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.

 

[kyle8921]

Yes, your approach to solving this problem is correct. To find the speed of the rock, you used the formula for linear velocity, which is equal to the circumference of the tire (2πr) multiplied by the number of revolutions per second (3). And to find the centripetal acceleration, you used the formula a=v^2/r, where v is the linear velocity and r is the radius of the tire. Keep in mind that this solution assumes the rock is moving in a circular path around the center of the tire, and there are no external forces acting on the system.