Question

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10.0 s period following the inital spin, the bike wheel undergoes 65.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ t s will it take the bike wheel to come to a complete stop?

Answer 1

Answer:

$\Delta t = 8 s$

Explanation:

As we know that the angular acceleration of the wheel due to friction is constant

so we can use kinematics

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

so we have

$(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)$

$130\pi = 180\pi + 50 \alpha$

$\alpha = -\pi rad/s^2$

now time required to completely stop the wheel is given as

$\omega_f = \omega_i + \alpha t$

$0 = (2\pi \times 9) + (-\pi) t$

$t = 18 s$

now time required to stop the wheel is given as

$\Delta t = 18 - 10$

$\Delta t = 8 s$