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Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity? |
Answer» Solution : Assume that conservation of energy of the rolling body. The potential energy lost by the body in rolling down the inclined PLANE is MGH. Where m is the mass of body, g is gravitational acceleration and h is height of slope. Equal kinetic energy gain by the body. The body start from the rest, the kinetic energy gained is equal to the final kinetic energy of the body. Hence final kinetic energy = `(1)/(2)Iomega^(2)+(1)/(2)mv^(2)` potential energy = final kinetic energy `mgh=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)` `mgh=(1)/(2)mk^(2)omega^(2)+(1)/(2)mv^(2) [because I=mk^(2)]` `:. gh=(1)/(2)k^(2)xx(v^(2))/(R^(2))+(1)/(2)v^(2) [because omega=(v)/(R )]` `:. gh=(1)/(2)v^(2)[1+(k^(2))/(R^(2))]` `therefore v^(2)=(2gh)/(1+(k^(2))/(R^(2)))` `therefore v=[(2gh)/(1+(k^(2))/(R^(2)))]^((1)/(2))""...(1)` Hence, the velocity is independent of the mass of the rolling body (1) For a ring : `k=R :. k^(2)=R^(2)` From eqn. (1), `therefore v_("ring")=[(2gh)/(1+(R^(2))/(R^(2)))]^((1)/(2))` `=[(2gh)/(1+1)]^((1)/(2))=[(2gh)/(2)]^((1)/(2))` `=SQRT(gh)` (2) For solid sphere : `k=(R)/(sqrt(2))impliesk^(2)=(R^(2))/(2)` `:. v_("cylinder")=[(2gh)/(1+(R^(2))/(2R^(2)))]^((1)/(2))=[(2gh)/(1+(1)/(2)]]^((1)/(2))` `=[(4gh)/(3)]^((1)/(2))` `=sqrt((4gh)/(3))` (3) For solid sphere : `k=Rxxsqrt((2)/(5))impliesk^(2)=(2R^(2))/(5)` `:. v_("solid sphere")=[(2gh)/(1+(2R^(2))/(5R^(2)))]^((1)/(2))=[(2gh)/(1+(2)/(5))]^((1)/(2))` `=[(2gh)/((7)/(5))]^((1)/(2))` `=sqrt((10gh)/(7))` Hence, the sphere has the greatest and the ring has the least velocity of the centre of mass at the BOTTOM of the inclined plane. So solid sphere reached at first to the bottom of slope. |
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