A rod of length 3m and its mass acting per unit length is directly proportional to distance x from its one end. The center of mass of the rod from that end will be at
(a) 1.5m (b)2m (c)2.5m (d) 3m
Answers
Answer:
That's your answer......
Answer:
The center of gravity from the end of the rod is 2 cm.
Explanation:
Given that,
Length = 3 m
Let us consider an elementary length dx.
Its mass acting per unit length is directly proportional to distance x from one of its end.
Its mass is
m=kxm=kx
Where, m=\dfrac{M}{L}m=
L
M
dm=kxdxdm=kxdx
We need to calculate the center of gravity of the rod
Using formula of center of gravity
x_{cm}=\dfrac{\int{x dm}}{\int{dm}}x
cm
=
∫dm
∫xdm
Put the value into the formula
x_{cm}=\dfrac{\int_{0}^{3}{x\timeskx dx}}{\int_{0}^{3}{kx dx}}x
cm
=
∫
0
3
kxdx
∫
0
3
x\timeskxdx
x_{cm}=\dfrac{(\dfrac{x^3}{3})_{0}^{3}}{(\dfrac{x^2}{2})_{0}^{3}}x
cm
=
(
2
x
2
)
0
3
(
3
x
3
)
0
3
x_{cm}=\dfrac{\dfrac{27}{3}}{\dfrac{9}{2}}x
cm
=
2
9
3
27
x_{cm}=2\ cmx
cm
=2 cm
Hence, The center of gravity from the end of the rod is 2 cm.
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