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Question

A torque of 50.0 N-m is applied to a grinding wheel (I = 20.0 kg-m²) for 20 s. (a) If it starts from rest, what is the angular velocity of the grinding wheel after the torque is removed? (b) Through what angle does the wheel move through while the torque is applied?

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a)\textbf{a)} we know that :

τ=Iα\tau = I \alpha

where :

  • II = 20 kg\cdotm2^2 is the moment of inertia of the flywheel .
  • τ\tau = 50 N\cdotm is the torque.

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

τ=Iαα=τI=5020=2.5\begin{align*} \tau &= I \alpha \\ \alpha&= \dfrac{\tau }{I}\\ &= \dfrac{50}{20}\\ &=2.5 \end{align*}

α=2.5   rad/s2\boxed{\alpha = 2.5\ \; \mathrm{rad/ s^2}}

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
  • Δθ\Delta \theta. is the angle.

From givens\textbf{givens} we know that : α=2.5   rad/s\alpha = 2.5\ \; \mathrm{rad/s} , because it starts from rest ωi=0   rad/s\omega_{i} = 0\ \; \mathrm{rad/s} and t=20   st = 20\ \; \mathrm{s} .

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

ωf=ωi+αt=0+2.5×20=50\begin{align*} \omega_{f}& = \omega_{i} + \alpha t \\ &= 0 + 2.5 \times 20 \\ &=50 \end{align*}

ωf=50   rad/s\boxed{\omega_{f} = 50\ \; \mathrm{rad/s}}

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