Answer to Question #108071 in Mechanics | Relativity for Valentine Agun

Question

Use conservation of energy to determine the angular speed of the spool shown in Figure P8.58 after the 3.00-kg bucket has fallen 4.00 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.

Answer 1

$\frac{mv^2}{2}+\frac{I\omega^2}{2}=mgh$

$v=\omega r$ and $I=\frac{1}{2}Mr^2$ . So,

$\frac{m(\omega r)^2}{2}+\frac{1/2\cdot Mr^2\omega^2}{2}=mgh$

$\omega=\sqrt{\frac{4mgh}{(2m+M)r^2}}=\sqrt{\frac{4\cdot 3\cdot 9.81\cdot 4}{(2\cdot3+5)0.6^2}}=10.9$ $rad/s$