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Lana71 [14]
3 years ago
7

A tire 0.500 m in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in t

he tread of the tire (on its outer edge).
Physics
1 answer:
arsen [322]3 years ago
8 0

Answer:

v = 10.47 m/s

a = 219.32 m/s²

Explanation:

The velocity of the stone can be given by the following formula:

v = r ω

where,

v = linear velocity of the stone = ?

r = radius of tire = 0.5 m

ω = angular velocity = (200 rev/min)(2π rad/1 rev)(1 min/60 s) = 20.94 rad/s

Therefore,

v = (0.5 m)(20.94 rad/s)

<u>v = 10.47 m/s</u>

<u></u>

The acceleration of the stone will be equal to the centripetal acceleration. Therefore,

a = v²/r

a = (10.47 m/s)²/0.5 m

<u>a = 219.32 m/s²</u>

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