Question 664: Centre of Mass and Rotational Motion

Question

A small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2,  the kinetic energy of the mass will

Accepted Answer

According to law of conservation of
angular momentum
mvr = mv'r'
$$vr = v'\left( {{r \over 2}} ight)$$
$$v' = 2v$$    ...(i)
$$ 	herefore $$ $${K \over {K'}} = {{{1 \over 2}m{v^2}} \over {{1 \over 2}mv{'^2}}} = {\left( {{v \over {v'}}} ight)^2}$$
$$ \Rightarrow {{K'} \over K} = {\left( {{{v'} \over v}} ight)^2} = {\left( 2 ight)^2}$$    (Using (i))
$$ \Rightarrow K' = 4K$$

Answer

decrease by a factor of 2

Answer

remain constant

Answer

increase by a factor of 2

Answer

increase by a factor of 4

Centre of Mass and Rotational Motion

A small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will

Choose the correct answer: