Question

A uniform hollow disk with inner radius 30cm and outer radius 60cm has two pieces of thin wire wrapped around its outer rim and is supported from the ceiling. Suddenly one of the wires breaks and the remaining wire does not slip as the disk rolls down. Use conservation of energy to find the speed of the center of this disk after it has fallen a distance of 2m.

Answer 1

Answer:

Speed of the center of the disc is 4.96 m/s

Explanation:

As we know that moment of inertia of the hollow disc is given as

$I = \frac{m(R_1^2 + R_2^2)}{2}$

now we know when one wire is break then disc will roll down so here loss in gravitational potential energy will convert into kinetic energy of the system

So we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

so we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R_2})^2$

so we have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}m(\frac{0.60^2 + 0.30^2}{2})\frac{v^2}{0.60^2}$

$(10)(2) = 0.5v^2 + 0.3125 v^2$

$20 = 0.8125 v^2$

$v = 4.96 m/s$