Triangular sheet of mass M and degree 45. find moment of Inertia
Answers
Answer:
Solution
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Correct option is A)
Assuming axis of rotation along Y-axis
Now consider an elemental section of mass dm, length=2y and of width dx as shown in the figure.
Moment of inertia of dI of the elemental section = dm×x
2
------(A)
As 2-D object mass of the elemental section dm=Area×density=2y×dx×ρ-------(B)
From figure y=
2
a
also DE=EB=
2
a
−x from similar triangle ΔAOB and ΔDEB
Therefore mass of elemental section dm=2(
2
a
−x)×dx×ρ
Putting the above value in A we get,
dI=2(
2
a
−x)×dx×ρ×x
2
Now integrating the above equation from 0 to
2
a
,we get
∫dI=∫
0
2
a
2ρ(
2
a
−x)x
2
dx
=
24
ρa
4
-----(C)
Now moment of inertia of full plate
= 2I =
12
ρa
4
= M
12
a
2
When the axis of rotation is shifted to corner as shown in fig 3, then moment of inertia
=
12
Ma
2
+
2
Ma
2
=
12
7Ma
2