Object slides in rolling cylinder

[Nathanael]

Homework Statement

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.

Homework 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##

The Attempt at a Solution

Net force in the horizontal (x) direction is friction:
f = MA+ma_x

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.

v_y 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.

 

[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.

 

[Nathanael]

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

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 X_M and the x-coordinate of the point mass is x_m. Also the y-coordinate of the point mass is y_m.

The torque on the system about this axis is:
##\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##
X_M and x_m 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.

The angular momentum about this axis is:
##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...

 

[Nathanael]

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

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...?

 

[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.

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

I meant the point on the ground, i.e. X_M=0.

 

[Nathanael]

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.

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).

I meant the point on the ground, i.e. X_M=0.

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

 

[haruspex]

I don't see any other conservation laws that can be used.

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

 

[Nathanael]

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

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

 

[TSny]

The Attempt at a Solution

Net force in the horizontal (x) direction is friction:
f = MA+ma_x

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

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##?

 

[haruspex]

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

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.

 

[Nathanael]

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##?

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+ma_x we get:
A = -0.5a_x(m/M) which implies V_f = -0.5(m/M)v_f

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?

 

[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 .

 

[Nathanael]

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 .

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

 

[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 .

 

[Nathanael]

Why would you use relative velocity ?

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

 

[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 ) .

 

[Nathanael]

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 ) .

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.

 

[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 .

 

[TSny]

So it should be f = -I₀A/R² = -MA which combined with the equation f = MA+ma_x we get:
A = -0.5a_x(m/M) which implies V_f = -0.5(m/M)v_f

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.)

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

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

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.

 

[andrewkirk]

You see, the rotation has no effect on the path of m (it is frictionless, so it might as well not be rotating.)
...
Only the CoM movement affects the path of m.

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.

 

[Nathanael]

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

I am getting ##F = mg(1+\frac{4M^2+2mM+m^2}{M(2M+m)})##

The only way I see to simplify it is to this: ##F = mg(3+\frac{m^2}{M(2M+m)})##

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.

Interesting idea. So you've chosen this 'effective mass' such that every ##\dot x## corresponds to the same energy in both situations. It seems like there is a deeper explanation as to why this must produce the same behavior.
(When I say "deeper" I am particularly thinking about the Lagrangian, from which the equations of motion somehow come from the energy at each state. I haven't studied the Lagrangian, though.)
When we eliminate friction in the original problem, we get the equation mv_x=-2MV which is equivalent to your equation (except we are using different coordinates) so it is indeed as if the cylinder had twice the mass. (This is a hind-sight explanation, though. The physical correlation between the two scenarios is foggy to me.)

Your coordinates are more suitable for finding the time-dependent motion, but if we only care about the time when it is at the lowest point, then we don't need to solve this.

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 .

The centripetal acceleration depends on the path of m, so you can't really separate your two cases. It is not about where the m is at, but where m is going.

If the CoM of the cylinder is moving (regardless of the cylinder's rotation) then in a short amount of time (when m transitions from your first-case to your second-case) the cylinder will have moved left a bit causing the mass m to curve upwards more quickly than if the cylinder were still (and so the radius of curvature is reduced).

Hope this helps.

 

[TSny]

I am getting ##F = mg(1+\frac{4M^2+2mM+m^2}{M(2M+m)})##

The only way I see to simplify it is to this: ##F = mg(3+\frac{m^2}{M(2M+m)})##

Using energy conservation, what did you get for v and V?

 

[Nathanael]

Using energy conservation, what did you get for v and V?

##v^2=\frac{4MgR}{2M+m}##
##V^2=\frac{m^2gR}{M(2M+m)}##

 

[Nathanael]

I got it now, sorry TSny for making you do the dirty work :\

It was a simple algebra mistake (I forgot a factor of 2 on the -vV term in (v-V)²) sorry and thank you

 

[TSny]

OK. Interesting problem.

 

[haruspex]

It is justified. That is, the net torque on the cylinder taken about the CoM of the cylinder is equal to IcαI_c \alpha even if the CoM is accelerating.

You're right - I'm cautious about this because I know it can lead to errors in assessing angular momentum.
To demonstrate that it also works taking moments about point of contact:
Let the puck be at ##\theta## below horizontal, and N be the normal force between puck and cylinder.
##R\alpha=A##
##\tau = NR\cos(\theta) = 2MR^2\alpha=2MRA##
Horizontal component of puck's acceleration=##N\cos(\theta)/m=A\frac{2M}m##.
Since the horizontal accelerations have a constant ratio and both initial velocities are zero, the horizontal velocities are also in that ratio.
Let the horizontal velocities when the puck is at ground level be v, V for m, M respectively, measured in opposite directions.
##v=V\frac{2M}m##
##mgR=\frac 12(mv^2+2MV^2)##
Relative velocity = ##v+V##.
Centripetal acceleration = ##\frac{(v+V)^2}R = g(2+\frac mM)##.
Normal force = ##mg(3+\frac mM)##.
Is that the result you got?

