if the mass per unit length of a rod of length L(=1m) varies as λ = ax+bx2 from x=0 to x=L.Locate the x-coordinate of the centre of mass of the rod.[a&b are constants]
Answers
Answer:
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)
Option A is correct.
Answer:
Given that,
Density of rod λ=λ
0
x
Length of rod = L
Now, center of mass
C.M=
M
1
0
∫
L
xdm
Now, total mass
M=
0
∫
L
dm
M=
0
∫
L
λ
0
dx
M=λ
0
0
∫
L
xdx
M=λ
0
[
2
x
2
]
0
L
M=
2
λ
0
L
2
....(I)
Now, center of mass
C.M=
M
1
0
∫
L
xdm
C.M=
M
1
0
∫
L
x(λdx)
C.M=
M
1
λ
0
0
∫
L
x(xdx)
C.M=
M
1
λ
0
[
3
x
3
]
0
L
C.M=
M
1
λ
0
3
L
3
Now, put the value of
M
λ
0
from equation (I)
C.M=
M
1
λ
0
3
L
3
C.M=
3
L
3
×
L
2
2
C.M=
3
2L
Now, the distance of center of mass of rod from the origin
Hence, The distance of center of mass of rod from the origin is
3
2L