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A spherical ball rolls on a table without slipping. Determine the percentage of its `K.E.` which is rotational. Moment of inertia of sphere `= (2//5) xx mass xx ("radius")^(2)`. |
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Answer» Total energy `= K.E.` of translation `+ K.E.` of rotation `(1)/(2)mv^(2) + (1)/(2)I omega^(2)` `E = (1)/(2)m v^(2) + (1)/(2) ((2)/(5) m r^(2)) omega^(2)` `= (1)/(2)m v^(2) + (1)/(5)m v^(2) = (7)/(10)m v^(2)` `(K.E. "of rotation")/("total energy") = (((1)/(2))I omega^(2))/(((7)/(10))m v^(2))` `=(((1)/(5))m v^(2))/(((7)/(10))m v^(2)) =(2)/(7) = (2)/(7) xx 100% = 28.57 %` |
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