How do I proof:

Textbook solution :

1 vote

I think the question is stating that the situation shown is a position of equilibrium for the centre of the small sphere. Which simply means that the resultant force on the small sphere is zero. However, the resultant torque about the CM is not zero. We must also assume that there is no slipping between sphere and cylinder, so the sphere accelerates about its COM.

Please explain the solution given by them: https://pasteboard.co/HXsICHVJ.png

I do not understand that solution either. Have you tried solving the problem yourself using my suggestion above?

No, but I derived that for a ball in the pure rotation along incline with inclination $\theta$, possess acceleration along incline given by $a=\dfrac{F/m}{1+\frac{k^2}{R^2}}$, where $F$ is a force along incline acting at CM (excluding frictional), $k$ is its radius of gyration. Using this I understand that $g\sin\theta$ term but why $\alpha r$?

Your images have disappeared. It may be best to upload again using Imgur. See http://physics.qandaexchange.com/?qa=253/how-to-upload-an-image.

1 vote

Best answer

I also do not understand the textbook solution.

In an equilibrium situation the ball is effectively rolling down a plane inclined at angle $\theta$ while the plane is itself accelerating upwards along the plane. In the ground frame of reference the centre of mass of the ball is stationary, so the resultant force on it must be zero. However, there is a non-zero torque which causes rotational acceleration down the plane.

The torque acting on the ball is $mgr\sin\theta$ where $r$ is its radius. Its acceleration down the plane, relative to the point of contact, is $r\beta$ where $\beta$ is its angular acceleration. Since there is no slipping at the point of contact, $r\beta$ must also equal the acceleration of the point of contact up the plane, which is $R\alpha$ where $R>r$ is the radius of the cylinder. That is : $$r\beta=R\alpha$$

The equation of motion for the angular acceleration of the ball is therefore $$mgr\sin\theta=I\beta=\frac25 mr^2 \frac{R}{r}\alpha$$ $$\sin\theta=\frac{2R\alpha}{5g}$$

If there is a finite coefficient of friction $\mu$ then we must also have that $$\tan\theta \le \mu$$

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