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A solid cylinder is rolling down on an inclined plane without slipping. the length of the plane is 30 m and its angle of inclination is 30 degrees . If the cylinder starts from rest then its vleocity at the bottom of the inclined plane is

A solid cylinder is rolling down on an inclined
plane without slipping. the length of the plane is
30 m and its angle of inclination is 30 degrees . If the
cylinder starts from rest then its vleocity at the
bottom of the inclined plane is

Grade:12th pass

2 Answers

Mahesh Koilada
35 Points
6 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.
Mahesh Koilada
35 Points
6 years ago
Equation of motion for the given problem will be as belowmg*sinx+u*mg*cosx = ma and this implies, a = g*(sinx + u*cosx)       …........... {u=coefficient of friction}#please ignore my above answer
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.

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