Question

An expensive carbon weave 5-spoke bike wheel has total mass M and radius R. If half the mass of the wheel is in the spokes, what is this wheel's moment of inertia?

Answer 1

Answer: I = $\frac{11}{6}MR^{2}$

Explanation: Moment of Inertia (I) is the opposition on a rotating body. Generally, is calculated by  $I=2mr^{2}$

A bycicle wheel is composed of numerous parts and each part has its own moment of inertia.

Moment of inertia for this 5-spoke bike wheel will be:

Inertia of ring: $I_{r}=m_{r}r^{2}$

For the 5-spoke wheel:

$I_{r}=MR^{2}$

Inertia of spokes: $I_{s}=\frac{1}{3}n.m_{s}r^{2}$

where

n is the number of spokes

For the 5-spoke wheel, half of the total mass is the spokes, then:

$I_{s}=\frac{1}{3}.5. \frac{M}{2}R^{2}$

$I_{s}=\frac{5}{6}MR^{2}$

Inertia of the wheel is the sum of both inertia:

$I=I_{r}+I_{s}$

$I=MR^{2}+\frac{5}{6}MR^{2}$

$I=\frac{11}{6}MR^{2}$

The moment of inertia of the 5-spoke bike wheel is $I=\frac{11}{6}MR^{2}$