At a time t = 3.10 s , a point on the rim of a wheel with a radius of 0.210 m ha

Question

At a time t = 3.10 s , a point on the rim of a wheel with a radius of 0.210 m has a tangential speed of 51.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.6 m/s2 .

Calculate the wheel's constant angular acceleration.  rad/s^2  

Calculate the angular velocity at t = 3.10 s. rad/s

Calculate the angular velocity at t=0. rad/s

Through what angle did the wheel turn between t=0 and t = 3.10 s? rad

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? s

Explanation / Answer

Given,

t = 3.1 s ; r = 0.21 m ; v = 51 m/s ; alpha = 10.6 m/s^2

we know that the linae and angular acceleration are related as:

a = r alpha

alpha = a/r = 10.6/0.21 = 50.48 rad/s^2

Hence, alpha = 50.48 rad/s^2

we know that,

v = r w => w = v/r

w = 51/0.21 = 242.86 rad/s

Hence, w = 242.86 rad/s

We know from eqn in circular motion

w = w0 + alpha t

w = 242.86 + 50.48 x 3.1 = 399.35 rad/s

Hence, w = 399.35 rad/s

we know that

theta = w0t + 1/2 alpha t^2

theta = 399.35 x 3.1 - 0.5 x 50.48 x 3.1^2 = 995.43 rad

Hence, theta = 995.43 rad

we know that,

a(radial) = v^2/r => v = sqrt (ar)

v = sqrt (9.81 x 0.21) = 1.44 m/s

w = v/r = 1.44/0.21 = 6.86 rad/s

we know that

w = w0 + alpha t

t = (w - w0)/alpha

t = (6.86 - 399.35)/-50.48 = 7.78 s

Hence, t = 7.78 s

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