Rotational Emotion

Question

A tire 2.0 ft in diameter is placed on a balancing machine where it is spun so that its tread is moving at a constant speed of 54 mph. A small stone is stuck in the tread of the tire. What is the acceleration of the stone as the tire is being balanced?

Answer 1

Hi Dominic!

 

If the stone is moving in uniform circular motion, then the net force on it must be the centripetal force appropriate for that circle, and the resultant acceleration must be the centripetal acceleration.  Thus:

 

a = a_c = v²/r

 

Here, v = 54 mph (the speed of the tread on the outside of the tire).  I am going to convert this to feet per second

 

54 miles per hour * (5280 ft/1 mile) * (3600 s/ 1 hr) = 79.2 ft/s

 

Then:

 

a_c = (79.2 ft/s)/(2 ft) = 39.6 ft/s²

 

I hope this helps!  If you need to convert it to m/s², just let me know.