Question

Fig shows a parabolic track y = kx² ; The part AO is frictional and part OB is smooth. A cylinder of radius R; mass M, is let go from rest from point P. It rolls down AO without slipping. It ascends on the other side of track up to point Q. There is no loss of energy when cylinder moves from part AO to OB at point O. Then

(1) \(h=\frac{2H+R}{3}\)

(2) \(h =\frac{H+R}{3}\)

(3) \(h=\frac{2(H+R)}{3}\)

(4) \(h=\frac{H}{3}+\frac{R}{2}\)

Answer 1

 (1) \(h=\frac{2H+R}{3}\)

Let v and ω be linear speed of C.M and angular speed of cylinder at O. From law of conservation of energy

Loss in gravitational P.E. = Gain in K.E.

Since there is no slipping v = Rω. Therefore

Since the part OB of track is smooth; as cylinder climbs up it losses translational part of its kinetic energy. 

The loss in translation K.E equals gain in gravitational P.E. Therefore

From Eqns (i) and (ii) we have