 

[Qwertywerty]

Haruspex , could you please explain to me why we take relative velocity for centripetal acceleration ?

 

[haruspex]

Haruspex , could you please explain to me why we take relative velocity for centripetal acceleration ?

Consider the point of the cylinder in contact with the ground. Even though it has no horizontal velocity it does have upward acceleration (##\frac {V^2}R##). So we can see that the speed of the cylinder will affect the acceleration of the puck.
We can take the cylinder's centre as a frame of reference. The cylinder's acceleration is always horizontal, so won't affect vertical forces (and at the time of interest, even that is zero). In this frame, the puck is traveling around a fixed circle radius R at speed v+V.

 

[Qwertywerty]

We can take the cylinder's centre as a frame of reference. The cylinder's acceleration is always horizontal, so won't affect vertical forces (and at the time of interest, even that is zero). In this frame, the puck is traveling around a fixed circle radius R at speed v+V.

So is this what happens ? - In the ground frame , the acceleration of the bottom part of the cylinder causes change in the radius of curvature , and so we shift to the Com frame .

Is it possible to calculate the radius of curvature in the ground frame ?

Thanks .

 

[haruspex]

Is it possible to calculate the radius of curvature in the ground frame ?

If my argument is correct, we can calculate it from ##\frac{(v+V)^2}R=\frac{v^2}{R'}##.
Feel free to try deriving it directly from geometry.

 

[Nathanael]

Normal force = ##mg(3+\frac mM)##.
Is that the result you got?

Yessir

Your method also uses an accelerating axis of rotation. I must say I'm a bit confused about when it is invalid to use accelerating axes of rotation.
Maybe it could be a topic for one of your insights-posts.

 

[haruspex]

Your method also uses an accelerating axis of rotation. I must say I'm a bit confused about when it is invalid to use accelerating axes of rotation.
Maybe it could be a topic for one of your insights-posts.

As I wrote in post #26, I withdraw my earlier objection to using such a frame for torque. The dangers of using an accelerating reference axis for using conservation of angular momentum is covered by one of my Insights posts (Moments).

 

[Nathanael]

As I wrote in post #26, I withdraw my earlier objection to using such a frame for torque. The dangers of using an accelerating reference axis for using conservation of angular momentum is covered by one of my Insights posts (Moments).

Okay, I didn't know you already covered it, I just knew you posted a lot about misconceptions and common tricky situations.

So torque is fine but cons.of angular momentum is not; that makes more sense, thanks!

 

[TSny]

The basic rotational dynamical law is net torque equals rate of change of angular momentum ##\vec{\tau}_{net} = d\vec{L}/dt##. This law is not generally valid if you take your origin to be a point that is accelerating relative to an inertial frame.

Taking torques about the point of contact, P, between the cylinder and floor can be ambiguous.

For example, you can think of this point as a fixed point of the floor that happens to coincide instantaneously with the point of contact. In this case, P is a fixed point in an inertial frame. Then, the net torque acting on the cylinder about P will equal the rate of change of angular momentum of the cylinder with respect to P.

However, if you consider point P as moving with the point of contact so that P is always underneath the center of the cylinder, then P is now a point that is generally accelerating relative to an inertial frame. Now it will not be true in general that the net torque about P will equal the rate of change of angular momentum about P. For example the rate of change of angular momentum of the cylinder relative to P would not be ##2MR^2 \alpha## (it would be ##MR^2 \alpha##). Setting up ##\vec{\tau}_{net} = d\vec{L}/dt## would lead to an incorrect answer.

As a trivial example consider a point particle falling freely as shown below. Let P be a fixed point in the inertial frame of the Earth that happens to be located at the same height as the particle at the time of interest. Then you can easily check that the torque due to gravity about P equals the rate of change of angular momentum of the particle about P.

Now let P accelerate with the particle so that P is always a distance d to the right of the particle. The torque about P is the same as before. But the particle always has zero angular momentum relative to P. So, relative to the accelerating point P, ##\vec{\tau}_{net} \neq d\vec{L}/dt##.

 

[haruspex]

This law is not generally valid if you take your origin to be a point that is accelerating relative to an inertial frame.

But it's ok if it's the CoM of the object in question, because the acceleration of the body does not correspond to a (virtual) torque about that point, right?

Taking torques about the point of contact, P, between the cylinder and floor can be ambiguous.

I did clarify somewhere in the thread that I meant a fixed point on the ground.