At a time t = 3.00 s , a point on the rim of a wheel with a radius of 0.250 m ha

Question

At a time t = 3.00 s , a point on the rim of a wheel with a radius of 0.250 m has a tangential speed of 45.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.0 m/s2 .

a)Calculate the wheel's constant angular acceleration.

b)Calculate the angular velocity at t = 3.00 s .

c)Calculate the angular velocity at t=0.

d)Through what angle did the wheel turn between t=0 and t = 3.00 s ?

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/r

alpha = 10/0.25 = 40 rad/s2

b)

at t = 3.0 sec

v = 45 m/s

so w = v/r = 45/0.25 = 180 rad/s

c)

w = w0 + alpha*t

180 = w0 + 40*3

w0 = 60 rad/s

d)

w2 = w02 + 2*alpha*theta

1802 = 602 + 2*40*theta

theta = 120*240/80 = 360 rad

e)

0 = u + a*t

t = u/g = 60*0.25/9.81 = 1.529 sec

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