# Rigid body kinetic energy+ constraints (upper level classical mechanics)

• fluidistic
In summary: This is the relative sliding velocity.What is the angular velocity of the cylinder \omega_{c} and what is the total energy of this system?
fluidistic
Gold Member

## Homework Statement

Using the corresponding constraints conditions, calculate the kinetic energy of
1)A homogeneous cylinder of radius a that rolls inside a cylindrical surface of radius R>a.

## Homework Equations

My toughts: I hope they meants "roll without slipping". Let's consider this case!
$T=\frac{mv ^2_{CM}}{2}+\frac{I_1\omega ^2 _x+I_2\omega ^2 _y+I_3 \omega ^2 _z}{2}$.

## The Attempt at a Solution

Ok I already found out that $I_1=I_2=\frac{m}{4} \left ( \frac{h^2}{3}+a ^2 \right )$ and $I_3=\frac{ma^2}{2}$.
The constraint is that the line of the rolling cylinder in contact with the cylindrical surface does not slide. Mathematically I'm not really sure how to write this down. First, I'd set my coordinate system in the center of the rolling cylinder. (I remember in introductory physics that we dealt with this by setting the relative speed of the cylinders to be 0 m/s, but here I'm not sure)
Considering the axis along the height of the rolling cylinder as the z-axis, I think that $\omega _x =\omega _y =\vec 0$. While $\omega _z \neq 0 \neq \text{constant}$. It would be worth $\omega _z (t)= \frac{d \theta (t)}{dt}$ but I'm not sure I should use "theta" as angle. I've read in Goldstein that I should use Euler angles for this, but I have absolutely no idea how.
If the cylinder starts rolling at $t=0$ from its maximum height, $\theta (0)=0$, which is an initial condition for the motion equation of the cylinder.

Any help is appreciated, as usual.

Edit: Just checked out the picture of Euler angles in wikipedia and I think that instead of theta I should use beta.

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using the circumference of the circular cross section of each cyclinder (outer & rolling) you should be able to relate the angular velocity of the centre of mass to to the angular velocity of the rotation of the inner cylinder itself

lanedance said:
using the circumference of the circular cross section of each cyclinder (outer & rolling) you should be able to relate the angular velocity of the centre of mass to to the angular velocity of the rotation of the inner cylinder itself

I appreciate your help very much. However I do not know how to do so.
I realize that the center of mass' motion is a circle of radius R-a. I know that the circumferences of the outer and inner cylinders are $2 \pi R$ and $2\pi a$ a respectively.
I know that the instant velocity of the inner cylinder surface with respect to the outer one is 0m/s since I considered no rolling.

It will help to consider 2 angular velocities, that of the CoM $\omega_{CM}$ and that of the smaller cylinder itself $\omega_{c}$

Now, how about first considering the case where the smaller cylinder is rigidly rotating about at the centre of the larger cylinder. In this case the surface slides. Imagine a rigid arm bolted across the smaller cylinder on a pin at the centre of the large cylinder.

Then if the centre of mass velocity is $v_{CM} = \omega_{CM}(r-a)$ then velocity of the small cylinder scales with distance form the inner cylinder centre and at the surface of the large cylinder it is $v_R = v_{CM} \frac{R}{R-a}$. This is the relative sliding velocity.

What is the angular velocity of the cylinder $\omega_{c}$
and what is the total energy of this system? Now when you move to the no sliding condition you have
$v_R = 0$

So there is an extra angular rotation of the body we must also account for . We want to relate the angular velocity of the CoM to the angular velocity of the small body itself.

2 equivalent ways to relate these
- use the zero velocity constraint to solve for the required body angular velocity
- use the ratio of circumferences to find out how many times the inner cylinder rotates in a single revolution of the larger cylinder

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Thanks for your patience and time!
lanedance said:
It will help to consider 2 angular velocities, that of the CoM $\omega_{CM}$ and that of the smaller cylinder itself $\omega_{c}$
Ok, perfect.

