Question 350: Work, Energy and Power & Circular Motion

Question

A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle $$	heta$$ at which the speed of the bob is half of that at A, satisfies,

Accepted Answer

The bob of mass m requires velocity to reach at a point with vertical circular motion.
$$V = \sqrt {5gl} $$
Then, $$h = L - L\cos 	heta  = L(1 - \cos 	heta )$$
Applying law of conservation of energy at point A & C
Kinetic energy (K.E)
A
+ Potential energy (U
A
) = Kinetic energy (K.E
C
) + Potential energy (U
C
)
$$ \Rightarrow {1 \over 2}m{V^2} + 0 = {1 \over 2}m{\left( {{{\sqrt {5gl} } \over 2}} ight)^2} + mgL(1 - 	heta )$$
$$ \Rightarrow {{5L} \over 2} = {{5L} \over 8} + L(1 - \cos 	heta )$$ ($$\because$$ $$V = \sqrt {5gl} $$)
$$ \Rightarrow {5 \over 2} - {5 \over 8} = 1 - \cos 	heta $$
$$ \Rightarrow {{20 - 5} \over 8} = 1 - \cos 	heta $$
$$ \Rightarrow \cos 	heta  = 1 - {{15} \over 8}$$
$$ \Rightarrow \cos 	heta  = {{ - 7} \over 8}$$
$$ \Rightarrow 	heta  = 152^\circ $$
$$	herefore$$ $${{3\pi } \over 4} < 	heta  < \pi $$

Answer

$$	heta  = {\pi  \over 4}$$

Answer

$${\pi  \over 4} < 	heta  < {\pi  \over 2}$$

Answer

$${\pi  \over 2} < 	heta  < {{3\pi } \over 4}$$

Answer

$${{3\pi } \over 4} < 	heta  < \pi $$

Work, Energy and Power & Circular Motion

A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle $\theta$ at which the speed of the bob is half of that at A, satisfies,

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