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 $R/4$ 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. $-\sqrt{\text{2gR}}$
• B. $\sqrt{\text{3gR}}$
• C. $-\sqrt{\text{4gR}}$
• D. $\sqrt{\text{5gR}}$

Answer 1

Answer: (D)

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

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