Question

A angular positio of a point on the rim of a rotating wheel is given by theta=4t-3t^(2)+t^(3) where theta is in radiuans and t is in seconds. What are the angualr velocities at (a).t=2.0 and (b). t=4.0s (c). What is the average angular acceleration for the time interval that begins at t=2.0s and ends at t=4.0s? (d). What are the instantaneous angular acceleration at the biginning and the end of this time interval?

Answer 1

Given the angular position of a point on the rim of a rotating wheel by θ = 4t - 3t² + t³.

a) Instantaneous angular velocity at any time = w = dθ/dt = 4 - 6t + 3t²

The Angular velocity at t = 2 seconds -

w = 4 - 6×2 + 3×4 = 4 rad/sec

b) The Angular velocity at t = 4 seconds -

w = 4 - 6×4 + 3×16 = 28 rad/sec

c) The Average angular acceleration between t = 2 seconds and t = 4 seconds -

α = (w(4) - w(2))/(4-2)

α = (28-4)/2 = 12 rad/s²

d) Instantaneous angular acceleration , α = dw/dt = -6+6t

The Angular acceleration at t = 2 seconds , α(2) = 6 rad/s²

The Angular acceleration at t = 4 seconds , α(4) = 18 rad/s²