Question

# A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme position and lowest position are equal. The angle $$\theta$$ of thread deflection in the extreme position will be:

A
tan1(2)
B
tan1(2)
C
tan1(12)
D
2tan1(12)

Solution

## The correct option is C $$2\tan ^{ -1 }{ \left( \cfrac { 1 }{ 2 } \right) }$$At highest point velocity is zero $$a_N =0$$                                                      $$a_T =gsin\theta$$At lowest point $$a_N = \dfrac{v^2}{l}$$                          $$a_T =0$$ Applying law of conventional energy       $$mgl(1-cos\theta) = \dfrac{1}{2}mv^2$$$$\dfrac{v^2}{l} = 2g(1-cos\theta)$$$$gsin\theta = 2g(1-cos\theta)$$$$2sin\dfrac{\theta}{2}cos\dfrac{\theta}{2} = 2 \times 2 {sin}^2\dfrac{\theta}{2}$$$$\theta = 2{tan}^{-1}\dfrac{1}{2}$$Physics

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