SATHEE: System of Particles and Rotational Motion

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 $ω_{0}$ is minimum, is given by

[2015 RS]

(a) $x = \frac{m_{1}}{m_{2}} L$

(b) $x = \frac{m_{2}}{m_{1}} L$

(d) $x = \frac{m_{1} L}{m_{1} + m_{2}}$

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Answer:

Correct Answer: 26. (c)

Solution:

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, $\frac{d I}{d x} = 0$

$\Rightarrow \frac{d I}{d x} = 2 m_{1} x + 2 m_{2} x - 2 m_{2} L$

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 $ω_{0}$ .

K.E. $= \frac{1}{2} I ω_{0}^{2}$

Work is minimum, when I is minimum. I is minimum about the centre of mass.

$∴ m_{1} x_{1} = m_{2} (L - x)$

$\Rightarrow m_{1} x = m_{2} L - M_{2} x$

$∴ x = \frac{m_{2} L}{m_{1} + m_{2}}$

System of Particles and Rotational Motion - Result Question 26

Answer:

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System of Particles and Rotational Motion - Result Question 26

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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