A solid sphere of mass M, radius R and having moment of inertia about as axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plance remains I. Then, radius of the disc will be.
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Answer:
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Length = \frac{L}{4}
4
L
Let the axis at length L/4 be AB.
We know that the moment of inertia of uniform rod with an axis passing through its middle is \frac{ {ML}^{2}}{12}.
12
ML
2
.
So, I_{AB} = I_{CD} + {ML}^{2}I
AB
=I
CD
+ML
2
I_{AB}= \frac{ {ML}^{2}}{12} +\frac{ {ML}^{2}}{16}I
AB
=
12
ML
2
+
16
ML
2
I_{AB}= \frac { {4ML}^{2} + {3ML}^{2}}{48}I
AB
=
48
4ML
2
+3ML
2
\boxed {I_{AB}= \frac {{7ML}^{2}}{48}}
I
AB
=
48
7ML
2
Explanation:
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