Question 510: Centre of Mass and Rotational Motion

Question

A thin uniform rod of mass ' $m$ ' and length ' $L$ ' is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above pivot and then released. The angular acceleration of the rod when it makes an angle ' $	heta$ ' with the vertical is [consider negligible friction at the pivot] ( $\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

Torque at angle $	heta$,
$$	au=M g \sin 	heta \frac{L}{2}$$
Also, we know , $	au=\mathrm{I} \alpha$
$$\Rightarrow \quad I \alpha=m g \sin 	heta \frac{L}{2}$$
$\because \quad$ For a thin uniform rod, the moment of inertia about an axis perpendicular to its length and passing through its one end is $\frac{\mathrm{ML}^2}{3}$, the above equation becomes:
$$\begin{aligned}
& \Rightarrow \quad \frac{\mathrm{ML}^2}{3} \cdot \alpha=\mathrm{Mg} \sin 	heta \frac{\mathrm{~L}}{2} \
& \Rightarrow \quad \frac{\mathrm{~L} \alpha}{2}=\mathrm{g} \frac{\sin 	heta}{2} \
& \Rightarrow \quad \alpha=\frac{3 \mathrm{~g} \sin 	heta}{2 \mathrm{~L}}
\end{aligned}$$

Answer

$\frac{3 g \sin 	heta}{2 L}$

Answer

$\frac{3 g \cos 	heta}{2 L}$

Answer

$\frac{2 g \sin 	heta}{3 L}$

Answer

$\frac{2 g \cos 	heta}{3 L}$

Centre of Mass and Rotational Motion

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