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Obtain the necessary condition v_(cm)=Romega for rolling body without stepping. |
Answer» Solution :When a rigid body (sphere, circular disc or WHEEL of vehicle) rolling without slipping then rolling motion exist.  As shown in figure a circular disc of radius R is rolling on its circular surface without slipping. Its centre of mass is on the geometric centre of the disc `V_(cm)` is the velocity of C. It is parallel to the surface level. The rotational motion of the disc is about axis. Which passes through centre. `P_(0),P_(1)andP_(2)` are the point on disc is shown in figure the velocity at these points, are obtained the sum of components of vectors. The resultant velocity at point `P_(1),vec(v_(1))=vec(v_(R))+vec(v_(cm))` where `vec(v_(R))` is the velocity at `P_(1)` point, `vec(v_(R))=Romega` `THEREFORE` The resultant velocity at point `P_(1),vec(v_(1))=Romega+Romega` [`because` are in same DIRECTION] [`because` velocity of cm, `vec(v_(cm))=Romega`] `therefore vec(v_(1))=2Romegaorvec(v_(cm))=2Romega` The position vector of `P_(2)` is `vecr`. So linear velocity at point `P_(2),vec(v_(r))=rvecomega`, which is perpendicular to position vector `vecr`. `therefore` The resultant velocity at point `P_(2)=vec(v_(2))=vec(v_(r))+vec(v_(cm))` `=vec(r_(omega))+vec(Romega)` `P_(0)` is instantaneously at rest and the velocity of this point `vec(v_(0))=vec(v_(R))+vec(v_(cm))`, here `v_(0)=0` `therefore 0=vec(v_(R))+vec(v_(cm))` but `vec(v_(R))andvec(v_(cm))` are in perpendicular direction when `|vec(v_(R))|=|-vec(v_(cm))|,P_(0)` becomes at rest. `therefore` Due to resultant velocity at point `P_(0)` zero, `v_(cm)=Romega`. This condition applies to all rolling bodies. |
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