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

$\tau = I \alpha$

where :

• $I$ = 20 kg$\cdot$m$^2$ is the moment of inertia of the flywheel .
• $\tau$ = 50 N$\cdot$m is the torque.

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

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

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

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

$\omega_{f} = \omega_{i} + \alpha t$

Where:

• $\omega_{i}$ is the initial angular velocity .
• $\omega_{f}$ is the final angular velocity .
• $\alpha$ is the angular acceleration .
• $t$ is the time
• $\Delta \theta$. is the angle.

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

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

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

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

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