Question

A uniform disc rotating at a constant speed in a vertical plane about a fixed horizontal axis, passing through the centre of the disc. A piece of disc from its rim detaches itself from the dist at the instant when it is on the same horizontal with the centre of the disc and moving upwards. Then about the fixed axis, the angular speed of the

• A. remaining disc remains unchanged
• B. remaining disc decreases
• C. remaining disc increases
• D. broken away piece decreases initially and later increases

Answer 1

Answer: (A), (D)

broken away piece decreases initially and later increases
remaining disc remains unchanged

$Conservingangularmomentum(I+m{r}^2)\omega =mvr+I{\omega }_{1}butinitiallyv={\omega }_{1}r\Rightarrow (I+m{r}^2)\omega =m{\omega }_1{r}^2+I{\omega }_1\Rightarrow {\omega }_1=\omega$
Since the piece initially was moving vertically up, it will keep moving up but its speed first decreases then it will increase as it is under gravity.