A cylinder , released from the top of an inclined plane rolls without sliding and reaches the bottom with a speed vr . Another identical cylinder released from the top of the same inclined plane slides without rolling and reaches the bottom with a speed vs then the relationship between r and vs is ?

L = distance that has to be travelled by the cylinder along the inclined plane

x = angle of inclination of the plane with horizontal

m = mass of the cylinder

g = acceleration due to gravity

u = coefficient of friction between cylinder and inclined plane

N = normal reaction-force between the cylinder & inclined plane

V1 = velocity of the cylinder at the bottom of the inclined plane in case-1 (pure rolling condition)

V2 = velocity of the cylinder at the bottom of the inclined plane in case-2 (pure sliding condition)

a1 = acceleration of the cylinder in case-1 (pure rolling condition)

a2 = acceleration of the cylinder in case-2 (pure sliding condition)

t1 = time taken by the cylinder to reach the bottom of the inclined plane in case-1 (pure rolling condition)

t2 = time taken by the cylinder to reach the bottom of the inclined plane in case-2 (pure sliding condition)

 

 

 

In this case, at the point of contact between inclined plane & cylinder, the-point-on-the-cylinder will have a tendency to move upwards along the inclined plane. But, becase of the imposed pure rolling condition (as stated in the problem), the relative velocity between that the-point-on-the-cylinder and a point exactly below this point on inclined plane, at any instant, will have to be zero. Therefore the friction force u*mg*cosx will always act downwards along the inclined plane (this is becase the friction force will always oppose the relative motion of the point at contact between interacting surfaces; in this case the-point-on-the-cylinder 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 mg along inclined plane will also act in the same direction. Therefore the final equation of motion will be, as below

 

mg*sinx+u*mg*cosx = ma1 and this implies, a1 = g*(sinx + u*cosx)

 

and we can write V1 = a1*t1 and L = ½*a1*t1^2

 

eliminating t1, we get V1 = sqrt [2gL(sinx + u*cosx)]

 

 

In this case, at the point of contact between inclined plane & cylinder, the-point-on-the-cylinder will have a tendency to move downwards along the inclined plane (as it’s pure sliding motion). Opposing this motion, friction acts upwards along the inclined plane and its magnitude will be u*mg*cosx. The equation of motion, in this case, will be as follows

 

mg*sinx-u*mg*cosx = ma2 and this implies, a2 = g*(sinx - u*cosx)

 

similarly, as above, we can write V2 = a2*t2 and L = ½*a2*t2^2

 

eliminating t2, we get V2 = sqrt [2gL(sinx – u*cosx)]

 

So, (by know the answer already !!) the relation between V1 and V2 is

 

V1 / V2 = sqrt [(sinx+u*cosx)/(sinx-u*cosx)]

 

All the best.

 

P.S. Please excuse me for any typo errors.