Question

The angular position of a point on the rim of a rotating wheel is given by ? = 6.0t - 2.0t2 + t3, where ? is in radians and t is in seconds. (a) What is the angular velocity at t = 2 s? rad/s (b)What is the angular velocity at t = 4.0 s? rad/s (c) What is the average angular acceleration for the time interval that begins at t = 2 s and ends at t = 4.0 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

Answer 1

here,

theta = 6 * t - 2 * t^2 + t^3

differentiating the equation for w

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

differentiating the equation for alpha

alpha(t) = - 4 + 6 * t

a)

at t = 2 s

the angular velocity , w(2) = 6 - 4 * 2 + 3 * 2^2 rad/s

w(2) = 10 rad/s

b)

at t = 4 s

w(4) = 6 - 4 * 4 + 3 * 4^2 rad/s

w(4) = 38 rad/s

c)

the average angular acceleration for the time interval that begins at t = 2 s and ends at t = 4.0 s , alpha = (w(4) - w(2))/t

alpha = ( 38 - 10)/2 = 14 rad/s^2

d)

the instantaneous angular acceleration at the beginning of this time interval , alpha(2) = - 4 + 6 * 2 = 8 rad/s

e)

instantaneous angular acceleration at the end of this time interval , alpha(4) = -4 + 6 * 4 = 20 rad/s