Question

A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?

Answer 1

Hi there!

We can begin by calculating the acceleration of the block and the wheel using the following equation:

d = vit + 1/2at², where initial velocity = 0 m/s

d = 1/2at²

2d/t² = a

2(1.5)/2² = 0.75 m/s²

Now, we can do a summation of torques:

∑τ =  rT

Rewrite using Newton's 2nd Law for rotation:

Iα = rT

Convert α to a using the relationship α = a/r:

I(a/r) = rT

Ia = r²T

I = r²T/a

Plug in the values:

I = (0.40²)(20)/(0.75) = 4.267 kgm²