- #1
kudoushinichi88
- 125
- 2
A rod has mass M and length L. Calculate the moment of inertia of the rod about an axis which is passing through its center of mass and forming an angle \theta to the rod.
I drew a diagram on an xy-plane where the rod is on the x-axis and the center of the rod is at the origin. Chopping the rod up into small portions of dm, they have a distance of x \sin \theta from the axis of rotation. Therefore,
<br /> I=\int r^2 dm<br />
Assuming the rod is uniform, \frac{dm}{M}=\frac{dx}{L}\Rightarrow dm=\frac{M}{L}dx
Therefore,
<br /> I=\int x^2 \sin^2\theta dm
=\frac{M}{L}\sin^2\theta\int_{\frac{-L}{2}}^{\frac{L}{2}}x^2 dx
=\frac{1}{12}ML^2\sin^2\theta
Is this correct? I just need someone to check my work because I have no solutions to refer to for this question...
I drew a diagram on an xy-plane where the rod is on the x-axis and the center of the rod is at the origin. Chopping the rod up into small portions of dm, they have a distance of x \sin \theta from the axis of rotation. Therefore,
<br /> I=\int r^2 dm<br />
Assuming the rod is uniform, \frac{dm}{M}=\frac{dx}{L}\Rightarrow dm=\frac{M}{L}dx
Therefore,
<br /> I=\int x^2 \sin^2\theta dm
=\frac{M}{L}\sin^2\theta\int_{\frac{-L}{2}}^{\frac{L}{2}}x^2 dx
=\frac{1}{12}ML^2\sin^2\theta
Is this correct? I just need someone to check my work because I have no solutions to refer to for this question...