Question

I can test a new wheel design by rolling it down a test ramp. I release a wheel of mass m=1.5 kg and radius r=0.36 m from rest at an initial height of h=8.0 m at the top of a test ramp. It rolls smoothly to the bottom without sliding. I measure the linear speed of the wheel at the bottom of the test ramp to be v=4.1 m/s. What is the rotational inertia of my wheel? Select one:

Answer 1

Using energy conservation between initial and final position:

KEi + PEi = KEf + PEf

KEi = 0, since starts from rest

PEf = 0, since at ground

PEi = m*g*h, where h = 8.0 m

KEf = KE_trans + KE_rot = (1/2)*m*V^2 + (1/2)*I*w^2

w = angular speed = V/R, So

0 + m*g*h = (1/2)*m*V^2 + (1/2)*I*(V/R)^2 + 0

m*V^2 + I*V^2/R^2 = 2*m*g*h

I = (2*m*g*h - m*V^2)*R^2/V^2

I = (2*m*g*h*R^2/V^2 - m*R^2)

I = m*R^2*(2*g*h/V^2 - 1)

Using given values:

I = 1.5*0.36^2*(2*9.81*8.0/4.1^2 - 1)

I = 1.62 kg.m^2

Let me know if you've any query.