At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m ha

Question

At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 46.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 11.0 m/s2 .

Part A

Calculate the wheel's constant angular acceleration.

Part B

Calculate the angular velocity at t = 2.80 s .

Part C

Calculate the angular velocity at t=0.

Part D

Through what angle did the wheel turn between t=0 and t = 2.80 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)alpha = a_t / r

= 11 m/s^2 / 0.230 m

= 47.8 rad/s^2

(B) w = v / r = (46/0.230) = 200 rad/s

(C) wf = wi + alpha t

200 = w0 - (47.8 x 2.80)

w0 = 333.8 rad/s

(D) wf^2 - wi^2 = 2 alpha theta

200^2 - 333.8^2 = 2(-47.8) (theta)

theta = 747 rad

(E) a_r = w^2 r

9.81 = w^2 (0.230)

w = 6.53 rad/s

6.53 = 333.8 + (-47.8)t

t = 6.85 sec

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