Question

A uniform circular disk of radius \(0.2\,\text{m}\) and mass \(1\,\text{kg}\) is pivoted at its top point \(C\) such that it can rotate freely around \(C\) in the \(XY\)-plane, as shown in the figure. Initially, when the disk is at rest, a particle of mass \(20\,\text{g}\), travelling along negative \(x\)-direction in the \(XY\)-plane with speed \(100\,\text{ms}^{-1}\), hits the circumference of the disk at a point \(P\). After collision the particle moves along negative \(y\)-direction at a speed of \(90\,\text{ms}^{-1}\). (Given: the acceleration due to gravity \(g = -10\hat{j}\,\text{ms}^{-2}\))

After the collision the disk starts to rotate around point \(C\) in the \(XY\)-plane. The maximum change in the height (in m) of its center \(O\) is:

Hint:

Angular momentum conservation must be applied about the fixed pivot $C$ to account for external impulse at the pivot.
Always verify if the rotational kinetic energy is sufficient for a full rotation ($KE > 2MgR$) or just a swing.
$I_C$ for a disk pivoted at the edge is $1.5 MR^2$.

Answer 1

Step 1: Understanding the Question:
A particle collides with a pivoted disk. We must find the disk's angular velocity using angular momentum conservation about the pivot $C$.
Then, determine if the disk rotates completely or just swings.
The point $P$ is on the circumference. Based on the $45^\circ$ angle, its coordinates relative to $O$ are $(R \sin 45^\circ, -R \cos 45^\circ)$.

Step 2: Key Formula or Approach:
Conservation of Angular Momentum about pivot $C$: $L_i = L_f$.
Moment of Inertia of disk about $C$: $I_C = I_O + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.
Energy conservation after collision: $\frac{1}{2} I_C \omega^2 = M g \Delta h$.

Step 3: Detailed Explanation:

• Parameters: $M=1\text{kg}, R=0.2\text{m}, m=0.02\text{kg}, v_i = 100\text{ms}^{-1}, v_f = 90\text{ms}^{-1}$.
$I_C = 1.5 \times 1 \times (0.2)^2 = 0.06\text{ kg m}^2$.

• Initial Angular Momentum ($L_i$):
$L_i = m v_i d_{\perp i}$. Distance of $v_i$ path from $C$ is $y_C - y_P = R + R \cos 45^\circ$.
$L_i = 0.02 \times 100 \times 0.2(1 + 1/\sqrt{2}) \approx 2 \times 0.2 \times 1.707 = 0.6828\text{ kg m}^2\text{s}^{-1}$.

• Final Angular Momentum ($L_f$):
$L_{particle, f} = -m v_f d_{\perp f}$. Distance of $v_f$ path from $C$ is $x_P = R \sin 45^\circ$.
$L_{pf} = -0.02 \times 90 \times (0.2/\sqrt{2}) \approx -0.2545\text{ kg m}^2\text{s}^{-1}$.
Conservation: $L_i = I_C \omega + L_{pf} \implies 0.6828 = 0.06 \omega - 0.2545 \implies \omega = 15.62\text{ rad/s}$.

• Check Energy:
$KE_{rot} = \frac{1}{2} \times 0.06 \times (15.62)^2 \approx 7.32\text{ J}$.
$PE$ needed to reach top position: $M g (2R) = 1 \times 10 \times 0.4 = 4\text{ J}$.
Since $KE > PE$, the disk completes full rotations.

Step 4: Final Answer:
The disk has enough energy to rotate completely around the pivot $C$.
The center $O$ moves from its lowest point ($R$ below $C$) to its highest point ($R$ above $C$).
Maximum height change $\Delta h = 2R = 0.4\text{ m}$.