A solid sphere is rolling ob a frictionless plane surface about its axis of symmetry. Find the rotational energy and the ratio of rotational energy to its total energy

Open in App

Solution

$T h e r o t a t i o n a l e n e r g y o f t h e s p h e r e i s g i v e n a s : E_{r o t} = \frac{1}{2} I_{c m} ω^{2} I_{c m} = m o m e n t o f i n e r t i a o f t h e s p h e r e a b o u t i t s c e n t r e o f m a s s ω = a n g u l a r v e l o c i t y {(I_{c m})}_{s p h e r e} = \frac{2}{5} M R^{2} ω = \frac{v_{c m}}{R} n o w, E_{r o t} = \frac{1}{2} \times \frac{2}{5} M R^{2} \times {(\frac{v_{c m}}{R})}^{2} E_{r o t} = \frac{1}{5} M {(v_{c m})}^{2} - - - (1) E_{t o t} = E_{r o t} + E_{t r a n s} E_{t r a n s} = \frac{1}{2} M {(v_{c m})}^{2} E_{t o t} = \frac{1}{5} M {(v_{c m})}^{2} + \frac{1}{2} M {(v_{c m})}^{2} E_{t o t} = \frac{7}{10} M {(v_{c m})}^{2} t h e r e f o r e, \frac{E_{r o t}}{E_{t o t}} = \frac{\frac{1}{5} M {(v_{c m})}^{2}}{\frac{7}{10} M {(v_{c m})}^{2}} = \frac{1}{5} \times \frac{10}{7} \frac{E_{r o t}}{E_{t o t}} = \frac{2}{7}$

Suggest Corrections

Similar questions

r

1 / 2 M R^{2}

24 J

Join BYJU'S Learning Program