# Dynamics of rotating cylinder

#### sergiokapone

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.
I have to think about it.

#### sergiokapone

Combination of $mr^2/2 \cdot \omega_0$ and $mr^2\omega_0$ leads to $3mr^2/2 \cdot \omega_0$ and then only in the case of pure rotation. How you got $mr^2/2 \cdot \omega_0$?

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#### TSny

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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.
Applying the $\tau$ = dL/dt equation to a point that is accelerating can get you in trouble. Think about the case of a cylinder that is given an initial velocity and no rotation at the bottom of a frictionless incline. It will slide up the incline without developing any rotation. You can see that the $\tau$ = dL/dt equation doesn't hold for a coordinate system with origin at the point of contact and moving with the point of contact.

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#### haruspex

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Applying the $\tau$ = dL/dt equation to a point that is accelerating can get you in trouble.
True.. so how to justify this method? It gives the right answer (i.e. the same I got by doing it the hard way). Keeping the point of reference fixed doesn't help, does it? Unless it's rolling the torque as measured from a point on the ramp does not match the torque on the cylinder.

#### TSny

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True.. so how to justify this method? It gives the right answer (i.e. the same I got by doing it the hard way). Keeping the point of reference fixed doesn't help, does it? Unless it's rolling the torque as measured from a point on the ramp does not match the torque on the cylinder.
The general theorem $\vec{\tau}= d\vec{L}/dt$ is true relative to any fixed point in an inertial frame. This relation is not true, in general, relative to a point that is accelerating relative to an inertial frame. The exception is the center of mass point. It can be shown that $\vec{\tau}_{cm} = d\vec{L}_{cm}/dt$ is true even if the center of mass is accelerating.

In this problem, the theorem as applied to a fixed point on the plane gives the correct equation:

$mgsin\alpha= -(I_{cm}d\omega /dt +mR\;dv_{cm}/dt)$ where R is the radius of the cylinder.

Here, the general relation $\vec{L} = \vec{L}_{cm} + M\vec{r}_{cm}\times\vec{v}_{cm}$ was used.

If you tried to use these relations relative to the point of contact P (which is accelerating) you would get the incorrect equation $mgsin\alpha= -I_{cm}\;d\omega /dt$.

If you happen to have rolling without slipping, then you can show that the equation relative to the fixed point on the plane can be written as $mgsin\alpha= -I_{P}\;d\omega /dt$ where $I_P$ is the moment of inertia relative to P. So, for rolling without slipping you will get the correct equation relative to point P if you use the moment of inertia relative to P.

It seems odd at first that in this problem the incorrect equation relative to P gives the correct result for the time it takes the cylinder to lose all of its initial angular momentum $L_0$. But I think this is due to the fact that the torque $\tau$ relative to P is the same as the torque relative to the fixed point. $\Delta t = |L_{0}/\tau|$ depends only on the torque and the initial angular momentum.

#### sergiokapone

In this problem, the theorem as applied to a fixed point on the plane gives the correct equation:

$mgsin\alpha= -(I_{cm}d\omega /dt +mR\;dv_{cm}/dt)$ where R is the radius of the cylinder.

Here, the general relation $\vec{L} = \vec{L}_{cm} + M\vec{r}_{cm}\times\vec{v}_{cm}$ was used.
This is should be the same as I used a separately the eqns of center mass motion and rotational motion in my initial post, but it is not.

#### TSny

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This is should be the same as I used a separately the eqns of center mass motion and rotational motion in my initial post, but it is not.
I'm not sure about your sign conventions when you wrote $I\dfrac{d\omega}{dt}=Fr$.

If you take positive values of $\omega$ to correspond to the direction of the initial angular velocity of the cylinder, then the equation would be $I\dfrac{d\omega}{dt}=-Fr$. Then I believe everything is ok.

#### sergiokapone

I'm not sure about your sign conventions when you wrote $I\dfrac{d\omega}{dt}=Fr$.

If you take positive values of $\omega$ to correspond to the direction of the initial angular velocity of the cylinder, then the equation would be $I\dfrac{d\omega}{dt}=-Fr$. Then I believe everything is ok.

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.
In the first post was a mistake in the signum.

#### TSny

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This is should be the same as I used a separately the eqns of center mass motion and rotational motion in my initial post, but it is not.
Do you see how your equations are consistent with what I wrote?

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

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

Multiply the first equation by r and then add it to the second equation. This will get rid of the friction force F and you will end up with the same equation that I wrote.