Question

A bicycle wheel, of radius 0.300 m and mass 1.47 kg (concentrated on the rim), is rotating at 4.00 rev/s. After 56.4 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces? Nm

Answer 1

Given,

The radius of the wheel, r=0.300 m

The mass of the wheel, m=1.47 kg

The initial angular frequency of the wheel, f=4.00 rev/s

The time it takes for the wheel to come to a stop, t=56.4 s

The final angular velocity of the wheel, ω=0 rad/s

The initial angular velocity of the wheel is

`\begin{gathered} \omega_0=2\pi f \\ =2\pi*4.00 \\ =25.13\text{ rad/s} \end{gathered}`

From the equation of motion,

`\omega=\omega_0+\alpha t`

Where α is the angular acceleration of the wheel.

On substituting the known values,

`\begin{gathered} 0=25.13+\alpha*56.4 \\ \Rightarrow\alpha=(-25.13)/(56.4) \\ =-0.45\text{ rad/s}^2 \end{gathered}`

The moment of inertia of the wheel whose mass is concentrated on the rim is given by,

`I=mr^2^{}`

The torque applied by the friction is given by,

`\begin{gathered} \tau=I\alpha \\ =mr^2\alpha \end{gathered}`

On substituting the known values,

`\begin{gathered} \tau=1.47*0.300^2*-0.45 \\ =-0.06\text{ Nm} \end{gathered}`

The negative sign of the torque indicates that the torque is applied to the wheel in the direction opposite to the direction of rotation of the wheel.

Thus the magnitude average torque due to frictional force is 0.06 Nm