Now, how about first considering the case where the smaller cylinder is rigidly rotating about at the centre of the larger cylinder. In this case the surface slides. Imagine a rigid arm bolted across the smaller cylinder on a pin at the centre of the large cylinder.
I've no problem with this, but here the English language is also a problem for me. I don't really understand the situation. Do you mean that a=R/2 (condition to be fixed on the center of the outer cylinder) and also that it's forced to slide over the outer cylinder's internal surface? Do you consider the outer cylinder to "swing" with respect to an external reference frame?

Then if the centre of mass velocity is $v_{CM} = \omega_{CM}(r-a)$
Ok I understand this
then velocity of the small cylinder scales with distance form the inner cylinder centre and at the surface of the large cylinder it is $v_R = v_{CM} \frac{R}{R-a}$. This is the relative sliding velocity.
Hmm, I don't know how to reach this.

What is the angular velocity of the cylinder $\omega_{c}$
Not sure. Maybe $\omega _c =v_R/a$?
and what is the total energy of this system?
I hope the outer cylinder isn't swinging. In this case it would be $\frac{mv^2_{CM} }{2}+\frac{I \omega _c }{2}+mg(h-a)$ where h is the height of the center of mass with respect to the lowest line in the outer cylinder.

Now when you move to the no sliding condition you have
$v_R = 0$
Fine.

So there is an extra angular rotation of the body we must also account for .
I don't understand why.
We want to relate the angular velocity of the CoM to the angular velocity of the small body itself.
In other words $\omega _{CM}$ in function of $\omega _c$?

2 equivalent ways to relate these
- use the zero velocity constraint to solve for the required body angular velocity
- use the ratio of circumferences to find out how many times the inner cylinder rotates in a single revolution of the larger cylinder
Ok. I'll try once I understand better all the previous things.

ok, that confused things more than i would have liked... let's just deal with the problem at hand... where the cylinder does not slide

lets start by working in the absolute reference frame and assuming we know $\omega_{CM}$,
R is radius of large cylinder
a that of small

then the velocity of the centre of mass of the small cylinder is
$$v_{CM} = \omega_{CM} (r-a)$$
the velocity of the surface (at r= a) is zero as it does not slip
$$v_{a} = 0$$

Now consider the dashed reference frame moving with the centre of mass,
$$v_{CM}' = 0$$
$$v_{a}' = -\omega_{CM} (r-a)$$

Can you find the angular velocity?

lanedance said:
ok, that confused things more than i would have liked... let's just deal with the problem at hand... where the cylinder does not slide

lets start by working in the absolute reference frame and assuming we know $\omega_{CM}$,
R is radius of large cylinder
a that of small

then the velocity of the centre of mass of the small cylinder is
$$v_{CM} = \omega_{CM} (r-a)$$
the velocity of the surface (at r= a) is zero as it does not slip
$$v_{a} = 0$$

Now consider the dashed reference frame moving with the centre of mass,
$$v_{CM}' = 0$$
$$v_{a}' = -\omega_{CM} (r-a)$$

Can you find the angular velocity?
I'm trying hard but can't make it.
I know that the small cylinder can make R/a rotations within the big cylinder before returning back to its original position.
So we have $v_a'=-\omega _{CM}(R-a)$. I'm not understanding very well the minus sign here though. Anyway I think that $v_a'=\omega _c a \Rightarrow \omega _{CM}=\omega _c \frac{a}{a-R}$. I'm totally stuck here.
P.S.:I realize my last equation isn't true. The modulus of $\omega _{CM}$ is greater than the one of $\omega _c$ which is senseless. By total intuition I would write $\omega _{CM}=\frac {\omega _ca}{R}$ but I've no idea if it's right nor how to derive it.

Edit: My intuition was wrong.

