Vertical Circular Motion!

Classical Mechanics Level 3

A bob of mass M M is suspended by mass-less string of length L L . The horizontal velocity at point A A is such that the tension at point A A is five times the weight of bob. The angle ' θ \theta ' at which the speed of bob is half of speed at point A A is:

120° 90° 60° 30° None of These

Nishant Rai by Nishant Rai , 25, India

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1 solution

Tanishq Varshney
Jun 7, 2015

Let T T be the tension in the string.

Given T = 5 M g T=5Mg

T − M g = M v 2 L T-Mg=M \large{\frac{v^2}{L}}

v 2 = 4 g L v^{2}=4gL ............................... ( 1 ) (1)

Applying conservation of energy

0.5 M v 2 = M g L ( 1 − cos ⁡ θ ) + 0.5 M ( v 2 ) 2 0.5Mv^2=MgL(1- \cos \theta)+0.5M \large{(\frac {v}{2})^{2}}

1 − cos ⁡ θ = 3 2 1- \cos \theta=\frac{3}{2}

cos ⁡ θ = − 0.5 \cos \theta=-0.5

θ = 12 0 o \theta= 120^{o}

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