Question 650: Centre of Mass and Rotational Motion

Question

A mass m moves in a circle on a smooth horizontal plane with velocity v
0
at a radius R
0
. 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
![](https://neetcare-image.cdn.examgoal.net/7k9eik78viq3g/c1c7a343-52f5-4028-b036-cfa572a766b1/e0e43dc0-5ba4-11ea-a4e3-ff1fddf1e89e/file-7k9eik78viq3h-316w.jpg)

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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