Question

A uniform circular disc has radius $R$ and mass $m$ . A particle, also of mass $m$ , is fixed at a point $A$ on the edge of the disc as shown in the diagram. The disc can rotate freely about a fixed horizontal chord $PQ$ that is at a distance $\frac{R}{4}$ from the centre $C$ of the disc. The line $AC$ is perpendicular to $PQ$ . Initially, the disc is held vertical with point $A$ at its highest position. It is then allowed to fall so that it starts rotating about $PQ$ . Find the linear speed of the particle as it reaches its lowest position.

• A. $-\sqrt{2 g R}$
• B. $\sqrt{3 g R}$
• C. $-\sqrt{4 g R}$
• D. $\sqrt{5 g R}$

Answer 1

Answer: (D)

As moment of inertia of a disc about a diameter is $\frac{1}{2}\left(\frac{1}{2} mR ^{2}\right)$, the moment of inertia of the disc about the chord $P Q$ by 'theorem of parallel axes' will be
$\left( I _{ D }\right)_{ PQ }=\frac{1}{4} mR ^{2}+ m \left(\frac{1}{4} R \right)^{2}=\frac{5}{16} mR ^{2}$
and as particle of mass $m$ is at a distance $[R+(R / 4)=(5 / 4) R]$ from $P Q$, the moment of inertia of the system about $P Q$
$I =\left( I _{ D }\right)_{ PQ }+\left( I _{ P }\right)_{ PQ }=\frac{5}{16} mR ^{2}+ m \left(\frac{5}{4} R \right)^{2}=\frac{15}{8} mR ^{2}$
Now if $\omega$ is the angular speed of the system when A reaches the lowest point $A^{\prime}$ on rotation about the axis $P Q$, by 'conservation of mechanical energy',
$\frac{1}{2} I ^{2}= mg \left( AA ^{\prime}\right)+ mgCC = mg [2 AD +2 CD ]$

$\text{i.e.,} \frac{1}{2} \times \frac{\text{15}}{8} \text{mR} ^{2} \left(\text{ω}\right)^{2} = 2 \text{mg} \left[\left(\text{R} + \frac{1}{4} \text{R}\right) + \frac{1}{4} \text{R}\right] \text{,}$
$\text{i.e.,} \text{ω} = 4 \sqrt{\frac{\text{g}}{\text{5R}}}$
$\text{so} \left(\text{v}\right)_{\left(\text{A}\right)^{'}} = \text{r} \text{ω} = \left(\text{R} + \frac{1}{4} \text{R}\right) \times 4 \sqrt{\frac{\text{g}}{\text{5R}}} = \sqrt{\text{5gR}}$