Question

A wagon wheel consists of 8 spokes of uniform diamter, each of mass m, and length L. The outer ring has a mass m rin. What is the moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring? Assume that each spoke extends from the center to the outer ring is of negligible thickness.

Answer 1

Answer:

$L^2(\dfrac{8m}{3}+m_r)$

Explanation:

m = Mass of each rod

L = Length of rod = Radius of ring

$m_r$ = Mass of ring

Moment of inertia of a spoke

$\dfrac{mL^2}{3}$

For 8 spokes

$8\dfrac{mL^2}{3}$

Moment of inertia of ring

$m_rL^2$

Total moment of inertia

$8\dfrac{mL^2}{3}+m_rL^2\\\Rightarrow L^2(\dfrac{8m}{3}+m_r)$

The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is $L^2(\dfrac{8m}{3}+m_r)$.