Question 601: Centre of Mass and Rotational Motion

Question

A thin uniform rod has mass $$M$$ and length $$L$$ The moment of inertia about an axis perpendicular to it and passing through the point at a distance $$\frac{L}{3}$$ from one of its ends, will be

Accepted Answer

Consider the figure shown below
A point at a distance $$\frac{L}{3}$$ from one end is at a distance of $$d=\frac{L}{2}-\frac{L}{3}=\frac{L}{6}$$ from the centre of rod.
The moment of inertia of rod about an axis passing through centre and perpendicular to it is
$$I_c=\frac{M L^2}{12}$$
Using parallel axis theorem,
$$\begin{aligned}
I_d & =I_c+M d^2 \
& =\frac{M L^2}{12}+M\left(\frac{L}{6}ight)^2=\frac{M L^2}{9}
\end{aligned}$$

Answer

$$\frac{M L^2}{12}$$

Answer

$$\frac{7}{8} M L^2$$

Answer

$$\frac{M L^2}{9}$$

Answer

$$\frac{M L^2}{3}$$

Centre of Mass and Rotational Motion

A thin uniform rod has mass $M and length L The moment of inertia about an axis perpendicular to it and passing through the point at a distance \frac{L}{3}$ from one of its ends, will be

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