# Object slides in rolling cylinder

1. Jul 19, 2015

### Nathanael

1. The problem statement, all variables and given/known data
A small puck of mass m is carefully placed onto the inner surface of the thin hollow thin cylinder of mass M and of radius R.

Initially, the cylinder rests on the horizontal plane and the puck is located at the height R above the plane as shown in the figure.
Find the interaction force between the puck and the cylinder at the moment when the puck passes the lowest point of its trajectory.
Assume that the friction between the puck and the inner surface of the cylinder is absent, and the cylinder moves on the plane without slipping. The free fall acceleration is g.

2. Relevant equations
Inertia about CoM of cylinder I0 = MR2
Force of friction $\equiv f$
I use capitals for the cylinder and lower case for the point-mass. For example:
Acceleration of point-mass $\equiv <a_x,a_y>$
Acceleration of cylinder's CoM $\equiv A$
Velocity of point-mass $\equiv <v_x,v_y>$
Velocity of cylinder's CoM $\equiv V$

3. The attempt at a solution
Net force in the horizontal (x) direction is friction:
f = MA+max

The only torque on the cylinder is friction:
fR = IA/R = MRA
f = MA

This implies that the point mass just falls straight downwards... not what I expected but I'll roll with it.

vy is zero at the lowest point, so conservation of energy gives:
mgR = 0.5(MV2+I0ω2) = MV2
V2 = mgR/M

If we move to the frame of the cylinder's CoM when the point mass is at the lowest point, then the point mass will be traveling in a circular arc (of radius R) at a speed which is the relative speed of the two objects. Since the point-mass is stationary, the relative speed is just the speed V of the cylinder's CoM. (No need to consider fictitious forces because the x-acceleration is momentarily zero.)

F-mg = mV2/R = m2g/M
F = mg(1+m/M)

This is not right.

2. Jul 19, 2015

### haruspex

It is not in general valid to take moments about an accelerating reference point. Try taking moments about point of contact on the ground.

3. Jul 19, 2015

### Nathanael

Suppose we have an axis perpendicular to V and g, which lies on the ground. Let our origin be on that axis. Let the the positive y direction be the direction of gravity, and the positive x direction be towards the contact point (between the cylinder and ground). Suppose the x-coordinate of the contact point (between the cylinder and ground) is XM and the x-coordinate of the point mass is xm. Also the y-coordinate of the point mass is ym.

$\tau = X_MN - X_MMg - x_mmg$
The normal force can be found in terms of the y acceleration of the point mass:
$(m+M)g - N = (m+M)\ddot y_{CoM} = m\ddot y_m \Rightarrow N = Mg+m(g-\ddot y_m)$
Which gives the torque:
$\tau = X_Mm(g-\ddot y_m) - x_mmg$
XM and xm can be related by $(x_m-X_M)^2 + (R-y_m)^2 = R^2$ but things are getting messy...

Perhaps finding the time derivative of angular momentum will make some things simplify.

$L = MVR + mv_xy_m - mv_yx_m$
The time derivative is:
$\dot L = MRA + m(a_xy_m + v_xv_y - v_xv_y -a_yx_m)=MRA + m(a_xy_m -a_yx_m)$

Somehow I don't think this is the path you meant...

4. Jul 19, 2015

### Nathanael

I just noticed you said through the point of contact. In my previous post I did it through an arbitrary point on the ground.

Isn't the point of contact also an accelerating reference point...?

5. Jul 19, 2015

### haruspex

Easiest may be to skip torques and forces for the generic position and use the conservation laws to find the velocities at the time of interest. Then you can find the forces that apply at that time.
I meant the point on the ground, i.e. XM=0.

6. Jul 19, 2015

### Nathanael

There are two unknowns, energy conservation gives one constraint, but I don't see any other conservation laws that can be used.

The momentum is almost conserved, except for what is imparted to Earth, which is ∫fdt, which gives us $MV+mv-\int fdt = 0$ which is just the integral of the force equation in the OP (with the initial condition added).

Okay, I thought you meant have the axis of rotation always be through the contact point, not just for the first instant.

7. Jul 19, 2015

### haruspex

If you take a fixed reference point on the ground, what external forces have a moment about that axis?

8. Jul 19, 2015

### Nathanael

Gravity and the normal force from the ground (which don't cancel)

9. Jul 20, 2015

### TSny

Consider the signs in your torque equation. Does positive angular acceleration $\alpha$ imply positive horizontal acceleration $A$?
That is, should you write $\alpha = A/R$ or $\alpha = -A/R$?

10. Jul 20, 2015

### haruspex

Hmm.. how true.

This problem looks really nasty. If theta is the angle of the puck's radius below starting point I get $(m+2M\sec^2(\theta))-m\dot\theta^2= g(2M+m)/R$.
Now, that might not be right in detail, but I think it will be about that nasty. Might need to bring other minds onto the problem.

11. Jul 20, 2015

### Nathanael

Nice catch if friction acts in the positive direction the resulting angular acceleration causes motion in the negative direction.

So it should be f = -I0A/R2 = -MA which combined with the equation f = MA+max we get:
A = -0.5ax(m/M) which implies Vf = -0.5(m/M)vf

It's still not coming out right.

The answer is supposed to be in the form of C1mg(C2+(m/M)/C3) where the Constants are to be determined. This was the form of my answer in the OP but the constants are not all supposed to be 1.

I am justified in taking the contact force to be ... F = mg+m(v-V)2/R ... right?
(Horizontal acceleration should be momentarily zero, so fictitious effects vanish.)

