Chapter 8, Problem 66A

To determine

The moment of inertia of the considered wheel.

Answer to Problem 66A

  $0.0498\text{kg}\cdot {\text{m}}^{\text{2}}$

Explanation of Solution

Given:

The radius of the bicycle wheel is, $r=38\text{cm = 0}\text{.38 m}$

The force applied on the edge of the wheel, $F=0.35\text{N}$

The angular acceleration of the wheel, $\alpha =2.67{\text{rad/s}}^{\text{2}}$

Formula used:

The torque can be obtained by:

  $\tau =Fr\sin \theta$

Where, F is the applied force, r is the distance of the point of application of force from the axis of rotation, and $\theta$ is the angle between F and r .

And, angular acceleration can be written as,

  $\alpha =\frac{{\tau }_{net}}{I}$

Where, ${\tau }_{net}$ is the net torque on a rotating object, and $I$ is the moment of Inertia of the object.

Calculation:

Since the force is applied on the edge of the wheel, thus the angle between the force and the radius of the wheel will be $90°$ .

Now, the torque generated on the wheel on account of applied force will be,

  $\begin{array}{c}{\tau }_{net}=Fr\sin 90°\\ =\left(0.35\right)\left(0.38\right)\\ =0.133\text{}\text{N}\cdot \text{m}\end{array}$

Then, it is known that the angular acceleration can be written as,

  $\alpha =\frac{{\tau }_{net}}{I}$

Using calculated values in above, the moment of inertia of the wheel will be,

  $\begin{array}{c}2.67=\frac{0.133}{I}\\ I=0.0498\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the moment of inertia of the wheel is $0.0498\text{kg}\cdot {\text{m}}^{\text{2}}$ .