Mahesh Koilada
Last Activity: 7 Years ago
along inclined plane will also act in the same direction. Therefore the final equation of motion will be, as belowmg will try to move upwards relative to the stationary inclined surface, so friction opposes this uoward motion and act in the opposite direction, that is downwards along the inclined plane). The component of the-point-on-the-cylinder will always act downwards along the inclined plane (this is becase the friction force will always oppose the relative motion of that point at contact between interacting surfaces; in this case u*mg*cosx. Therefore the friction force zero and a point exactly below this point on inclined plane, at any instant, will have to be the-point-on-the-cylinder condition (as stated in the problem), the relative velocity between that pure rolling will have a tendency to move upwards along the inclined plane. But, becase of the imposed the-point-on-the-cylinderAt the point of contact between inclined plane & cylinder,
mg*sinx+u*mg*cosx = ma and this implies, a = g*(sinx + u*cosx) …........... {u=coefficient of friction}
L=1/2*a*t^2.............................{L=length of the plane, t=time taken for the cylinder to roll down the plane upto bottom}
therefore t=sqrt(2L/a)
and we know V=at=a*sqrt(2L/a)=sqrt(2aL)=sqrt(2*g*(sinx+u*cosx)*L)
P.S. u has to be specified to get the final numerical answer.
All the best.