The angular position of a point on the rim of a rotating wheel is given by ? = 5

Question

The angular position of a point on the rim of a rotating wheel is given by ? = 5.0 t + (-7.5)t2 + (1.0)t3, where ? is in radians and t is given in seconds. What is the angular velocity at t = 2.0s?
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 is the instantaneous angular accelerations at the end of this time interval?

Explanation / Answer

here ,,

theta = 5t - 7.5t^2 + t^3

Now , w = d(theta/dt)

w = 5 - 15t + 3t^2

at t = 2

w(2) = -13 rad/s

the angular velocity at t = 2 is -13 rad/s

at t = 4 s

w(4) = 5 - 15*4 + 3*4^2

w(4) = -7 rad/s

the angular velocity at t = 4 s is -7 rad/s

Now ,

avaerage acceleration = (-7 + 13)/2

avaerage acceleration =3 rad/s^2

the avaerage acceleration is 3 rad/s^2

angular acceleration ,a = -15 + 6t

at t = 2 s

a = -15 + 2*6

a(2) = -3 rad/s^2

the angular acceleration at t=2 is -3 rad/s^2

at t = 4 s

a = -15 + 4*6

a(2) = 9 rad/s^2

the angular acceleration at t =4 s is 9 rad/s^2

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