| Moment of inertia of a rod of mass M and length L that rotated about its centre along an axis, which makes angle of \[30{}^\circ \] with the length of rod will be |
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A) \[\frac{M{{L}^{2}}}{12}\]
B) \[\frac{M{{L}^{2}}}{3}\]
C) \[\frac{M{{L}^{2}}}{48}\]
D) \[\frac{M{{L}^{2}}}{16}\]
Correct Answer: C
Solution :
| Consider an element of length \[dx\] at a distance x from centre. |
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| \[dM=\frac{M}{L}dx\] |
| \[r=x\sin \theta =\frac{x}{2}\] |
| \[dI=dM\cdot {{r}^{2}}=\frac{M}{L}\frac{{{x}^{2}}}{4}\cdot dx\] |
| \[\therefore \]\[I=\int_{-L/2}^{L/2}{\frac{M}{L}\cdot \frac{{{x}^{2}}}{4}dx=\frac{1}{48}M{{L}^{2}}}\] |
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