Rotational Motion Question 51
Question: A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be:
[AIPMT 2003]
Options:
A) $ \frac{K^{2}}{K^{2}+R^{2}} $
B) $ \frac{R^{2}}{K^{2}+R^{2}} $
C) $ \frac{K^{2}+R^{2}}{R^{2}} $
D) $ \frac{K^{2}}{R^{2}} $
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Answer:
Correct Answer: A
Solution:
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Key Idea: In rolling without slipping, total energy of ball is the sum of its translational and rotational energy.
Kinetic energy of rotation is
$ K _{rot}=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}MK^{2}\frac{v^{2}}{R^{2}} $
where K is radius of gyration
Kinetic energy of translation is
$ K _{trans}.=\frac{1}{2}Mv^{2} $
Thus, total energy
$ E=K _{rot}+{K _{trans.}} $
$ =\frac{1}{2}MK^{2}\frac{v^{2}}{R^{2}}+\frac{1}{2}Mv^{2} $
$ =\frac{1}{2}Mv^{2}( \frac{K^{2}}{R^{2}}+1 ) $
$ =\frac{1}{2}\frac{Mv^{2}}{R^{2}}(K^{2}+R^{2}) $
Hence,
$ \frac{K _{rot}.}{{K _{trans.}}}=\frac{\frac{1}{2}MK^{2}\frac{v^{2}}{R^{2}}}{\frac{1}{2}\frac{Mv^{2}}{R^{2}}(K^{2}+R^{2})}=\frac{K^{2}}{K^{2}+R^{2}} $
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