29) A ring rolls on a horizontal surface without sliding The velocity of the centre is v It - Physics - System Of Particles And Rotational Motion

Question

29) A ring rolls on a horizontal surface without sliding The
velocity of the centre is v It encounters a step of height 0 3 R
where R is the radius of the ring Calculate the angular velocity of
the ring just after the impact Assume that the ring does not return
back (and there is a sufficient friction to avoid slipping ) Find
the minimum value of 'v' so that the ring ascends the step

Sanjay Singh · Accepted Answer

Dear student
As the torque acting on the ring about point P is zero so applying conservation of angular momentum about point P just befre and after collision
Angular momentum before collision is Lip=Icmω+rcm×mvcm=mr2vr+mv×(r-0
3r)=1 7mvr
1  (rcm×mvcm=mvcm×lever arm that is perpendicular line dropped from P on linear momentum line)After collision ring rotates about axis passing through point P so angular momentum after  collision is Lfp=(moment of inertia about point P)×ωLfp=(moment of inertia about point P)×ω=(Icm+mr2)×ω1=(mr2+mr2)×ω1=2mr2×v1r=2mv1r
2equating eqn 1 and 21 7mvr=2mv1rv1=1 7v2=0
85vNow applying conservation of energy 12m×v21+mg(0
7r)=mgr (for v minimum it will climb with negligible speed so  final kinetic energy is taken as zero)12m×v21=1
3mgrv21=1 6grv1=1 6grRegards0