Chapter 8, Problem 28P

To determine

To Find:The moment of the inertia of bicycle.

Answer to Problem 28P

The moment of inertia of a bicycle wheel is $0.139{\text{kg·m}}^2$ .

Explanation of Solution

Given information:

Diameter of wheel, $D=66.7\text{cm}$

The combined mass of rim and tire, $M=1.25\text{kg}$

Formula used:

The equation for the radius of the wheel is,

  $r=\frac{D}{2}$

Here,

D = diameter of the wheel

The equation for the moment of inertia of the bicycle wheel is,

  $I=M{r}^2$

Here,

M = combined mass of the rim and tire

r = radius of the wheel

Calculation:

Substitute 66.7 cm for D in above equation.

  $\begin{array}{l}r=\frac{\text{66}\text{.7cm}\left(\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1}\text{cm}}\right)}{\text{2}}\\ r=0.3335\text{m}\end{array}$

According to the question, the mass of the hub can be ignored, substitute 1.25 kg for M and 0.335 m for radius to calculate inertial.

  $\begin{array}{l}I=\left(1.25\text{kg}\right){\left(0.3335\text{m}\right)}^2\\ I=0.139\text{kg}{\text{.m}}^2\end{array}$

Therefore, the moment of inertia of a bicycle wheel is $0.139{\text{kg·m}}^2$

Conclusion:

The moment of inertia of a bicycle wheel is $0.139{\text{kg·m}}^2$ .