Solving for Speed & Acceleration of a Rock in a Tire

[Jacob87411]

Just curious if this was the right way to do this problem.

A tire is going at 180 rev/min. There is a rock in the tire (on the outside) and the tire has a radius of .5 meters. What is the speed and acceleration of the rock?

180 rev/min = 3 rev / s
3 rev/s means its going 3 circumferences every second so that's 2(3.14)(.5) which is 3.14*3. So the rock is going 9.42 m/s for the speed. Then the acceleration is centripetal which is 9.42^2/.5?

 

[Astronuc]

You are correct.

http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq

At a constant angular velocity, the tangential angular acceleration is zero, but there is still centripetal acceleration.

The rotational frequency, f, in rpm or rps is related to the angular frequency, \omega = 2 \pi f, and the period of rotation, T = 1/f.