At a time t = 2.60 s , a point on the rim of a wheel with a radius of 0.200 m ha

Question

At a time t = 2.60 s , a point on the rim of a wheel with a radius of 0.200 m has a tangential speed of 55.0 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

2). Calculate the angular velocity at t = 2.60 s .

3). Calculate the angular velocity at t=0.

4). Through what angle did the wheel turn between t=0 and t = 2.60 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 ?

Explanation / Answer

GIVEN DATAS

t = 2.60 s

radius = 0.200 m

tangential speed = 55.0 m/s

tangential acceleration of constant magnitude 11.0 m/s2 .

1. angular accel in rad/s^2 = tangential accel / r = 11m/s^2 / 0.2m = 55 rad/s^2

2. angular velocity in rad/s = tangential speed / r = 55.0m/s / 0.2m = 275 rad/s

3. accel is negative (slowing down)
so velocity at t=0 = u = v + a*t = 275 + (55)*2.6 rad/s = 418 rad/s

4. angle s = u*t - 0.5*a*t^2
s = (418*2.6) - 0.5*55*(2.6)^2 = 900.9 radians

5. radial accel = v^2/r = 9.81m/s^2
v = sqrt(9.81*0.2) = 1.4 m/s = 1.4 / 0.2 rad/s = 7 rad/s
This occurs when t = (u - v) / a = (418 - 7) / 55 = 7.47s

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