Question

A light rope is wrapped several times around a large wheel with a radius of 0.405 m . The wheel rotates in frictionless bearings about a stationary horizontal axis, as shown in the figure (Figure 1) . The free end of the rope is tied to a suitcase with a mass of 13.0 kg . The suitcase is released from rest at a height of 4.00 m above the ground. The suitcase has a speed of 3.15 m/s when it reaches the ground.

Answer 1

Using Energy Conservation,
Initial potential Energy = Final kinetic Energy + Final Rotational Energy
m g h = 1/2 m v^2 + 1/2 * I*ω2
Where,
ω = v/r where v is tangential speed of wheel rim which is the same as the linear v of the suitcase
ω = 3.15/0.405
ω = 7.78 rad/s
subtituing Values,
13.0*(9.8)*(4.00) = 1/2 (13.0)*(3.15)^2 + 1/2 *I *(7.78)^2
I = 14.7  kgm^2