(D) The moment of inertia of a uniform rod about an axis passing through its centre is $I = \frac{ML^2}{12}$.At the lowest point,the rod has maximum k

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(D) The moment of inertia of a uniform rod about an axis passing through its centre is $I = \frac{ML^2}{12}$.At the lowest point,the rod has maximum kinetic energy,which is given by $K.E. = \frac{1}{2} I \omega^2$.Substituting the value of $I$,we get $K.E. = \frac{1}{2} \times \frac{ML^2}{12} \times \omega^2 = \frac{ML^2 \omega^2}{24}$.As the rod swings,this kinetic energy is converted into gravitational potential energy at the maximum height $h$.By the law of conservation of energy,$P.E. = K.E.$$Mgh = \frac{ML^2 \omega^2}{24}$.Solving for $h$,we get $h = \frac{L^2 \omega^2}{24g}$.

Class 11 Physics System of Particles and Rotational Motion Mix Example - System of Particles and Rotational Motion

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$A$ 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 [where $g$ is gravitational acceleration]:

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A
$\frac{\omega^2 L^2}{12 g^2}$
B
$\frac{\omega^2 L^2 g}{6}$
C
$\frac{\omega^2 g}{12 L^2}$
D
$\frac{\omega^2 L^2}{24 g}$

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Mix Example - System of Particles and Rotational Motion

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MediumAIPMT 2015

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$A$ 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 [where $g$ is gravitational acceleration]:

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$A$ uniform rod $AB$ of mass $2 \ kg$ and length $30 \ cm$ is at rest on a smooth horizontal surface. An impulse of force $0.2 \ Ns$ is applied to end $B$. The time taken by the rod to turn through a right angle will be $\frac{\pi}{X} \ s$,where $X = \text{ . . . . . . }$.

$A$ thin uniform straight rod of mass $2 \, kg$ and length $1 \, m$ is free to rotate about its upper end when at rest. It receives an impulsive blow of $10 \, Ns$ at its lowest point,normal to its length as shown in the figure. The kinetic energy of the rod just after impact is ........ $J$.

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