Rotational Dynamics of Discs on a Platform
Let us analyze the motion of the two small discs A and B tied to the nail P on a rotating platform. The problem requires us to find the equilibrium conditions and the variation of the angle $\theta$ between the threads as the angular velocity $\omega$ increases.
Figure 1: Top-down view of the platform. $O$ is the center of rotation, $P$ is the nail.
1. System Geometry and Force Analysis
Let the origin be at the center of the platform $O$. The nail is at point $P$ such that the vector $\vec{OP}$ lies along the x-axis with magnitude $l$. The threads connecting the discs to the nail also have length $l$.
We analyze the motion in the rotating frame of the platform. In this frame, each disc experiences a centrifugal force. To find the equilibrium position, it is most convenient to calculate torques about the nail $P$, as the tension force in the string passes through $P$ and exerts zero torque.
Derivation of Centrifugal Torque about P
Let’s derive the expression $\tau_P = m\omega^2 l^2 \sin \phi$.
Coordinate Setup:
- Let $O$ be at origin $(0,0)$.
- Let the nail $P$ be at position $(l, 0)$.
- Let the string make an angle $\phi$ with the extension of the line $OP$ (the positive x-axis).
Figure 1: Geometry for torque calculation. $\phi$ is the angle between the string and the outward radial extension $OP$.
The position vector of the mass $m$ relative to $P$ is:
$$ \vec{r}_{m/P} = l \cos \phi \, \hat{i} + l \sin \phi \, \hat{j} $$The position of the mass relative to the center $O$ is:
$$ \vec{r}_{m/O} = \vec{r}_{P/O} + \vec{r}_{m/P} = (l + l \cos \phi) \hat{i} + (l \sin \phi) \hat{j} $$The centrifugal force acts radially outward from $O$:
$$ \vec{F}_{cf} = m \omega^2 \vec{r}_{m/O} = m \omega^2 [ (l + l \cos \phi) \hat{i} + (l \sin \phi) \hat{j} ] $$The torque $\vec{\tau}$ about $P$ is defined as $\vec{r}_{m/P} \times \vec{F}_{cf}$.
$$ \vec{\tau}_P = (l \cos \phi \, \hat{i} + l \sin \phi \, \hat{j}) \times m \omega^2 [ (l + l \cos \phi) \hat{i} + (l \sin \phi) \hat{j} ] $$Using the cross product rule ($\hat{i} \times \hat{i} = 0$, $\hat{j} \times \hat{j} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{i} = -\hat{k}$):
$$ \vec{\tau}_P = m \omega^2 [ (l \cos \phi)(l \sin \phi)\hat{k} + (l \sin \phi)(l + l \cos \phi)(-\hat{k}) ] $$Factoring out $m \omega^2 l^2$:
$$ \tau_z = m \omega^2 l^2 [ \cos \phi \sin \phi – \sin \phi (1 + \cos \phi) ] $$ $$ \tau_z = m \omega^2 l^2 [ \cos \phi \sin \phi – \sin \phi – \sin \phi \cos \phi ] $$ $$ \tau_z = – m \omega^2 l^2 \sin \phi $$The negative sign indicates the torque tends to decrease $\phi$ (rotating clockwise in the diagram above), pushing the mass towards the x-axis ($\phi=0$).
Magnitude of Torque:
$$ \tau_P = m \omega^2 l^2 \sin \phi $$This rotational tendency is opposed by friction. The maximum frictional torque available is: $$ \tau_{friction} = f_{max} \cdot l = (\mu m g) l $$
The disc will start to slip when the centrifugal torque exceeds the maximum frictional torque: $$ m\omega^2 l^2 \sin\phi > \mu m g l \implies \omega^2 l \sin\phi > \mu g $$
2. Calculating Critical Angular Velocities
Initial State:
The problem states the threads initially make angles $\theta_0 = 45^\circ$ with the line perpendicular to $OP$.
- Disc A: $\phi_A = 90^\circ$. $\sin\phi_A = 1$.
- Disc B: $\phi_B = 135^\circ$ (since it is $45^\circ$ further back). $\sin\phi_B = \sin(135^\circ) = \frac{1}{\sqrt{2}}$.
Finding $\omega_1$ (A starts to slip):
Condition for A: $\omega_1^2 l (1) = \mu g$.
$$ \omega_1 = \sqrt{\frac{\mu g}{l}} = \sqrt{\frac{0.4 \times 10}{0.25\sqrt{2}}} = \sqrt{8\sqrt{2}} \approx 3.36 \text{ rad/s} $$
Finding $\omega_2$ (B starts to slip):
Disc B holds until: $\omega_2^2 l \sin(135^\circ) = \mu g$.
$$ \omega_2^2 l \frac{1}{\sqrt{2}} = \mu g \implies \omega_2 = \sqrt{\frac{\sqrt{2}\mu g}{l}} = \sqrt{\frac{\sqrt{2} \times 4}{0.25\sqrt{2}}} = 4 \text{ rad/s} $$
3. Expression for Angle $\theta$
Let $\theta$ be the angle between threads $PA$ and $PB$. $\theta = \phi_B – \phi_A$.
Phase 1 ($0 < \omega < \omega_1$): Static friction holds both discs. $\theta = 135^\circ – 90^\circ = 45^\circ = \pi/4$.
Phase 2 ($\omega_1 < \omega < \omega_2$): Disc A slips. Its new angle $\phi_A$ is determined by equilibrium: $$ \sin\phi_A = \frac{\mu g}{l\omega^2} $$ Disc B remains fixed at $\phi_B = 135^\circ = 3\pi/4$. $$ \theta(\omega) = \frac{3\pi}{4} – \sin^{-1}\left( \frac{\mu g}{l\omega^2} \right) = \frac{\pi}{4} + \cos^{-1}\left( \frac{\mu g}{l\omega^2} \right) $$
Phase 3 ($\omega > \omega_2$): Disc B slips. Both discs are now unstable and will slide to align with the radius. $\theta \to 0$.
Final Result & Graph
The angle $\theta$ vs $\omega$ is defined as:
$$ \theta = \begin{cases} \frac{\pi}{4} & 0 < \omega \le \omega_1 \\ \frac{\pi}{4} + \cos^{-1}\left( \frac{\mu g}{l\omega^2} \right) & \omega_1 < \omega < \omega_2 \\ 0 & \omega_2 \le \omega \end{cases} $$