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A solid sphere rolls on horizontal surface without slipping. What is the ratio of its rotational to translation kinetic energy. |
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Answer» Consider a symmetric rigid body, like a sphere or a whell or a disc, rolling without slipping on a horizontal plane surface with friction along a straight path. The rolling motion of the body can be considered as a combination of translation of the centre of mass and rotation about the centre of mass. Hence, the kinetic nergy of a rolling body is `E=E_(tran)+E_(rot)" "`.....(1) where `E_(tran)` and `E_(rot)` are the kinetic energies associated with translation of the centre of mass and rotation about an axis through the centre of mass, respectively. Let M and R be the mass and radius of the body. Let `omega`, k and I be the angular speed, radius of gyration and moment of inertia for rotation about an axis thorugh its centre of mass, and v be the translational speed of the centre of mass. `:. v=omegaR` and `I=Mk^(2)" "`.........(2) `:. E_(tran)=(1)/(2)Mv^(2)` and `E_(rot)=(1)/(2)Iomega^(2)" "`........(3) `:. E=(1)/(2)Mv^(2)+(1)/(2)Iomega^(2)=(1)/(2)Mv^(2)+(1)/(2)Mk^(2)(v^(2))/(R^(2))` `=(1)/(2)Mv^(2)(1+(k^(2))/(R^(2)))" "`......(4) For a solid sphere, `k=sqrt((2)/(5))R" "`......(5) `:. E=(1)/(2)Mv^(2)(1+(2)/(5))=(1)/(2)Mv^(2)((7)/(5))=(7)/(10)Mv^(2)" "`....(6) |
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