Question

A child is pushing a playground merry-go-round. The angle through which the merry-go-round has turned varies with time according to θ(t)=γt+βt3, where γ=0.400rad/s and β=0.0120rad/s3.
Part A Calculate the angular velocity of the merry-go-round as a function of time. Express your answer in radians per second in terms of γ, β, and t. ω(t) = rad/sec

Part Part B What is the initial value ω0 of the angular velocity? Express your answer in radians per second. ω0 = rad/s

Part C Calculate the instantaneous value of the angular velocity ω(t) at time t=5.00s. Express your answer in radians per second. ω(5.00) = rad/s

Part D Calculate the average angular velocity ωav for the time interval t=0 to t=5.00 seconds. Express your answer in radians per second. ωav = rad/s

Answer 1

Answer:

A. $\gamma + 3\beta t^2 rad/s$

B. 0.4 rad/s

C. 1.3 rad/s

D. 0.7 rad/s

Explanation:

Part A. We can take the derivative of the motion function in order to get the angular velocity function with respect to time

$\omega(t) = \theta^{'}(t) = (\gamma t + \beta t^3)^{'} = \gamma + 3\beta t^2 rad/s$

Part B. The initial value of angular velocity is when t = 0

$\omega_0 = \omega(0) = \gamma + 3\beta 0^2 = \gamma = 0.4 rad/s$

Part C. The instantaneous value of the angular velocity at t = 5s

$\omega(5) = \gamma + 3\beta 5^2 = \gamma + 75\beta = 0.4 + 75*0.012 = 1.3 rad/s$

Part D. The angle distance it travels during the first t = 5 seconds is

$\theta(5) = \gamma * 5 + \beta * 5^3 = 5\gamma + 125\beta = 5*0.4 + 125*0.012 = 3.5rad$

So the average angular velocity during this 5 seconds would be the total angle traveled divided by the time

$\omega_a = \theta(5) / t = 3.5 / 5 = 0.7 rad/s$