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Question

A meter stick of mass $$400$$ gm is pivoted at one end and displaced through an angle $$60$$ from vertical The increase in its angular speed is:-


Solution

We know initial potential energy $${E}_{P1}=mg\dfrac{h}{2}$$ ---------(A)
Given m=400 g=.4 kg and h=1 m and g=10$$m/{s}^{2}$$
Putting the above values in A we get 
$${E}_{P1}=0.4\times 10\dfrac{1}{2}=2J$$
Now when the stick is displaced with an angle $${60}^{0}$$
Potential Energy $${E}_{P2}=mg\dfrac{h}{2}\cos{60}^{0}$$
$${E}_{P2}=0.4\times 10\times \dfrac{1}{2} \times \dfrac{1}{2}=1 J$$
Thus change in potential energy $$=2-1$$
$$=1 J$$
  

Physics

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