Is it true that the torque equation is unjustified because the cylinder's CoM is accelerating?

Last edited: Jul 20, 2015
12. Jul 20, 2015

### Qwertywerty

I may be wrong but from what I have read , when writing W-E theorem , you do not include KE of block , but when you write centripetal acceleration , you make use of m's velocity .

13. Jul 20, 2015

### Nathanael

In the centripetal acceleration term, I am not using m's velocity (v) I am using the relative velocity (v-V) but I incorrectly-concluded v=0 which is why I wrote mV2/R

14. Jul 20, 2015

### Qwertywerty

Why would you use relative velocity ?

When at the bottom , velocity of m is only , say v , and you should use that for both W-E theorem and writing centripetal acceleration .

15. Jul 20, 2015

### Nathanael

We have to switch to a reference frame in which the cylinder's CoM is not moving, or else the radius of curvature of the path of m will not be R.
(Imagine both the cylinder and the object m moving... the path of m is not a circle...)

The velocity of m in a reference frame where the cylinder's CoM is stationary is just the relative velocity v-V

16. Jul 20, 2015

### Qwertywerty

The block's path is not a circular motion , but at the instant at which the block is at the bottom , the radius of curvature should be R , since the bottom of the cylinder is instantaneously at rest and it becomes a simple case of circular motion with velocity v ( at that instant ) .

17. Jul 20, 2015

### Nathanael

I don't agree.

Consider a case where the cylinder's CoM to be at rest, but the cylinder is rotating. You see, the rotation has no effect on the path of m (it is frictionless, so it might as well not be rotating.)

It seems irrelevant whether one part of the cylinder is instantaneously at rest w.r.t. the ground. Only the CoM movement affects the path of m.

(I know this is not a good explanation, but I don't know how to explain it well because my understanding comes from visualizing the situation.)

I am fairly confident the radius of curvature will be reduced in the original reference frame.

18. Jul 20, 2015

### Qwertywerty

Consider the instant at which m is at the bottom and the case in which it is ever so slightly above the bottom and to the right of it .

In the second case , the moving cylinder exerts a normal that sort of pushes the block along with the cylinder . In the first , the block is not at all affected as in the second cas , by the normal .

Thus in the first case m's centripetal acceleration is independent of M .

*Thanks for being so patient .

19. Jul 20, 2015

### TSny

I don't see any error here.
So, are you getting C1 = C3 = 1 and C2 = 3? (Or something equivalent)

It is justified. That is, the net torque on the cylinder taken about the CoM of the cylinder is equal to $I_c \alpha$ even if the CoM is accelerating. Here $I_c$ is the moment of inertia of the cylinder about the CoM of the cylinder.

Last edited: Jul 20, 2015
20. Jul 20, 2015

### andrewkirk

I think that's the key,

Since the rotational KE of a rolling, hollow, thin-walled cylinder is equal to the translational energy (so the total KE is double the translational energy), and the cylinder-puck interaction is frictionless, we can solve the problem by replacing the problem with one where the cylinder's mass is doubled and the horizontal surface is frictionless. It's then just like a puck sliding down a sloping, sliding block, except the block has a curved surface.

With this revised problem, we regain the ability to use conservation of momentum.
Let $\theta$ be the puck's angle below the horizontal, which is initially zero, and let $x$ be the distance of the centre of the cylinder from its initial position.
Then conservation of energy gives us:

$$mg\sin\theta=\frac{1}{2}(2M)\dot{x}^2+\frac{1}{2}mv^2$$

Where $v$ is the speed of the puck relative to the ground. This is obtained by vector-adding its velocity $\dot{\theta}R$ relative to the cylinder to the velocity $\dot{x}$ of the cylinder. From a diagram and the cosine rule this is:
$$v^2=\dot{x}^2+(\dot{\theta}R)^2-2\dot{x}\dot{\theta}R\cos(90-\theta)=\dot{x}^2+(\dot{\theta}R)^2-2\dot{x}\dot{\theta}R\sin\theta$$

So the conservation of energy equation is

$$mg\sin\theta=\frac{1}{2}(2M+m)\dot{x}^2+\frac{1}{2}m(\dot{\theta}R)^2-m\dot{x}\dot{\theta}R\sin\theta$$

Momentum is conserved in the horizontal direction, as follows:
$$(2M)\dot{x}+m(\dot{x}-(\dot{\theta}R)\sin\theta)=0$$
where the second term inside the bracket is the horizontal component of the puck's motion relative to the cylinder.
That is:
$$\dot{x}=\frac{m\dot{\theta}R\sin\theta}{2M+m}$$

Substituting for $\dot{x}$ into the energy equation we get:

\begin{align*}mg\sin\theta&=\frac{1}{2}(2M+m)(\frac{m\dot{\theta}R\sin\theta}{2M+m})^2+\frac{1}{2}m(\dot{\theta}R)^2-\frac{m\dot{\theta}R\sin\theta}{2M+m}m\dot{\theta}R\sin\theta\\
&=\frac{1}{2}mR^2\dot{\theta}^2-\frac{1}{2}(\dot{\theta}\sin\theta)^2\frac{(mR)^2}{2M+m}
\end{align*}

This is an ODE of the form

$$a\sin\theta=b\dot{\theta}^2-c(\dot{\theta}\sin\theta)^2$$

I don't know if it can be solved analytically. If not, it could be solved numerically, given values of $R,M,m$, and then $\dot{x}$ can be calculated by the equation higher up.

Last edited: Jul 20, 2015