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I think you are on the right track

the minus sign they are in different direction hopefully you have drawn a good picture

first of the 2 cylinders together, working in the global ref frame (large cylinder is stationary)
- make the contact point at the far right (positive x direction)
- say $\omega_{CM}$ is CCW
- this makes $v_{CM}=\omega_{CM}(R-a)$ in the positive y direction
- mark the contact point as v_a=0

now draw the single cylinder alone, still in the same ref frame, and mark on the same velocities

finally draw the the single cylinder alone, now in the ref frame of the small cylinders CoM velocity, but not rotating
- $v_{CM}'=0$ by def'n
- $v_{a}'=v_a-v_{CM} = 0-\omega_{CM}(R-a)$
so writing the
$$v_{a}'=\omega_c a = -\omega_{CM}(R-a)$$

you get as you had (i prefer considering the little cylinder)
$$\omega_c = \omega_{CM}\frac{(R-a)}{a}$$

now to convince yourself it is a reasonable answer

first imagine the large cylinder was instantaneously removed
- this says the small cylinder would move off at speed $v_{CM =}\omega_{CM}(R-a)$ and rotating at $\omega_c = \omega_{CM}\frac{(R-a)}{a}$
seem reasonable

examine the limiting behaviour
$$\omega_c = \omega_{CM}\frac{(R-a)}{a}$$

when $a<<R [/tex] the little cylinder cylinder spins really fast when [itex]a \approx R [/tex] the little cylinder cylinder barely rotates try and convince yourself this all makes sense (and feel free to question, though it look good to me everyone makes mistakes no and then;) Last edited: the other thing to consider as mentioned before is the ratio of circumference imagine the large cylinder is unrolled flat, it has a length of [itex] 2 \pi R$

if you rolled the small cylinder along the unrolled length without slipping it would rotate $2 \pi \frac{R}{a}$ radians

when you roll the cylinder back up, this is not quite true and i think you actually rotate only $2 \pi (\frac{R}{a}-1)$ radians, as you remove whole rotation as you go round the cylinder, but you may want to check that (consider the limiting case where R~a)

Thanks once again lanedance.
Actually I didn't understand how it's possible to get $\omega _c =\frac{\omega _{CM} (R-a) }{a}$. I get $\omega _c =\frac{\omega _{CM} (a-R) }{a}$. Since a-R<0, it means that my answer says that if the cylinder making a clock-wise swing inside the big cylinder, then it is rotating on itself counter-clowisely. While your answer implies that $\omega _c$ and $\omega _{CM}$ have the same direction and hence if the cylinder swings clock-wise then it also rotate on itself clockwise. Am I right on this?

Also I didn't get the point of the ratio circumference. It's supposed to help me to find $\omega _c$, thus $\omega _{CM}$ and thus $V _{CM}$ I guess... but how?

you're right, whilst i mentioned the rotation directions i dropped a minus sign - good pick up

the ratio of circumferences was (for me) an interesting insight to the problem, however if its not helping or confusing the issue, don't worry about it...

Ok :)
Now how do I continue?
Once I've $\omega _c$, $\omega {CM}$ or $V_{CM}$ I'm done.

fluidistic said:

## Homework Statement

Using the corresponding constraints conditions, calculate the kinetic energy of
1)A homogeneous cylinder of radius a that rolls inside a cylindrical surface of radius R>a.

## Homework Equations

My toughts: I hope they meants "roll without slipping". Let's consider this case!
$T=\frac{mv ^2_{CM}}{2}+\frac{I_1\omega ^2 _x+I_2\omega ^2 _y+I_3 \omega ^2 _z}{2}$.

