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Initially a sphere of radius r is rotating with an angular velocity omega about its own horizontal axis. When the sphere falls on a surface (coefficient of friction mu), then it begins to skid first and then starts rotating without skidding. What will be the final linear velocity of its centre of mass? |
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Answer» Solution :Let the mass of the sphere be m. Then its moment of INERTIA about the AXIS of rotation, `I= (2)/(5) mr^(2)` Fig. Moment of frictional force `(mumg)` resists the rotational motion of the sphere. If the angular retardation is `alpha` then `mu mgr =Ialpha = (2)/(5) mr^(2)alpha` or, `alpha =(5mug)/(2r) "" cdots (1)` Due to this it the angular velocity of the sphere BECOMES `omega`. in time t, then `omega. = omega-alphat = omega- (5 mu"gt")/(2r)` The speed of a rotating point on the upper surface of the sphere, `v. = omega.r= (omega-(5mu"gt")/(2r))r ""cdots(2)` Again, due to frictional force `mu`mg if the sphere skids over the surface with an acceleration a, then `mu`mg= ma or, a =`mu`g `:.` The LINEAR velocity of the centre of mass of the sphere in time t, v = 0 +at = `mu`gt `""cdots(3)` The condition of rotational motion of the sphere with out skidding is v= v.. If the values of these two velocities become the same in time t, the sphere will undergo pure rotation. From equations (2) and (3) we get, `mu "gt"=(omega-(5mu"gt")/(2r))r=omegar-(5mu"gt")/(2)` or,`(7)/(2)mu"gt"=omegaror,t=(2omegar)/(7mug)` `:.` From equation (3) we get , `v = mugxx(2omegar)/(7mug)=(2)/(7)omegar`. 
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