Angular motion of a falling cylinder problem

Question

The shank of a cotton reel, of mass M, is a cylinder of radius r and the moment of inertia of the real about its axis is mk^2 . The free end of the cotton is held, and the reel falls, unwinding the cotton, the axis of the reel remaining in a ?xed horizontal direction, Show that the acceleration is gr^2/ (r^2 + k^2) and find the tension of the cotton. Show also that the upward acceleration that must be given to the free end of the cotton to keep the centre of gravity of the reel at rest is gr^2/k^2

Answer 1

Let $T$ be the tension in the string, $m$ be the mass of the reel, the moment of inertia is $mk^2$, the radius is $r$ and the angular acceleration of the reel is $\ddot{\theta}$.

We assume that the cotton is light and inextensible and remains vertical. We also assume that the cotton does not slip on the reel, and this means that the linear acceleration of the centre of the reel is $r\ddot{\theta}$.

Newton's second law gives $$mg-T=mr\ddot{\theta}$$

Taking moments about the centre of the reel gives $$Tr=mk^2\ddot{\theta}$$

Eliminating $T$, we get the linear acceleration of the reel as $$r\ddot{\theta}=\frac{gr^2}{k^2+r^2}$$ as anticipated.

We can then substitute back and get the tension in the string as $$T=\frac{mgk^2}{k^2+r^2}$$

For the last part, using the same notation as before, the centre of the cotton reel is not accelerating, so $$T=mg$$

Taking moments about the centre of the reel we get $$Tr=mk^2\ddot{\theta}$$ as before. Solving these leads to $$\ddot{\theta}=\frac{gr}{k^2}$$

Once again, the cotton is not slipping, so the linear acceleration, $r\ddot{\theta}$, of every point on the rim of the reel is the same as the linear acceleration of the cotton string itself, and hence the result.