Question

1. Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. Radius of tire is 50 cm. What angle did the tire move through in those 5 secs

Answer 1

Answer:

$\theta=65.18rad$

Explanation:

The angle in rotational motion is given by:

$\theta=\frac{w_o+w_f}{2}t$

Recall that the angular speed is larger than regular frequency (in rpm) by a factor of $2\pi$, so:

$\omega_f=2\pi f\\\omega_f=2\pi*250rpm\\\omega_f=1570.80 \frac{rad}{min}$

The wheel spins from rest, that means that its initial angular speed is zero($\omega_o$). Finally, we have to convert the given time to minutes and replace in the first equation:

$t=5s*\frac{1min}{60s}=0.083min\\\theta=\frac{\omega_f}{2}t\\\theta=\frac{1570.800\frac{rad}{min}}{2}(0.083min)\\\theta=65.18rad$