Question 692: Centre of Mass and Rotational Motion

Question

A stone is tied to a string of length $$l$$ and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity  as it reaches a position where the string is horizontal (g being acceleration due to gravity) is

Accepted Answer

The total energy at A = the total energy at B
$$ \Rightarrow {1 \over 2}m{u^2} = {1 \over 2}m{v^2} + mgl$$
$$ \Rightarrow v = \sqrt {{u^2} - 2gl} $$
The change in magnitude of velocity = $$\sqrt {{u^2} + {v^2}} $$
$$ = \sqrt {2\left( {{u^2} - gl} \right)} $$

Answer

$$\sqrt {2\left( {{\mu ^2} - gl} \right)} $$

Answer

$$\sqrt {{u^2} - gl} $$

Answer

$$u - \sqrt {{u^2} - 2gl} $$

Answer

$$\sqrt {2gl} $$

Centre of Mass and Rotational Motion

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