Recent content by aditya~

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 )

solving the first part click here