Solution to Question 20
Physics Solution: Rotational Dynamics with Air Resistance
Step 1: Calculating the Retarding Torque
Consider a small element of area $dA$ at a distance $r$ from the axis of rotation. The speed of this element is $v = \omega r$.
The excess pressure is given by $p = kv = k\omega r$.
The force on this element is $dF = p \cdot dA = k\omega r \cdot dA$.
Torque ($d\tau$) on the element about the axis:
$$ d\tau = r \cdot dF = r (k\omega r dA) = k\omega r^2 dA $$
Total Torque ($\tau$):
$$ \tau = \int d\tau = \int k\omega r^2 dA = k\omega \int r^2 dA $$
The term $\int r^2 dA$ relates to the Moment of Inertia $I$.
Since the lamina is uniform with mass $M$ and area $S$, the surface mass density is $\sigma = M/S$.
The Moment of Inertia is defined as $I = \int r^2 dm = \int r^2 (\sigma dA) = \sigma \int r^2 dA$.
Therefore, $\int r^2 dA = \frac{I}{\sigma} = \frac{I \cdot S}{M}$.
Substituting this back into the torque equation:
$$ \tau = k\omega \left( \frac{I S}{M} \right) $$
Step 2: Solving the Equation of Motion
Using Newton’s Second Law for rotation $\tau_{net} = I \alpha$:
$$ – \frac{k\omega I S}{M} = I \alpha $$
$$ \alpha = – \frac{kS}{M} \omega $$
Using the chain rule $\alpha = \omega \frac{d\omega}{d\theta}$:
$$ \omega \frac{d\omega}{d\theta} = – \frac{kS}{M} \omega $$
$$ d\omega = – \frac{kS}{M} d\theta $$
Step 3: Finding Total Revolutions
Integrate from initial velocity $\omega_0$ to stop ($0$):
$$ \int_{\omega_0}^{0} d\omega = – \frac{kS}{M} \int_{0}^{\theta_{total}} d\theta $$
$$ -\omega_0 = – \frac{kS}{M} \theta_{total} $$
$$ \theta_{total} = \frac{M \omega_0}{kS} $$
The number of revolutions $n$ is $\frac{\theta_{total}}{2\pi}$:
$$ n = \frac{M \omega_0}{2 \pi k S} $$
