A flywheel of radius R = 0.8 m rotates with constant angular…

A flywheel of radius $R = 0.8\ \mathrm{m}$ rotates with constant angular velocity $\omega_0 = 7.5\ \mathrm{rad/s}$. At some moment the motor driving it stops, and the flywheel keeps turning by inertia for $t = 24\ \mathrm{s}$ before it stops. Find the angular deceleration of the flywheel and the tangential acceleration of a point on its rim during braking.

Explanation of Method

A coasting flywheel: angular and tangential deceleration.

Step 1: Angular deceleration

Assuming it slows down uniformly from $\omega_0$ to zero in 24 s: $\alpha = \frac{0 - \omega_0}{t} = -\frac{7.5}{24}$.

\[\alpha = -0.3125\ \mathrm{rad/s^2}\]

Step 2: Tangential acceleration on the rim

\[ a_t = \alpha R = -0.3125 \cdot 0.8. \]

\[a_t = -0.25\ \mathrm{m/s^2}\]

The minus sign means it points against the rim's velocity.

Check

Rim speed falls from $ \omega_0 R = 6\ \mathrm{m/s} $ to 0 in 24 s: $ a_t = -\tfrac{6}{24} = -0.25\ \mathrm{m/s^2}\ \checkmark $

$\omega\ (\mathrm{rad/s})$ $t\ (\mathrm{s})$ $0$ $12$ $24$ $7.5$ $90\ \mathrm{rad} \approx 14.3\ \mathrm{rev}$ $\alpha = -0.3125\ \mathrm{rad/s^2}$ $\vec v$ $\vec a_t$ $R$ $a_t = \alpha R = -0.25\ \mathrm{m/s^2}$

Left: the angular velocity falls uniformly to zero in 24 s; the area is the angle turned while coasting. Right: a point on the rim has a tangential acceleration pointing against its velocity.

Result

\[\boxed{\,\alpha = -0.3125\ \mathrm{rad/s^2}\,}\]

\[\boxed{\,a_t = -0.25\ \mathrm{m/s^2}\,}\]

Remark

Extra. While coasting, the flywheel turns through $\theta = \tfrac{\omega_0}{2}\,t = 3.75 \cdot 24 = 90\ \mathrm{rad} \approx 14.3$ revolutions. At the start of braking a rim point also has a large normal acceleration, $\omega_0^2 R = 45\ \mathrm{m/s^2}$, which shrinks to zero as the wheel stops.