Question

A solid cylinder of mass ($M=1\text{kg}$) and radius $R=0.5\text{m}$ is pivoted at its centre The axis of rotation of the cylinder is horizontal. Three small particles of mass ($m=0.1\text{kg}$) are mounted along its surface as shown in figure. The system is initially at rest.

The angular speed of cylinder, when it has rotated through ${90}^{\circ }$ in anticlockwise direction.
(Take $g=10{\text{m/s}}^2$)

• A. $\sqrt{5}\text{rad/s}$
• B. $2\sqrt{3}\text{rad/s}$
• C. $\sqrt{10}\text{rad/s}$
• D. $4\sqrt{2}\text{rad/s}$

Answer 1

The correct option is A $\sqrt{5}\text{rad/s}$

M.O.I. of system (cylinder + masses) about axis passing through centre of solid cylinder is,
$I=I_0+I_1+I_2+I_3$
$\Rightarrow I=\frac{M{R}^2}{2}+m_1{R}^2+m_2{R}^2+m_3{R}^2$
$\Rightarrow I=R^2(\frac{M}{2}+m_1+m_2+m_3)$
or, $I=\left(0.5{)}^2\right(0.5+0.3)=0.20{\text{kg m}}^2$


From change in configuration of system, loss in P.E has occured.
Applying mechanical energy conservation between initial and final position.
$\Rightarrow$ Loss in P.E. = Gain in rotational K.E.
$\Rightarrow mgR=\frac{1}{2}I{w}^2$
$\Rightarrow \left(0.1\right)\times 10\times 0.5=\frac{1}{2}\times 0.20\times w^2$
$\Rightarrow w^2=\frac{10\times 0.1\times 0.5}{\frac{1}{2}\times 0.20}=5$
$\therefore w=\sqrt{5}\text{rad/s}$