Question

Moment of inertia Wheel If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.72rev/s, friction in the bearings causes the wheel to stop in just 12s. If the moment of inertia of the wheel about its axle is 0.30kg*m2, what is the magnitude of the frictional torque? Please show your work

Answer 1

Answer:

0.113097335529 Nm

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

I = Moment of inertia = 0.3 kgm²

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-0.72\times 2\pi}{12}\\\Rightarrow \alpha=-0.376991118431\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.3\times (-0.376991118431)\\\Rightarrow \tau=-0.113097335529\ Nm$

The magnitude of torque is 0.113097335529 Nm