Question 650: Centre of Mass and Rotational Motion

Question

A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. The mass is attached to a string which passes through a smooth hole in the plane as shown. the tension in the string is increased gradually and finally m moves in a circle of radius $${{{R_0}} \over 2}$$.
The final value of the kinetic energy is

Accepted Answer

According to law of conservation of
angular momentum
mvr = mv'r'
$${v_0}{R_0} = v\left( {{{{R_0}} \over 2}} ight);v = 2{v_0}$$   Ã¢ÂÂ¦(i)
$$ 	herefore $$ $${{{K_0}} \over K} = {{{1 \over 2}mv_0^2} \over {{1 \over 2}m{v^2}}} = {\left( {{{{v_0}} \over v}} ight)^2}$$
$$ \Rightarrow $$ $${K \over {{K_0}}} = {\left( {{v \over {{v_0}}}} ight)^2} = {\left( 2 ight)^2}$$    (Using (i))
K = 4K
0
= 2$${mv_0^2}$$

Answer

2mv$$_0^2$$

Answer

$${1 \over 2}$$mv$$_0^2$$

Answer

mv$$_0^2$$

Answer

$${1 \over 4}$$mv$$_0^2$$

Centre of Mass and Rotational Motion

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