

\theta

and

\omega

to the point of interest.
Equations of motion
For two-dimensional rotational motion (constant

\hat L

), Newton's second law can be adapted to describe the relation between torque and angular acceleration:

{\tau} = I\ {\alpha}

where

{\tau}

is the total torque exerted on the body, and

I

is the mass moment of inertia of the body.
Constant acceleration
For all constant values of the torque,

{\tau}

{\alpha} = \frac{\tau}{I}.

Non-constant acceleration
For any non-constant torque, the angular acceleration of an object will change with time. The equation becomes a differential equation instead of a constant value. This differential equation is known as the equation of motion of the system and can completely describe the motion of the object. It is also the best way to calculate the angular velocity.
----
<Cross Product> wiki 16-Aug-2014)

Definition

The cross product of two vectors a and b is defined only in three-dimensional space and is denoted by a × b. In physics, sometimes the notation a ∧ b is used,^[2] though this is avoided in mathematics to avoid confusion with the exterior product.
The cross product a × b is defined as a vector c that is perpendicular to both a and b, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.
The cross product is defined by the formula^[3]^[4]

\mathbf{a} \times \mathbf{b} = \left\| \mathbf{a} \right\| \left\| \mathbf{b} \right\| \sin \theta \ \mathbf{n}

Geometric meaning

Figure 1. The area of a parallelogram as a cross product

The magnitude of the cross product can be interpreted as the positive area of the parallelogram having a and b as sides (see Figure 1):

A = \left\| \mathbf{a} \times \mathbf{b} \right\| = \left\| \mathbf{a} \right\| \left\| \mathbf{b} \right\| \sin \theta. \,\!

Because the magnitude of the cross product goes by the sine of the angle between its arguments, the cross product can be thought of as a measure of ‘perpendicularity’ in the same way that the dot product is a measure of ‘parallelism’. Given two unit vectors, their cross product has a magnitude of 1 if the two are perpendicular and a magnitude of zero if the two are parallel. The opposite is true for the dot product of two unit vectors. (bold typed by sptt)

From wiki <Centripetal force> (05-July-2014)

Centripetal force

Centripetal force (from Latin centrum "center" and petere "to seek"^[1]) is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion.
In simple terms, centripetal force is defined as a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the centre.^[2]^[3] The mathematical description was derived in 1659 by Dutch physicist Christiaan Huygens.^[4] Isaac Newton's description was: "A centripetal force is that by which bodies are drawn or impelled, or in any way tend, towards a point as to a centre."^[5]

Formula
The magnitude of the centripetal force on an object of mass m moving at tangential speed v along a path with radius of curvature r is:^[6]

F = ma_c = \frac{m v^2}{r}

where

a_c

is the centripetal acceleration. The direction of the force is toward the center of the circle in which the object is moving, or the osculating circle, the circle that best fits the local path of the object, if the path is not circular.^[7] The speed in the formula is squared, so twice the speed needs four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force. This force is also sometimes written in terms of the angular velocity ω of the object about the center of the circle:

F = m r \omega^2. \,

Expressed using the period for one revolution of the circle, T, the equation becomes:

F = m r \frac{4\pi^2}{T^2}.

^[8]

In particle accelerators, velocity can be very high (close to the speed of light in vacuum) so the same rest mass now exerts greater inertia (relativistic mass) thereby requiring greater force for the same centripetal acceleration, so the equation becomes:

F = \frac{\gamma m v^2}{r}

Where the Lorentz factor is defined as:

\gamma = \frac{1}{\sqrt{1-v^2/c^2}}

Sources of centripetal force

A body experiencing uniform circular motion requires a centripetal force, towards the axis as shown, to maintain its circular path.

