The density of a rod of length L varies linearly with position along its length, such that it increases to thrice its value from one end to the other. This rod is placed on a frictionless horizontal surface and a horizontal force of magnitude F is applied to a point on the rod. At what point should this force be applied such that the rod slides on the surface without rotating?
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Answer:
Consider an element dx at a distance x from one end of the rod of length L.
The center of mass of the rod is X
cm
=
∫
0
L
λdx
∫
0
L
xλdx
or X
cm
=
∫
0
L
(A+Bx)dx
∫
0
L
x(A+Bx)dx
=
[Ax+Bx
2
/2]
0
L
[Ax
2
/2+Bx
3
/3]
0
L
=
AL+BL
2
/2
AL
2
/2+BL
3
/3
=
6
3AL
2
+2BL
3
×
2AL+BL
2
2
=
3L(2A+BL)
L
2
(3A+2BL)
=
3(2A+BL)
L(3A+2BL)
Explanation:
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