# Dynamics of rotating cylinder

1. Homework Statement

Rotating with angular velocity $ω_0$ solid homogeneous cylinder of radius $r$ placed without starting forward speed at the bottom of the inclined plane, forming an angle $\alpha$ with the horizontal plane, and starts to roll in up. Determine the time during which the cylinder reaches the highest position on the inclined plane.

2. Homework Equations

$m\dfrac{dv}{dt}=F-mg\sin\alpha$

$I\dfrac{d\omega}{dt}=Fr$

3. The Attempt at a Solution
From integrating of the above eqns, I get:
$\dfrac{\omega r}{2}-v=g\sin\alpha \cdot t-const$
Where the constant we can determine from the initial conditions, e.g.
$const=-\dfrac{mr\omega_0}{2}$
So, I get eqn
$\dfrac{\omega r}{2}-v=g\sin\alpha \cdot t-\dfrac{\omega_0 r}{2}$

I think that at the start of pure rotation, the cylinder has to stop, ie, it reaches the maximum height. How is it possible to argue?

Then $v=-\omega r$, and $\omega=0$

$0=g\sin\alpha \cdot t-\dfrac{\omega_0 r}{2}$
$t=\dfrac{\omega_0 r}{2g\sin\alpha}$

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mfb
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I am missing some reason for the cylinder to roll upwards. Friction? If yes, how much? This will certainly influence the motion.

I think that at the start of pure rotation, the cylinder has to stop
The cylinder can keep rolling upwards afterwards.

I am missing some reason for the cylinder to roll upwards. Friction?
Yes, friction $F$.

If yes, how much? This will certainly influence the motion.
Interesting quastion.

I think the friction force, which raises the body, must be greater than the rolling friction, which is neglected in this problem. It turns out that an arbitrarily small sliding friction force can pull the body. But it also means that the acceleration is negative and the velocity of the center of mass must decrease with raising of body. [strike]When friction is included at the start, the body instantly appears nonzero velocity of the center of mass. I wonder what it is then?[/strike]

The cylinder can keep rolling upwards afterwards.
Yes, by inertia, I realized this already.

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Chestermiller
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You need to express your second equation in terms of m, dv/dt, and r. What is the moment of inertia of a cylinder expressed in terms of m and r? What do you get if you multiply dω/dt by r? From the second equation, what is F expressed in terms of m, r, and dv/dt? Take this result, and substitute it into your first equation.

You need to express your second equation in terms of m, dv/dt, and r. What is the moment of inertia of a cylinder expressed in terms of m and r? What do you get if you multiply dω/dt by r? From the second equation, what is F expressed in terms of m, r, and dv/dt? Take this result, and substitute it into your first equation.
I'm it's all done in the first post. Is not it?
Strange moment in this problem, it is that the time does not depend on the coefficient of friction. But it must, I think.

haruspex
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Oh, this is a nasty question. It defies intuition because it requires unlimited coefficient of friction. The instant the cylinder is placed on the slope there will be an impact. This means work is not conserved. You have to use conservation of angular momentum to figure out the rate of spin immediately after contact.
The easiest way is to take moments about the point of contact. That way, the impulse itself (being along the line of the ramp) has no moment.

Chestermiller
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I'm it's all done in the first post. Is not it?
Strange moment in this problem, it is that the time does not depend on the coefficient of friction. But it must, I think.
In the first equation, F has the wrong sign. F is pointing down the incline. It is harder to accelerate or decelerate a rotating cylinder than one that is sliding without rotation. The equivalent mass should be 3m/2, not m/2.

In the first equation, F has the wrong sign. F is pointing down the incline. It is harder to accelerate or decelerate a rotating cylinder than one that is sliding without rotation. The equivalent mass should be 3m/2, not m/2.
But then what is the force begins drag up of the cylinder to the top? I think that only the friction force, so it must be pointing up the incline. I was wrong only in the sign in the second equation, since the friction slows the rotation.

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haruspex
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All this discussion of F is irrelevant. Once rolling contact is established, and you know the speed at that point, you can use energy to calculate the height reached. Since deceleration will be constant, you can use the usual constant acceleration formulae to find the time.
But the first step is to find the initial speed up the ramp.

