The density of a non-uniform rod of length 1m is given by rho (x) = a(1+bx 2) where a and b are constants and 1 ox ≤≤ . The centre of mass of the rod will be at
(a)3(2+b)/4(3+b)(b)4(2+b)/3(3+b)(c)3(3+b)/4(2+b)(d)4(3+b)/3(2+b)explain how to solve?
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Answered by
0
Answer:
i dont know
Explanation:
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Answered by
2
Answer:
your answer is below
Explanation:
Consider a differential part at the rod at a distance x
Then,
l=
dx
dm
=a(l+bx
2
)
⇒dm=a(1+bx
2
)dx
We know,
X
COM
=
∫
0
1
(dm)
∫
0
1
(dm)x
=
∫
0
1
a(1+bx
2
)dx
∫
0
1
a(1+bx
2
)x.dx
X
COM
=
a∫
0
1
(1+bx
2
).dx
a∫
0
1
(x+bx
3
)dx
=
[x+
3
bx
3
]
0
1
[
2
x
2
+
4
bx
]
0
1
X
COM
=
1+
3
b
2
1
+
4
b
=
4(3+b)
3(2+b)
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