A ball of radius R rolls without slipping. Find the fraction of total energy associated with its rotational energy, if the radius of the gyration of the ball about an axis passing through its centre of mass is K .

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Correct answer: A. (K^2)/(K^2+R^2)

Kinetic energy of rotation is K_ rot =(1)/(2) l ω^2=(1)/(2) M K^2 (v^2)/(R^2) where, k is radius of gyration. Kinetic energy of translation is K_ trans =(1)/(2) M v^2 Thus total energy, aligned E & =K_ rot +K_ trans & =(1)/(2) M K^2 (v^2)/(R^2)+(1)/(2) M v^2 & =(1)/(2) M v^2((K^2)/(R^2)+1) & =(1)/(2) (M v^2)/(R^2)(K^2+R^2) aligned Hence, aligned K_ rot Total energy, E & =((1)/(2) M K^2 (v^2)/(R^2))/((1)/(2) (M v^2)/(R^2)(K^2+R^2)) & =(K^2)/(K^2+R^2) aligned

A ball of radius R rolls without slipping. Find the fraction of...

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