The angular position of a point on the rim of a rotating wheel is given by = 5.0

Question

The angular position of a point on the rim of a rotating wheel is given by = 5.0t - 2.0t2 + t3, where is in radians and t is in seconds.

(a) What is the angular velocity at t = 4 s?
rad/s

(b)What is the angular velocity at t = 6.0 s?
rad/s

(c) What is the average angular acceleration for the time interval that begins at t = 4 s and ends at t = 6.0 s?
rad/s2

(d) What is the instantaneous angular acceleration at the beginning of this time interval?
rad/s2

(e)What is the instantaneous angular acceleration at the end of this time interval?
rad/s2

Explanation / Answer

here ,

angle , theta = 5t - 2t^2 + t^3

a)

as angular velocity , w = d(theta)/dt

w = d/dt( 5t - 2t^2 + t^3 )

w = 5 - 4t + 3 t^2

Now, at t = 4 s

w = 5 - 4 * 4 + 3 * 4^2

w = 37 rad/s

the angular velocity is 37 rad/s

b)

at t = 6 s

w = 5 - 4t + 3 t^2

w = 5 - 4 * 6 + 3 * 6^2

w = 89 rad/s

the angular velocity is 89 rad/s

c)

average angular acceleration = w(6) - w(4) /(6 - 4)

average angular acceleration = (89 - 37)/2

average angular acceleration = 26 rad/s^2

the average angular acceleration is 26 rad/s^2

d)

as angular acceleration = dw/dt

angular acceleration = d/dt(5 - 4t + 3 t^2 )

angular acceleration = -4 + 6t

at t = 0

angular acceleration = -4 rad/s^2

the angular acceleration is -4 rad/s^2

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