## The Attempt at a Solution

Ok I already found out that $I_1=I_2=\frac{m}{4} \left ( \frac{h^2}{3}+a ^2 \right )$ and $I_3=\frac{ma^2}{2}$.
The constraint is that the line of the rolling cylinder in contact with the cylindrical surface does not slide. Mathematically I'm not really sure how to write this down. First, I'd set my coordinate system in the center of the rolling cylinder. (I remember in introductory physics that we dealt with this by setting the relative speed of the cylinders to be 0 m/s, but here I'm not sure)
Considering the axis along the height of the rolling cylinder as the z-axis, I think that $\omega _x =\omega _y =\vec 0$. While $\omega _z \neq 0 \neq \text{constant}$. It would be worth $\omega _z (t)= \frac{d \theta (t)}{dt}$ but I'm not sure I should use "theta" as angle. I've read in Goldstein that I should use Euler angles for this, but I have absolutely no idea how.
If the cylinder starts rolling at $t=0$ from its maximum height, $\theta (0)=0$, which is an initial condition for the motion equation of the cylinder.

Any help is appreciated, as usual.

Edit: Just checked out the picture of Euler angles in wikipedia and I think that instead of theta I should use beta.

first of all i must say the question is askin energy not momentum or velocity so you need not go into vectors and coordinate system as energy is scaler...yes u can do it with cooerdinate but it will complicate solution

i will try to do it with superposition principle

"cylinder is moving inside the big cyllinder that means the work done by the normal force is zero":- implies KE is constant:- again IMPLIES :if we remove big cyllinder KE remains same(as force was perpendicular to motion and the friction doesn't work in rolling)

after removing big cyll small cyll moves lenearly and also rotates

now we have to find 'w' (angular vel about its axis afterwards)

and lenear velocity 'v' (afterwards)

NOW the solution:
lets us cosider small cylinders angular velocity(abouts its own axis)= 'w' {remains same}
its speed in circular motion be 'v'= w*(R-a) {after removing big Cyll it is its lenear velocity}
'a' is the radius of small cylinder

moment of inertia along small cyllinders centre axis=1/2(m*a^2)

now total energy= translation KE+ rotational KE
=1/2(m*v^2) + 1/2(I*w^2)

this is the correct solution

darkxponent said:
now total energy= translation KE+ rotational KE
=1/2(m*v^2) + 1/2(I*w^2)

this is the correct solution

Thanks for the reply :D Well, I don't really understand why you consider the case when we remove the big cylinder... I mean it's not stated in the problem.
Shouldn't the solution be =1/2(m*v'^2) + 1/2(I*w'^2)? (notice the primes)Edit: Also I don't agree with you when you say that the kinetic energy remains constant. When the small cylinder is at its max height, its kinetic energy is null.

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fluidistic said:
Thanks for the reply :D Well, I don't really understand why you consider the case when we remove the big cylinder... I mean it's not stated in the problem.
Shouldn't the solution be =1/2(m*v'^2) + 1/2(I*w'^2)? (notice the primes)

Edit: Also I don't agree with you when you say that the kinetic energy remains constant. When the small cylinder is at its max height, its kinetic energy is null.

i did not consider gravity...it is no where mentioned in the question

you shud post full question then

fluidistic said:
Ok :)
Now how do I continue?
Once I've $\omega _c$, $\omega {CM}$ or $V_{CM}$ I'm done.

should be pretty much there - can you take it from here?

darkxponent said:
i did not consider gravity...it is no where mentioned in the question

you shud post full question then
My bad, for some reason I thought there was gravity in this problem. You are right, there is none. However I didn't really understand the point of removing the big cylinder. I understand that we can do it since there's no external force "eating" the kinetic energy of the small cylinder, but I don't understand the point of doing it. Where does this simplify the equations?

lanedance said:
should be pretty much there - can you take it from here?
My bad... I had a big mental block lasting for days. I think I get it now.
$T=\frac{ma^2 \omega _c ^2 }{2} \left ( \frac{1}{(a-R)^2 }+1 \right )$.
I thought that $\omega _c$ was unknown and was a property of the cylinders, which is totally false. It's a "given" value. I hope I didn't make any algebra error.

fluidistic said:
My bad, for some reason I thought there was gravity in this problem. You are right, there is none. However I didn't really understand the point of removing the big cylinder. I understand that we can do it since there's no external force "eating" the kinetic energy of the small cylinder, but I don't understand the point of doing it. Where does this simplify the equations?

