i am facing problem in attempting the peoblem attached below
if i work in the non inertial frame (the rotating)
shouldn't the acceleration of m_a after the catch is removed
be simply =\omega2ra
thanks in advance
i think the work energy theorem itself is a manifestation of Newton's secomd law
we write
F(net)=ma
say the motion is 1 dimensional in the x-axis
F(x)=mvdv/dx
take the dx to the other side and integrate
integral(F(x)dx) =delta kinetic energy
so if u have done by work...
the second part can also be proved:
after m hits the table ,the string becomes lose
find the velocity of M at that instant using work energy theorem
and then equate this K.E of M to its max potential energy(that corresponds to max height )