Chestermiller
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But then what is the force begins drag up of the cylinder to the top? I think that only the friction force, so it must be pointing up the incline. I was wrong only in the sign in the second equation, since the friction slows the rotation.
The incline is exerting a downward tangential force on the cylinder, also causing it to rotate. There is no force causing the cylinder to go up the incline. The cylinder goes up the incline as a result of its prior inertia. But it slows down as a result of the tangential gravitational component, and the tangential non-slip force component F.

All this discussion of F is irrelevant. Once rolling contact is established, and you know the speed at that point, you can use energy to calculate the height reached. Since deceleration will be constant, you can use the usual constant acceleration formulae to find the time.
But the first step is to find the initial speed up the ramp.
I am now so do the calculations, of course. But then where is the guarantee that the disc does not slip when climbing up? If it slips, the energy is not conserved, and the proposed method does not work you.

The cylinder goes up the incline as a result of its prior inertia. But it slows down as a result of the tangential gravitational component, and the tangential non-slip force component F.
The cylinder was no forward movement, but there were only rotating, so the center of mass can not move prior inertia. The friction force transferred energy of rotational motion into energy of translational motion, then it did the work.

haruspex
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I am now so do the calculations, of course. But then where is the guarantee that the disc does not slip when climbing up? If it slips, the energy is not conserved, and the proposed method does not work you.
Reread both my posts. Since you're not given the coefficient of friction, the problem cannot be solved unless you assume friction is adequate to prevent slipping at all times. (In the extreme, with no friction, it would spin forever at the bottom.) That splits the problem into two steps:
- what happens at initial contact? How can the rotation turn into a combination of rotation and linear movement 'instantly'? (see my first post)
- after that instant, there is rolling contact and you can use conservation of work (my 2nd post)

- what happens at initial contact? How can the rotation turn into a combination of rotation and linear movement 'instantly'? (see my first post)
You have to use conservation of angular momentum to figure out the rate of spin immediately after contact.
The easiest way is to take moments about the point of contact.
From the conservation of angilar momentum about the point of contact looks like:
$\dfrac{3mr^2\omega_0}{2}=mv_0 r$ It is Ok?

TSny
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Since you're not given the coefficient of friction, the problem cannot be solved unless you assume friction is adequate to prevent slipping at all times.
Unless I'm overlooking something (very possible), I think you can allow slipping at the beginning. I believe the only requirement is to assume that the kinetic friction force is enough to initially make the cylinder climb the slope. Otherwise, the amount of friction is not important.

Using your idea of taking moments about a fixed point (in the earth frame) that coincides with the initial point of contact makes the problem surprisingly simple.

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haruspex
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From the conservation of angilar momentum about the point of contact looks like:
$\dfrac{3mr^2\omega_0}{2}=mv_0 r$ It is Ok?
No. The value on the LHS would be true if it were already rolling, i.e. if it had a linear component of motion rω0. If you subtract out the moment that would have about the point of contact then you'll get the right value.
Conversely, on the RHS, you only have the moment due to linear motion. You need to add in the rotational component.

haruspex
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I think you can allow slipping at the beginning. I believe the only requirement is to assume that the kinetic friction force is enough to initially make the cylinder climb the slope.
Not sure what you mean by 'initially climb the slope'. If you mean rolling contact immediately then there is no slipping. If there's any slipping at all it will sap energy. You could assume some finite coefficient of friction and see what it leads, but I'd be astonished if the assumed unknown magically disappeared from the equations.

EDIT: I'm suitably astonished - see below.

