I am impressed by the amount of interest (confusion?) this question has generated but hats off to all the mentors (Tsny , haruspex and tiny tim ) for explaining it scientifically.
Yes I took the torque due to gravity to be zero in comparison to the impulsive force at that moment because simply...
@echild , the energy is NOT conserved because it can be clearly seen that collision is inelastic.
@haruspex, its not mv^2/r its mv2×r ie new linear velocity of cm.×r
Actually I ve figured it out and got the answer.
Firstly taking angular momentum about edge of the step:
Mv×3r/4 +(ang...
No, the angular momentum about the step IS conserved because whenever we decide to conserve ang momentum we see if net EXTERNAL torque acting on that point is zero and Mg is internal to that cylinder .(Hence the torque is zero)
I am sorry that I didnt say anything about the assumptions. .
I have...
1. Homework Statement :
Cylinder of mass m, radius r is rolling without slipping with angular velocity ω and Vcm is ωr.
Cylinder comes across a step of height R /4. The angular velocity just after the collision is...
Assuming cylinder remains in contact and no slipping occurs on the edge of the...