For a satellite in orbit around a planet, the centripetal force is supplied by gravity. Some sources, including Newton, refer to the entire force as a centripetal force, even for eccentric orbits, for which gravity is not aligned with the direction to the center of curvature.^[9]
The centripetal force acts from the center of mass of the rotating object, on an object a distance "r" from its center; If both objects are rotating they will affect each other; for circular orbits, the center of mass is the center of the circular orbits. For non-circular orbits or trajectories, only the component of force directed orthogonally to the path (toward the center of the osculating circle) is termed centripetal; the remaining component acts to speed up or slow down the satellite in its orbit.^[10] For an object swinging around on the end of a rope in a horizontal plane, the centripetal force on the object is supplied by the tension of the rope. For a spinning object, internal tensile stress provides the centripetal forces that make the parts of the object trace out circular motions.
The rope example is an example involving a 'pull' force. The centripetal force can also be supplied as a 'push' force such as in the case where the normal reaction of a wall supplies the centripetal force for a wall of death rider.
Another example of centripetal force arises in the helix which is traced out when a charged particle moves in a uniform magnetic field in the absence of other external forces. In this case, the magnetic force is the centripetal force which acts towards the helix axis.
Analysis of several cases
Below are three examples of increasing complexity, with derivations of the formulas governing velocity and acceleration.
Uniform circular motion

Uniform circular motion refers to the case of constant rate of rotation. Here are two approaches to describing this case.
Calculus derivation
In two dimensions the position vector

\textbf{r}

which has magnitude (length)

r

and directed at an angle

\theta

above the x-axis can be expressed in Cartesian coordinates using the unit vectors

\hat{x}

and

y-hat

:^[11]

\textbf{r} = r \cos(\theta) \hat{x} + r \sin(\theta) \hat{y}.

Assume uniform circular motion, which requires three things.

The object moves only on a circle.

The radius of the circle $r$ does not change in time.

The object moves with constant angular velocity $\omega$ around the circle. Therefore $\theta = \omega t$ where $t$ is time.

Now find the velocity

\textbf{v}

and acceleration

\textbf{a}

of the motion by taking derivatives of position with respect to time.

\textbf{r} = r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y}

\dot{\textbf{r}} = \textbf{v} = - r \omega \sin(\omega t) \hat{x} + r \omega \cos(\omega t) \hat{y}

(*)

\ddot{\textbf{r}} = \textbf{a} = - r \omega^2 \cos(\omega t) \hat{x} - r \omega^2 \sin(\omega t) \hat{y}

\textbf{a} = - \omega^2 (r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y})

"

(*) Sptt Note

Please note the rotation matrix.

In two dimensions (from wiki)

A counterclockwise rotation of a vector through angle θ. The vector is initially aligned with the x-axis.

In two dimensions every rotation matrix has the following form:

R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}

This rotates column vectors by means of the following matrix multiplication:

\begin{bmatrix} x' \\ y' \\ \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix}

So the coordinates (x',y') of the point (x,y) after rotation are:

x' = x \cos \theta - y \sin \theta\,

y' = x \sin \theta + y \cos \theta\,

The direction of vector rotation is counterclockwise if θ is positive (e.g. 90°), and clockwise if θ is negative (e.g. -90°). Thus the clockwise rotation matrix is found as:

R(-\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{bmatrix}\,

Note that the two-dimensional case is the only non-trivial (e.g. one dimension) case where the rotation matrices group is commutative, so that it does not matter the order in which multiple rotations are performed. An alternative convention uses rotating axes,^[1] and the above matrix also represents a rotation of the axes clockwise through an angle θ.

"

Notice that the term in parenthesis is the original expression of

\textbf{r}

in Cartesian coordinates. Consequently,

\textbf{a} = - \omega^2 \textbf{r}.

Vector relationships for uniform circular motion; vector Ω representing the rotation is normal to the plane of the orbit with polarity determined by the right-hand rule and magnitude dθ /dt.