what does the big cyllindrical surface doing?? :-it is providing 1) normal force 2)friction to support rolling.

for me:i don't know the big cyllinder
i only know the two forces

now situation: "a cyllinder being acted upon by two forces which don't work"

now as both forces do not do any work on the little cyllinder...removing these forces would
not change KE of the cyllinder

so as soon as we remove these forces the motion of cyllinder is linear as well as rotating about its axis this

so the solutin remains same (provided there is no gravity)

yeah so as darkxponent has said, the energy of the little cylinder is conserved

now that you know how the angular velocities, you should be able to write the total energgy in terms of only the mass, radii & angular rotational speed of the CoM

if required incorporating gravity would be a simple exercise, just adding a gravitational
potential term

darkxponent said:
what does the big cyllindrical surface doing?? :-it is providing 1) normal force 2)friction to support rolling.

for me:i don't know the big cyllinder
i only know the two forces

now situation: "a cyllinder being acted upon by two forces which don't work"

now as both forces do not do any work on the little cyllinder...removing these forces would
not change KE of the cyllinder

so as soon as we remove these forces the motion of cyllinder is linear as well as rotating about its axis this

so the solutin remains same (provided there is no gravity)
Ah thanks... I think I understand. So in the situation of the little cylinder alone with linear motion, it would rotate on itself, not because of friction (since you removed it) but from its inertia.

this is superposition principle dude...try to change a complicated situation into a simpler one

Whoops, I missed your post (I was typing my post while you were sending yours)
lanedance said:
yeah so as darkxponent has said, the energy of the little cylinder is conserved

now that you know how the angular velocities, you should be able to write the total energgy in terms of only the mass, radii & angular rotational speed of the CoM

if required incorporating gravity would be a simple exercise, just adding a gravitational
potential term
I almost did this. Instead of $\omega _{CM}$ I took $\omega _c$. Is this still correct?
Is my solution in post 17 correct?

Edit: It should be: $T=\frac{ma^2 \omega _c ^2 }{2} \left ( \frac{1}{(a-R)^2 }+ \frac {1}{2} \right )$.

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lanedance said:
yeah so as darkxponent has said, the energy of the little cylinder is conserved

now that you know how the angular velocities, you should be able to write the total energgy in terms of only the mass, radii & angular rotational speed of the CoM

if required incorporating gravity would be a simple exercise, just adding a gravitational
potential term

But when we consider gravity then KE is not constant...so question must provide info about initial positions and time...and also the solution would be a function of time or a function of theeta

I've just seen that the problem is from Landau & Lifgarbagez's mechanics book. The answer given in the book is $\frac{3}{4} \mu (R-a)^2 \omega _c ^2$. I don't know what mu is though. I've been looking through the book and it doesn't seem to be the reduced mass. It might just be the mass?
So it means my answer is wrong... Where did I go wrong?!

here it is...you are doing some calculation mistake...this question was meant to be done thru superpostion principal only

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fluidistic said:
Edit: It should be: $T=\frac{ma^2 \omega _c ^2 }{2} \left ( \frac{1}{(a-R)^2 }+ \frac {1}{2} \right )$.

how did you get this answer? This is Dimensionally incorrect. Didnt you notice it?

Thanks for the reply and for showing me that mu is the mass, for Landau. Actually I think I made a mistake in the answer given by Landau. It should be $T=\frac{3}{4} (R-a)^2 \omega ^2 _{CM}$. In terms of $\omega _c$ I get $T= \frac{3}{4} a^2 \omega _c ^2$.
Now I understand that it's irrelevant to write (R-a) or (a-R) since you have to square it and it gives the same.
So I reached the answer thanks to you guys. But I must admit I'm having an extremely hard time to "understand" the step when I have to relate the 2 omega's. My brain just can't seem to "simulate" the situation even though I have the expression under my eyes.
I think that $\frac{\omega _c }{\omega _{CM} }=\frac{ (a-R) }{a}$ is a bit more natural to me. I realize I was wrong when I said this implies that $\omega _c$ is smaller than $\omega _{CM}$. In fact there's a critical value for the radius a so that $\omega _c$ becomes smaller than $\omega _{CM}$, something my brain still have trouble with. This occurs when $a=R/2$. For smaller values of a, $\omega _c$ is greater than $\omega _{CM}$ (which my brain is OK with) while for $R/2<a<R$, $\omega _c$ is smaller than $\omega _{CM}$ which seems weird to me.

