At a time t = 2.70 s, a point on the rim of a wheel with a radius of 0.170 m has

Question

At a time t = 2.70 s, a point on the rim of a wheel with a radius of 0.170 m has a tangential speed of 48.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.6 m/s^2 Calculate the wheel's constant angular acceleration. Calculate the angular velocity at t = 2.70 s Calculate the angular velocity at t = 0. Through what angle did the wheel turn between t = 0 and t = 2.70 s? 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/s^2

Explanation / Answer

Part A :

Angular accel in rad/s^2 = tangential accel / r = 10.6 m/s^2 / 0.170 m = 62.35 rad/s^2

Part B:

Angular velocity in rad/s = tangential speed / r = 48.0 m/s / 0.170 m = 282.35 rad/s

Part C:

Acceleration is negative (slowing down)

so velocity at t=0 is

u = v + a*t = 282.35 + (62.35 )*2.7 = 450.7 rad/s

Part D:

Angle

s = u*t - 0.5*a*t^2

=450.7 * 2.7 - 0.5 *62.35 *2.7^2

= 987.73 radians

Part E:

Radial accel = v^2/r = 9.81m/s^2

v = sqrt(9.81*0.170) = 1.29 m/s = 1.29/0.170=7.59 rad/s

This occurs when t = (u - v) / a = (450.7 - 7.59 ) / 62.35 = 7.1

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