A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of
cylinder when it reaches the bottom of the plane will be
Answers
Answer:
Your answer is------ Initial Velocity
Answer:
first of all let's assume that there is an incline plane with height h and a solid cylinder of radius R is placed over its top point and it has a radius R and and will have moment of inertia (MR^2)/2 . and now find energies at different point , first of all energy at point A will only be Kinetic and potential energy which is equal to MV^2/2 + Mgh and on point B will be only kinetic energy + rotational kinetic energy which is equal to {let's assume it has velocity at point B u and angular velocity w}
thus total every at pont B= Mu^2/2 +Lw^2/2 and since the body is going pure rolling it must have u= wR and and on putting the values of u and moment of inertia in the equation of total energy at point B and equating it to energy at point A(law of conservation of mechanical energy) you can get velocity at point B as your answer (2(v^2+ 3gh)/3) square root .(there will always also be additional potential energy MgR in both cases but only equating the equations it will be cancel out therefore I have neglected it)