All we have to do is to consider that point of contact remains at rest which gives the required equations between Vcm, Wcm and Wc.

fluidistic said:
for $R/2<a<R$, $\omega _c$ is smaller than $\omega _{CM}$ which seems weird to me.

It is not weird. Just try to visualise it.

darkxponent said:
It is not weird. Just try to visualise it.

I still can't. I mean, for a=R/2 or a quarter of the diameter, the 2 angular velocities are the same, and only for this radius. So weird to me.

okay

NOW let's visualise:

fisrt i forgot 'Wc' . Just forget that it exists in the question.

i only know Wcm and Vcm:

NOW according to pure rolling Vcm = Wcm * a;

now the only twist is that this cyllinder is rolling not on a plane but in cyllindrical surface

We Find time period(Tc) of the cyllinder = 2*pi*(R-a)/Vcm

THIS GIVES Wc as::

Wc = 2*pi/Tc

= Vcm/(R-a)

=Wcm *a/(R-a)

darkxponent said:
okay

NOW let's visualise:

fisrt i forgot 'Wc' . Just forget that it exists in the question.

i only know Wcm and Vcm:

NOW according to pure rolling Vcm = Wcm * a;

now the only twist is that this cyllinder is rolling not on a plane but in cyllindrical surface

We Find time period(Tc) of the cyllinder = 2*pi*(R-a)/Vcm

THIS GIVES Wc as::

Wc = 2*pi/Tc

= Vcm/(R-a)

=Wcm *a/(R-a)
I understand everything if I assume that Vcm = Wcm * a. Could you explain a bit more how do you get it?
You considered a cylinder of radius a rolling over a plane (with rolling and not slipping?)? In those case I don't understand the meaning of $\omega _{CM}$. I mean the cylinder rotates around an axis that passes through the center of mass over a plane. Since the center of mass does not rotate, omega of center of mass is worth 0... What am I misunderstanding?

i took Wcm is not the angular velocity of centre of mass. It is angular velocity of the cyllinder about its centre of mass and similarly Wc is the angular velocity if the cyllinder about the centre of the cyllindrical surface

## 1. What is the definition of rigid body kinetic energy?

Rigid body kinetic energy refers to the energy possessed by a rigid body due to its motion. It is a form of mechanical energy and is calculated as the sum of the translational and rotational kinetic energies of the body.

## 2. How is the kinetic energy of a rigid body affected by constraints?

Constraints can limit the motion of a rigid body, which in turn affects its kinetic energy. For example, if a constraint restricts the rotation of a body, its rotational kinetic energy will be reduced.

## 3. Can constraints change the total kinetic energy of a rigid body?

No, constraints cannot change the total kinetic energy of a rigid body. The sum of the translational and rotational kinetic energies will remain constant, but the distribution of energy between these two forms may be affected by constraints.

## 4. How do you incorporate constraints into the equations for rigid body kinetic energy?

Constraints are incorporated into the equations for rigid body kinetic energy by using Lagrange multipliers. These multipliers act as constraints on the motion of the body and are included in the equations of motion to account for the effects of constraints on the body's kinetic energy.

## 5. What are some real-life examples of rigid body kinetic energy and constraints?

One example is a spinning top, where the constraint of the tip touching the ground limits its rotational motion. Another example is a pendulum, where the constraint of the string restricts the motion of the bob to a circular path. In both cases, the constraints affect the distribution of kinetic energy in the system.

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