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

`v=0.57(m)/(s)`

`a_c=10.83(m)/(s^2)`

Step-by-step 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=(2\pi r)/(T)`

The period is inversely proportional to the frequency:

`T=(1)/(f)`

So, we have:

`v=(2\pi r)/((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(m)/(s)`

The centripetal acceleration is defined as:

`a_c=(v^2)/(r)\\a_c=((0.57(m)/(s))^2)/(30*10^(-2)m)\\\\a_c=10.83(m)/(s^2)`