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Question

A flywheel (I = 50 kg-m²) starting from rest acquires an angular velocity of 200.0 rad/s while subject to a constant torque from a motor for 5 s. (a) What is the angular acceleration of the flywheel? (b) What is the magnitude of the torque?

Solution

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a) From the kinematics of the rotational motion\textbf{the kinematics of the rotational motion} we knw that :

ωf=ωi+αt\omega_{f} = \omega_{i} + \alpha t

where:

  • ωi\omega_{i} is the initial angular velocity .
  • ωf\omega_{f} is the final angular velocity .
  • α\alpha is the angular acceleration .
  • tt is the time .

From givens\textbf{givens} we know that : ωf=200   rad/s\omega_{f} = 200\ \; \mathrm{rad/s} , ωi=0   rad/s\omega_{i} = 0\ \; \mathrm{rad/s} and t=5   st = 5 \ \; \mathrm{s} .

Plugging\textbf{Plugging} known information into our equation we get :

ωf=ωi+αtαt=ωfωiα=ωfωit=20005=40\begin{align*} \omega_{f}& = \omega_{i} + \alpha t \\ \alpha t &=\omega_{f} - \omega_{i} \\ \alpha&=\dfrac{\omega_{f} - \omega_{i}}{t}\\ &= \dfrac{200 - 0}{5}\\ &=40 \end{align*}

ωf=40   rad/s\boxed{\omega_{f} = 40\ \; \mathrm{rad/s}}

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