Rolling Dynamics of a Rotating Cylinder on an Inclined Plane

[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]

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.

I have already written

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]

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.