A wheel of mass 10 kg and radius 20 cm is spinning at an angular...

Given the initial angular speed of the wheel:

$\omega_0 = 100 \text{ rev/min} = 100 \times \dfrac{2\pi}{60} = \dfrac{10\pi}{3} \text{ rad/s}$

The final angular speed is $\omega = 0$.

The angular displacement to bring the wheel to a halt is:

$\theta = 10 \text{ rev} = 10 \times 2\pi = 20\pi \text{ rad}$

Using the kinematic equation for rotational motion:

$\omega^2 = \omega_0^2 + 2\alpha\theta$

Substituting the values:

$0 = \left(\dfrac{10\pi}{3}\right)^2 + 2\alpha(20\pi)$

$0 = \dfrac{100\pi^2}{9} + 40\pi\alpha$

$\alpha = -\dfrac{100\pi^2}{9 \times 40\pi} = -\dfrac{5\pi}{18} \text{ rad/s}^2$

Assuming the wheel to be a solid disc, its moment of inertia is:

$I = \dfrac{1}{2}mR^2$

The magnitude of the required torque is:

$\tau = I|\alpha| =$\left(\dfrac{1}{2}mR^2\right) \left(\dfrac{5\pi}{18}\right)<br><br>Since the torque is provided by a tangential forceF$, we have$\tau = F \times R$. Therefore:<br><br>$F \times R = $\dfrac{1}{2}$mR^2 \times $\dfrac{5\pi}{18}<br><br>$F = $\dfrac{1}{2}$mR \times $\dfrac{5\pi}{18}<br><br>Substituting$m = 10 \text{ kg}$and$R = 20 \text{ cm} = 0.2 \text{ m}$:<br><br>$F = $\dfrac{1}{2} \times$10 \times 0.2 \times $\dfrac{5\pi}{18}<br><br>$F = 1 \times $\dfrac{5\pi}{18} = \dfrac{5\pi}{18} \text{ N}<br><br>$F \approx $\dfrac{5 \times 3.1416}{18} \approx$0.87 \text{ N}$<br><br>Answer:$0.87 \text{ N}