At a time t = 2.70 s , a point on the rim of a wheel with a radius of 0.240 m ha

Question

At a time t = 2.70 s , a point on the rim of a wheel with a radius of 0.240 m has a tangential speed of 49.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 11.0 m/s2 .

1. Calculate the wheel's constant angular acceleration. I already got that =

-45.8

2. Calculate the angular velocity at t = 2.70 s .

3. Calculate the angular velocity at t=0.

4. Through what angle did the wheel turn between t=0 and t = 2.70 s ?

5. 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?

-45.8

Explanation / Answer

a. angular accel in rad/s^2 = tangential accel / r = 11.0m/s^2 / 0.24m = 45.8 rad/s^2

b. angular velocity in rad/s = tangential speed / r = 49.0m/s / 0.24m = 204.1 rad/s

c. accel is negative (slowing down)
so velocity at t=0 = u = v + a*t = 204.1 + (45.8)*2.7 rad/s = 327.7 rad/s

d. angle s = u*t - 0.5*a*t^2
s = (327.7*2.7) - 0.5*45.8*(2.7)^2 = 717.8 radians

e. radial accel = v^2/r = 9.81m/s^2
v = sqrt(9.81*0.24) = 1.53 m/s = 1.53 / 0.24 rad/s = 6.39 rad/s
This occurs when t = (u - v) / a = (327.7 - 6.39) / -45.8 = 7.01s

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