Question 516: Centre of Mass and Rotational Motion

Question

A thin uniform metal rod of mass ' $M$ ' and length ' $L$ ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' $\omega$ '. Its centre of mass rises to a maximum height of ( $\mathrm{g}=$ Acceleration due to gravity)

Accepted Answer

By conservation of energy, $\frac{1}{2} \mathrm{I} \omega^2=\mathrm{Mgh}$
$$	herefore \quad \mathrm{h}=\frac{\mathrm{I} \omega^2}{2 \mathrm{Mg}}=\mathrm{h}\quad	ext{.... (i)}$$
The M.I of a uniform rod about an axis passing through its centre is $\frac{\mathrm{ML}^2}{12}$.
As the axis passing through the end, using parallel axis theorem,
$$\mathrm{I}=\frac{\mathrm{ML}^2}{12}+\mathrm{M}\left(\frac{\mathrm{~L}}{2}ight)^2=\frac{\mathrm{ML}^2}{3}\quad	ext{.... (ii)}$$
Putting (ii) into (i),
$$h=\frac{L^2 \omega^2}{6 g}$$

Answer

$\frac{\mathrm{L}^2 \omega^2}{3 \mathrm{~g}}$

Answer

$\frac{\mathrm{L}^2 \omega^2}{2 \mathrm{~g}}$

Answer

$\frac{\mathrm{L}^2 \omega^2}{6 \mathrm{~g}}$

Answer

$\frac{\mathrm{L}^2 \omega^2}{4 \mathrm{~g}}$

Centre of Mass and Rotational Motion

A thin uniform metal rod of mass ' M ' and length ' L ' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' \omega '. Its centre of mass rises to a maximum height of ( \mathrm{g}= Acceleration due to gravity)

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