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

Answer 1

Answer: (D)

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

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