Question 528: Centre of Mass and Rotational Motion

Question

A thin uniform rod of length ' $L$ ' and mass ' $M$ ' is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is ' $\omega$ '. Its centre of mass rises to a maximum height of [ $\mathrm{g}=$ gravitational acceleration]

Accepted Answer

To determine the maximum height to which the center of mass of the swinging rod rises, we need to apply the principle of conservation of energy.
Initially, the entire energy of the system is kinetic because the rod is swinging with its maximum angular speed $\omega$ around its center. The moment of inertia $I$ of a thin uniform rod of length $L$ about an axis through its center is given by:
$$ I = \frac{1}{12} M L^2 $$
The initial kinetic energy ($K_i$) of the rod is thus:
$$ K_i = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{12} M L^2\right) \omega^2 = \frac{1}{24} M L^2 \omega^2 $$
At the highest point of the swing, all this kinetic energy is converted to potential energy ($U_f$), assuming no energy is lost:
$$ U_f = Mgh $$
where $h$ is the maximum height attained by the center of mass and $g$ is the gravitational acceleration. By conservation of energy:
$$ K_i = U_f $$
$$ \frac{1}{24} M L^2 \omega^2 = Mgh $$
Solving for $h$, we get:
$$ h = \frac{1}{24} \frac{L^2 \omega^2}{g} $$
Thus, the maximum height of the center of mass is:
Option D:
$$ \frac{\omega^2 L^2}{24 \mathrm{~g}} $$

Answer

$\frac{\omega^2 \mathrm{~L}^2}{12 \mathrm{~g}^2}$

Answer

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

Answer

$\frac{\omega^2 g}{12 \mathrm{~L}^2}$

Answer

$\frac{\omega^2 L^2}{24 \mathrm{~g}}$

Centre of Mass and Rotational Motion

A thin uniform rod of length ' L ' and mass ' M ' is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is ' \omega '. Its centre of mass rises to a maximum height of [ \mathrm{g}= gravitational acceleration]

Choose the correct answer: