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 the tread of the tire (on its outer edge).

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)

v = 10.47 m/s

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

a = 219.32 m/s²

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )

the rearrangement of particles in a physical change. hope this