A wagon wheel consists of 8 spokes of uniform diameter, each of mass m_s and length L cm. The outer ring has a mass m_ring. What is the moment of inertia of the wheel? Assume that each spoke extends from the center to the other ring and the ring is of negligible thickness.

Question

Quizplus · Accepted Answer

The moment of inertia of one spoke is given by:

Ispoke = (1/12)mL^2

Since there are eight spokes, the moment of inertia of all the spokes is:

Ispokes = 8(1/12)mL^2 = (2/3)mL^2

The moment of inertia of the outer ring can be calculated as that of a thin hoop of uniform density:

Iring = mR^2

where R is the radius of the ring. Since the ring has a mass mring and the same density as the spokes, its radius is:

R = L/2

Therefore:

Iring = mring(L/2)^2 = (1/4)mringL^2

The moment of inertia of the entire wheel is the sum of the moments of inertia of the spokes and the outer ring:

I = Ispokes + Iring = (2/3)mL^2 + (1/4)mringL^2

Substituting mring = 8m and simplifying gives:

I = (33/12) mL^2

which is closest to option B.

A Wagon Wheel Consists of 8 Spokes of Uniform Diameter