Question

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of 22 N is applied tangentially to a sprocket of radius 6 cm for 6 s, what angular speed does the wheel achieve, assuming it rolls without slipping?

Answer 1

Answer:

The angular speed after 6s  is $\omega = 1466.67s^{-1}$.

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$, with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$, at a distance $d = 6cm =0.06m$, to a ring with the moment of inertia of $I =mr^2$; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I}$

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2}$

$\alpha = 244.44\: s^{-2}$

Therefore, the angular speed $\omega$ which is

$\omega = \alpha t$

after 6 seconds is

$\omega = 244.44 \: s^{-2}* 6s$

$\boxed{\omega = 1466.67s^{-1}}$