Question

The drive chain in a bicycle is applying a torque of 0.850 nm to the wheel of the bicycle. treat the wheel as a hoop with a mass of 0.750 kg and a radius of 33.0 cm. what is the angular acceleration of the wheel

Answer 1

The equivalent of Newton's second law for rotational objects is given by:
$\tau = I \alpha$
where
$\tau$ is the net torque acting on the object
$I$ is its moment of inertia
$\alpha$ is its angular acceleration

For a hoop rotating around its perpendicular axis, the moment of inertia is
$I=mr^2$
where m is the mass and r the radius. By using the data of the wheel, m=0.750 kg and r=33.0 cm=0.33 m, we find
$I=mr^2 = (0.750 kg)(0.33 m)^2=0.082 kg m^2$

and since the torque is $\tau=0.850 Nm$, the angular acceleration of the wheel is
$\alpha= \frac{\tau}{I}= \frac{0.850 Nm}{0.082 km m^2}=10.37 rad/s^2$