Question 693: Centre of Mass and Rotational Motion

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

Accepted Answer

Rotational energy =$${1 \over 2}I{\left( \omega  ight)^2} = {1 \over 2}\left( {m{K^2}} ight){\omega ^2}$$
Linear kinetic energy = $${1 \over 2}m{\omega ^2}{R^2}$$
$$ 	herefore $$ Required fraction
$$ = {{{1 \over 2}\left( {m{K^2}} ight){\omega ^2}} \over {{1 \over 2}\left( {m{K^2}} ight){\omega ^2} + {1 \over 2}m{\omega ^2}{R^2}}} = {{{K^2}} \over {{K^2} + {R^2}}}$$

Answer

$${{{K^2} + {R^2}} \over {{R^2}}}$$

Answer

$${{{K^2}} \over {{R^2}}}$$

Answer

$${{{K^2}} \over {{K^2} + {R^2}}}$$

Answer

$${{{R^2}} \over {{K^2} + {R^2}}}$$

Centre of Mass and Rotational Motion

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

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