System of Particles and Rotational Motion - Result Question 26
26. Point masses $m_1$ and $m_2$ are placed at the opposite ends of a rigid rod of length $L$, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point $P$ on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity $\omega_0$ is minimum, is given by
[2015 RS]
(a) $x=\frac{m_1}{m_2} L$
(b) $x=\frac{m_2}{m_1} L$
(c) $x=\frac{m_2 L}{m_1+m_2}$
(d) $x=\frac{m_1 L}{m_1+m_2}$
Show Answer
Answer:
Correct Answer: 26. (c)
Solution:
- (c) From figure,
Total moment of inertia of the rod,
$I=m_1 x^{2}+m_2(L-x)^{2}$
$I=m_1 x^{2}+m_2[L^{2}+x^{2}-2 L x]$
$I=m_1 x^{2}+m_2 L^{2}+m_2 x^{2}-2 m_2 L x$
As I is minimum, so, $\frac{dI}{dx}=0$
$\Rightarrow \frac{dI}{dx}=2 m_1 x+2 m_2 x-2 m_2 L$
Using equation (i)
$\Rightarrow 2 m_1 x+2 m_2 x-2 m_2 L=0$
$\Rightarrow x=\frac{m_2 L}{(m_1+m_2)}$
When I is minimum, the work done by rotating a rod will be minimum.
Work required to set the rod rotating with angular velocity $\omega_0$.
K.E. $=\frac{1}{2} I \omega_0^{2}$
Work is minimum, when I is minimum. I is minimum about the centre of mass.
$\therefore m_1 x_1=m_2(L-x)$
$\Rightarrow m_1 x=m_2 L-M_2 x$
$\therefore x=\frac{m_2 L}{m_1+m_2}$
