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 frcition on a vertical rod fixed at a distance d from the support. Initally, the spring makes an angle of 37 o
with the horizontal as shown in the figure, The system is released from rest. The speed of the ring at the same angle substended downward will be
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Explanation:
ANSWER
If l is the stretched length of the spring, then from figure
l
d
=cos37
∘
=
5
4
,i.e.,l=
4
5
d
So, the stretch y=l−d=
4
5
d−d=
4
d
and h=lsin37
∘
=
4
5
d×
5
3
=
4
3
d
Now, taking point B as reference level and applying law of conservation of mechanical energy between A and B,
E
A
=E
B
or mgh+
2
1
ky
2
=
2
1
mv
2
[as for B, h=0 and y=0]
or
4
3
mgd+
2
1
k(
4
d
)
2
=
2
1
mv
2
[as for A, h=
4
3
dandy=
4
1
d]
or v=d
2d.
3g
+
16m
h
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