At a time t = 3.30 s , a point on the rim of a wheel with a radius of 0.180 m ha

Question

At a time t = 3.30 s , a point on the rim of a wheel with a radius of 0.180 m has a tangential speed of 51.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.7 m/s2 .

Part A

Calculate the wheel's constant angular acceleration.

Part B

Calculate the angular velocity at t = 3.30 s .

Part C

Calculate the angular velocity at t=0.

Part D

Through what angle did the wheel turn between t=0 and t = 3.30 s ?

Part E

Prior to the wheel coming to rest, at what time will the radial acceleration at a point on the rim equal g = 9.81 m/s2 ?

Explanation / Answer

a) angular acceleration = tangential acceleration / r

= 10.7 / 0.180 = 59.44 rad/s2

b) angular velocity = tangential speed / r

= 51 / 0.18 = 283 rad/s

c) velocity at t = 0

u = v + a t = 283 + (59.44 * 3.3 ) = 479 rad/s

d) s = u t - 0.5 * a * t2

s = (479 * 3.3) - 0.5 * 59.44 * 3.32 = 1257 rad

e) v = sqrt (9.81 * 0.18) = 1.33 m/s = 1.33 / 0.18 rad/s = 7.39 rad/s

t = (u - v) / a = (479 - 7.39) / 59.44

= 7.93 s

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