The angular position of a point on the rim of a rotating wheel is given by ? = 7

Question

The angular position of a point on the rim of a rotating wheel is given by ? = 7.5 t + (-4.0)t2 + +(1.2)t3, where ? is in radians if t is given in seconds. What is the angular velocity at t = 2.0 s? What is the angular velociy at t = 4.0 s?   What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s? What is the instantaneous angular acceleration at the beginning of this time interval? What are the instantaneous angular accelerations at the end of this time interval?

Explanation / Answer

Here ,

theta = 7.5t - 4t^2 + 1.2 *t^3

angular velocity ,w = d(theta)/dt

w = 7.5 - 8t + 3.6t^2

Now, at time t = 2

w(2) = 7.5 - 8*2 + 3.6 * 2^2

w(2) = 5.9 rad/s

Now, at time t= 4 s

w(4) = 7.5 - 8*4 + 3.6 * 4^2

w(4) = 33.1 rad/s

Now, average angular acceleration = (33.1 - 5.9)/2

average angular acceleration =13.6 rad/s

the average angular acceleration is 13.6 rad/s

Now ,

instantaneous acceleration , a =dw/dt

a = -8 + 7.2t

at time , t = 2 s

a(2) = -8 + 7.2 * 2

a(2) = 6.4 rad/s^2

the acceleration in the begining is 6.4 rad/s

at t = 4 s

a(4) = -8 * 7.2 * 4

a(4) = 20.8 rad/s^2

the acceleration at the end of this inteval is 20.8 rad/s^2

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