Mechanics
Q6. One end of a light spring of natural length d and spring constant k (= mg/d) is fixed on a rigid support and the other end is fixed to a smooth ring of mass m which can slide without friction on a vertical rod fixed at a distance d from the support. Initially, the spring makes an angle of 37° with the horizontal as shown in the figure.The system is released from rest. The speed of the ring at the same angle subtended downward will be
(A) (B)
(C) (D)
Dear Student
Cos37= h/l
l= h/cos37
l=1.25h
extension = l-h
=1.25h- h= 0.25h
equalling KE and PE
1/2mv2= 1/2k(0.25h)2
v2= k/m x h2/16
v= h/4 √k/m
l= h/cos37
l=1.25h
extension = l-h
=1.25h- h= 0.25h
equalling KE and PE
1/2mv2= 1/2k(0.25h)2
v2= k/m x h2/16
v= h/4 √k/m
Regards