Question

A chain with negligible mass is wrapped around a gear of radius R = 2.0 meters, and it cannot slip. The gear rotates without friction about an axis through its center. A weight of mass M = 14 kg a suspended from the end of the chain. When the system is released from rest it is observed that the weight descends 10 meters in 2.0 seconds. What is the moment of inertia j of the gear (assuming it is a disk with i = 1/2 MR^2)?

Answer 1

here,

R = 2 m

m = 14 kg

height descends , h = 10 m

t = 2 s

let the accelration be a

h = u*t + 0.5 * a * t^2

10 = 0 + 0.5 * a * 2^2

a = 5 m/s^2

let the mass of the gear be M

a = ( net force) /( effective mass)

a = ( m * g) /( m + (I) /R^2)

5 = (14 * 9.8 ) /( 14 + 0.5 * M)

M = 26.88 kg

the moment of inertia of gear , I = 0.5 * M * R^2

I = 0.5 * 26.88 * 2^2

I = 53.76 kg.m/s^2

the moment of inertia of the gear is 53.76 kg.m/s^2