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- after that instant, there is rolling contact and you can use conservation of work (my 2nd post)
$mgh=\dfrac{mr^2}{4}\dfrac{v_0^2}{r^2}$ -concervation energy low - I'm not sure

$h=\dfrac{3r^2}{4g}\omega_0^2$

Distance along the incline $x=\dfrac{h}{\sin\alpha}=\dfrac{v_0^2}{2a}$
$h=\dfrac{v_0^2\sin\alpha}{2a}$

from angular momentum conservation $\dfrac{mr^2}{2}\omega_0=mv_0r$ - I'm not sure

$v_0=\dfrac{\omega_0r}{2}$

$a=2g\sin\alpha$

$0=\dfrac{\omega_0r}{2}-2g\sin\alpha \cdot t$

$t=\dfrac{\omega_0 r}{4g\sin\alpha}$

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TSny
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Not sure what you mean by 'initially climb the slope'. If you mean rolling contact immediately then there is no slipping. If there's any slipping at all it will sap energy. You could assume some finite coefficient of friction and see what it leads, but I'd be astonished if the assumed unknown magically disappeared from the equations.
You can let the cylinder slip initially so that it's a force of kinetic friction that starts the cylinder moving up the slope. [EDIT: To be clear, Even though the cylinder slips when placed on the incline, it nevertheless starts accelerating up the incline immediately.] Eventually, that will change to rolling without slipping. You can go through a detailed calculation where you break it into parts and find the time for the slipping part and the additional time for the rolling w/o slipping. But, following your lead regarding taking moments about the initial point of contact, you don't need to go through all that.

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haruspex
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following your lead regarding taking moments about the initial point of contact, you don't need to go through all that.
You're quite right - it can all be solved in terms of angular momentum. My thought experiment with no friction misled me. If the friction is very low, the initial movement will be down the ramp. If we allow the ramp to be as long as necessary, and the friction to be any nonzero value, the time taken to stop rotating is constant.
sergiokapone, your latest answer is almost right, but it's out by a factor of 2.
What do you think the angular momentum is about the point of contact before contact is made? What is the torque about the point of contact subsequently?

Correction: it's not a matter of whether friction is nonzero. As TSny said (took me a while to catch up), you have to assume it moves up the ramp at least a little bit. That will be true if F > mg sin(α). Otherwise the answer is zero time.

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sergiokapone, your latest answer is almost right, but it's out by a factor of 2.
What do you think the angular momentum is about the point of contact before contact is made? What is the torque about the point of contact subsequently?
I do not understand how can I calculate the angular momentum of a rotating body around a different axis, which rotates about its axis. Before contact, spin is $\dfrac{mr^2}{2}\omega_0$, but it is the spin, i.e. angular momentum about cetnter of mass axis.

Again, if the body has already fallen. At the point of contact friction torque acts $-F \cdot r$, which slows the rotation. And, of course, prior the torque the spin is not conserved.

Of course, if the law of conservation writing relative to the points of contact, the friction force does not create torque, and one can write the conservation of angular momentum, but I do not understand how?

haruspex
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I do not understand how can I calculate the angular momentum of a rotating body around a different axis, which rotates about its axis. Before contact, spin is $\dfrac{mr^2}{2}\omega_0$, but it is the spin, i.e. angular momentum about cetnter of mass axis.
A body rotating about its mass centre has the same angular momentum about all points. (It's analogous to the fact that a "screw" - a pair of equal and opposite forces acting along different lines - has the same torque about every point.) Here's one way to see this: consider a body rotating about some point P distance r from its mass centre C. By the parallel axis theorem, the moment about P is (I+mr2)ω. But you can think of the body's movement at an instant as a combination of rotating about C plus a linear movement perpendicular to PC. The latter has angular momentum mr2ω about P, so rotation of the body about C must have angular momentum Iω about P.
Again, if the body has already fallen. At the point of contact friction torque acts $-F \cdot r$, which slows the rotation. And, of course, prior the torque the spin is not conserved.

Of course, if the law of conservation writing relative to the points of contact, the friction force does not create torque, and one can write the conservation of angular momentum,
Exactly. You can either fix on the point of initial contact, in which case you have to consider the moments of the normal force and gravity, or allow it to move as the point of contact move (so the only force with moment is gravity). It will lead to the same equation. So: what is the torque about the point of contact?

So: what is the torque about the point of contact?
$-mgr\sin\alpha$

Eqn of rotational movement:
$\dfrac{mr^2}{2}\dfrac{d\omega}{dt}=-mgr\sin\alpha$

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haruspex
$mgr\sin\alpha$
From integrating of the eqn of rotational motion $t=\dfrac{r\omega_0}{2g\sin\alpha}$