When tires are installed or reinstalled on a car, they are usually first balance

Question

When tires are installed or reinstalled on a car, they are usually first balanced on a device that spins them to see if they wobble. A tire with a radius of 0.380 m is rotated on a tire balancing device at exactly 460 revolutions per minute. A small stone is embedded in the tread of the tire. What is the magnitude of the centripetal acceleration experienced by the stone? When tires are installed or reinstalled on a car, they are usually first balanced on a device that spins them to see if they wobble. A tire with a radius of 0.380 m is rotated on a tire balancing device at exactly 460 revolutions per minute. A small stone is embedded in the tread of the tire. What is the magnitude of the centripetal acceleration experienced by the stone?

Explanation / Answer

We already know that,

a = v²/r
Also, v = r
--> plugging into a, we get

a = (r)²/r = ²r

You just need to find (which has to be in radians)...

1 revolution is 2 radians, so 460 rpm is 460 * 2 radians per minute:

If we just use what we have, our units will be meters/minute²:

(460 * 2)² * 0.380 ~ 3,174,380.60273 meters/minute²

Let's change to seconds though (since I suspect that's what you want):

460 * 2 rad/min * 1 minute / 60 seconds = 460 * 2 / 60 rad/s

-->

a = (460 * 2 / 60)² * 0.380 ~ 881.772389648 m/s²

Therefore, the magnitude of the centripedal acceleration experienced by the stone is 881.772389648 m/s².

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