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Question

A meter stick is held vertically with one end on the floor and is allowed to fall. The speed of the other end when it hits the floor assuming that the end at the floor does not slip is $$(g=9.8{m/s}^{2})$$


A
3.2 m/s
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B
5.4 m/s
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C
7.6 m/s
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D
9.2 m/s
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Solution

The correct option is B $$5.4\ m/s$$
When meter stick allowed to fall from verticle postion, its potential energy is converted to rotational energy 
Lets height of the meter stick is L and mass is M.
center of mass of meter stick fall from height $$\frac{1}{2}L$$
using energy conservation 
$$Mg\frac{1}{2}L=\frac{1}{2}I\omega^2$$ ..eq(1)
where I is a moment of inertia from one end of the meter stick $$I=\frac{1}{3}ML^2$$ and $$\omega$$ is angular velocity .
from eq(1)
$$\frac{MgL}{2}=\frac{1}{2}\times\frac{ML^2}{3}\omega^2=\frac{ML^2}{6}\omega^2$$
$$\Rightarrow \omega={\sqrt{\frac{3g}{L}}}$$
speed of other end is $$v=\omega L$$
$$\Rightarrow v=\sqrt{3gL}$$
where L=1m
$$v=\sqrt{3g}=5.4ms^{-1}$$
Hence B option is correct.


Physics

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