Question

•• Your roommate is working on his bicycle and has the bike upside down. He spins the 60-cm-diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. What are the pebble’s speed and acceleration?

Answer 1

Answer:

$v=0.57\frac{m}{s}$

$a_c=10.83\frac{m}{s^2}$

Explanation:

We have an uniform circular motion, therefore, the pebble’s speed is given by the distance traveled in a revolution $(2\pi r)$ and the period (T), since this is the time pebble’s takes to complete a revolution:

$v=\frac{2\pi r}{T}$

The period is inversely proportional to the frequency:

$T=\frac{1}{f}$

So, we have:

$v=\frac{2\pi r}{\frac{1}{f}}\\v=2\pi rf$

Recall that the radius is the half of the diameter and one revolution per is equal to one Hz:

$v=2\pi (30*10^{-2}m)(3Hz)\\v=0.57\frac{m}{s}$

The centripetal acceleration is defined as:

$a_c=\frac{v^2}{r}\\a_c=\frac{(0.57\frac{m}{s})^2}{30*10^{-2}m}\\\\a_c=10.83\frac{m}{s^2}$