A uniform cylindrical rod of mass m and density po is having a point mass m' at its bottom end. m' is just the minimum value of mass so that, the complete arrangement is just found to be in stable equilibrium at the position shown in diagram. The value of ratio of 'density of liquid' to 'that rod' under the given condition, is
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Answer:
By parallel axis Theorem, MOL of rod
about hinge is
I
0
=
12
mL
2
+m(
4
L
)
2
=
48
7mL
2
When rod reaches the vertical position
the center of mass of rod falls by
L/4
By consecration of energy
mg(
4
L
)=
2
1
×I
0
w
2
4
mgL
=
2
1
×
48
7mL
2
w
2
∴w=
7L
24g
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