In the figure, a wheel of radius r = 0.107 m is mounted on a frictionless axle.
ID: 1463758 • Letter: I
Question
In the figure, a wheel of radius r = 0.107 m is mounted on a frictionless axle. A massless cord is wrapped around the wheel and attached to a box of mass m = 2.75 kg that slides on a frictionless incline at an angle ? = 33.2° with the horizontal. The box accelerates down the surface with a constant acceleration of a = 2.24 m/s2. What is the rotational inertia (in kg?m2) of the wheel about its axle?
The answer is 0.0439 kg?m2 , can you please show the steps
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved. halliday_10e_fig_10 50Explanation / Answer
Applying Newton’s second law to the mass,
Fnet = ma = mgsin - T-------------------(1)
The rotational inertia of the wheel = I = MR²/2
By law of conservation of the torque
net = I = F*R
MR²/2(a/R) = T *R
Ma/2 = T----------------------(2)
we don’t know mass (M) of the wheel.
Plugging (2) into (1) and solving for M,
ma = mgsin - Ma/2
M = [2m(gsin - a)] / a
= [2(2.75kg)(9.80m/s²sin33.2° - 2.24m/s²)] /2.24m/s²
= 7.67 kg
The rotational inertia (moment of inertia) is:
I = (7.67kg)(0.107m)² / 2 = 0.0439kgm²