The image at right shows the vector relationships for uniform circular motion. The rotation itself is represented by the angular velocity vector Ω, which is normal to the plane of the orbit (using the right-hand rule) and has magnitude given by:

|\mathbf{\Omega}| = \frac {\mathrm{d} \theta } {\mathrm{d}t} = \omega \ ,

with θ the angular position at time t. In this subsection, dθ/dt is assumed constant, independent of time. The distance traveled dℓ of the particle in time dt along the circular path is

\mathrm{d}\boldsymbol{\ell} = \mathbf {\Omega} \times \mathbf{r}(t) \mathrm{d}t \ ,

which, by properties of the vector cross product, has magnitude rdθ and is in the direction tangent to the circular path.
Consequently,

\frac {\mathrm{d} \mathbf{r}}{\mathrm{d}t} = \lim_{{\Delta}t \to 0} \frac {\mathbf{r}(t + {\Delta}t)-\mathbf{r}(t)}{{\Delta}t} = \frac{\mathrm{d} \boldsymbol{\ell}}{\mathrm{d}t} \ .

In other words,

\mathbf{v}\ \stackrel{\mathrm{def}}{ = }\ \frac {\mathrm{d} \mathbf{r}}{\mathrm{d}t} = \frac {\mathrm{d}\mathbf{\boldsymbol{\ell}}}{\mathrm{d}t} = \mathbf {\Omega} \times \mathbf{r}(t)\ .

Differentiating with respect to time,

\mathbf{a}\ \stackrel{\mathrm{def}}{ = }\ \frac {\mathrm{d} \mathbf{v}} {d\mathrm{t}} = \mathbf {\Omega} \times \frac{\mathrm{d} \mathbf{r}(t)}{\mathrm{d}t} = \mathbf{\Omega} \times \left[ \mathbf {\Omega} \times \mathbf{r}(t)\right] \ .

Lagrange's formula states:

\mathbf{a} \times \left ( \mathbf{b} \times \mathbf{c} \right ) = \mathbf{b} \left ( \mathbf{a} \cdot \mathbf{c} \right ) - \mathbf{c} \left ( \mathbf{a} \cdot \mathbf{b} \right ) \ .

Applying Lagrange's formula with the observation that Ω • r(t) = 0 at all times,

\mathbf{a} = - {|\mathbf{\Omega|}}^2 \mathbf{r}(t) \ .

In words, the acceleration is pointing directly opposite to the radial displacement r at all times, and has a magnitude:

|\mathbf{a}| = |\mathbf{r}(t)| \left ( \frac {\mathrm{d} \theta}{\mathrm{d}t} \right) ^2 = r {\omega}^2\

where vertical bars |...| denote the vector magnitude, which in the case of r(t) is simply the radius r of the path. This result agrees with the previous section, though the notation is slightly different.
When the rate of rotation is made constant in the analysis of nonuniform circular motion, that analysis agrees with this one.
A merit of the vector approach is that it is manifestly independent of any coordinate system.

Centripetal Force

Any motion in a curved path represents accelerated motion, and requires a force directed toward the center of curvature of the path. This force is called the centripetal force which means "center seeking" force. The force has the magnitude

Swinging a mass on a string requires string tension, and the mass will travel off in a tangential straight line if the string breaks.
The centripetal acceleration can be derived for the case of circular motion since the curved path at any point can be extended to a circle.

Note that the centripetal force is proportional to the square of the velocity, implying that a doubling of speed will require four times the centripetal force to keep the motion in a circle. If the centripetal force must be provided by friction alone on a curve, an increase in speed could lead to an unexpected skid if friction is insufficien

Centripetal Acceleration

The centripetal acceleration expression is obtained from analysis of constant speed circular motion by the use of similar triangles. From the ratio of the sides of the triangles:

-------

sptt

Lecture on Physics by Dr Yukawa

Friday, August 8, 2014

My Note - 2) Rotation

Angular acceleration

Mathematical definition

Equations of motion

Constant acceleration

Non-constant acceleration

No comments:

Post a Comment

About Me