Solution to Q28
1. Analysis of Motion in the Steady State
The disc is subjected to a force of constant magnitude $F$ which rotates with a constant angular velocity $\omega$. In the steady state, the velocity vector $\vec{v}$ of the disc will rotate with the same angular velocity $\omega$, maintaining a constant magnitude $v$ and a fixed orientation relative to the driving force $\vec{F}$.
Since the velocity vector rotates with angular speed $\omega$ while maintaining a constant magnitude, the acceleration $\vec{a}$ is purely centripetal (perpendicular to the velocity vector). The magnitude of this acceleration is given by:
The forces acting on the disc are:
- The external driving force $\vec{F}$.
- The resistive force $\vec{f} = -k\vec{v}$, which opposes the velocity.
2. Application of Newton’s Second Law
Applying Newton’s Second Law, $\vec{F}_{net} = m\vec{a}$: $$ \vec{F} – k\vec{v} = m\vec{a} \quad \Rightarrow \quad \vec{F} = k\vec{v} + m\vec{a} $$
The vectors $k\vec{v}$ (resistive force magnitude) and $m\vec{a}$ (mass times centripetal acceleration) are perpendicular because acceleration is normal to velocity. The applied force $\vec{F}$ is the vector sum of these two components. By the Pythagorean theorem:
Substituting $a = \omega v$: $$ F^2 = (kv)^2 + (m \omega v)^2 $$ $$ F^2 = v^2 [k^2 + (m\omega)^2] $$
3. Calculation
Solving for speed $v$: $$ v = \frac{F}{\sqrt{k^2 + (m\omega)^2}} $$
Given values: $m = 0.1 \text{ kg}$, $k = 0.04 \text{ kg/s}$, $F = 0.06 \text{ N}$, $\omega = 0.3 \text{ rad/s}$.
First, compute the inertial impedance term $m\omega$: $$ m\omega = 0.1 \times 0.3 = 0.03 \text{ kg/s} $$
Substitute values into the velocity equation: $$ v = \frac{0.06}{\sqrt{(0.04)^2 + (0.03)^2}} $$ $$ v = \frac{0.06}{\sqrt{0.0016 + 0.0009}} = \frac{0.06}{\sqrt{0.0025}} $$ $$ v = \frac{0.06}{0.05} $$