(D) The rod rotates about an axis passing through its end. The moment of inertia of the rod about this axis is $I = \frac{1}{3}ML^2$.The rotational ki

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(D) The rod rotates about an axis passing through its end. The moment of inertia of the rod about this axis is $I = \frac{1}{3}ML^2$.The rotational kinetic energy of the rod at its lowest point (maximum angular speed) is $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{3}ML^2)\omega^2 = \frac{1}{6}ML^2\omega^2$.At the maximum height $h$,the centre of mass of the rod rises by $h$. The potential energy gained by the rod is $U = Mgh$.By the law of conservation of energy,the rotational kinetic energy at the bottom equals the potential energy at the maximum height:$\frac{1}{6}ML^2\omega^2 = Mgh$.Solving for $h$:$h = \frac{L^2\omega^2}{6g}$.

Class 11 Physics System of Particles and Rotational Motion Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

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

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System of Particles and Rotational Motion

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Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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$A$ thin uniform rod $AB$ of mass $m$ and length $l$ is hinged at one end $A$ to the ground level. Initially, the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its $B$ end strikes the ground is $(g = \text{acceleration due to gravity})$

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$A$ uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta$,its angular velocity $\omega$ is given as

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The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

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$A$ uniform rod of mass $M$ and length $L$ is pivoted at one end and is free to rotate in a vertical plane. The rod is released from rest in a horizontal position. What is the angular velocity of the rod when it reaches the vertical position?

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Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$

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Column-$I$	Column-$II$
$(1)$ $W = F \Delta x$	$(a)$ $P = \tau \omega$
$(2)$ $P = Fv$	$(b)$ $W = \tau \Delta \theta$
	$(c)$ $L = I \omega$

$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

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$A$ uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta$,its angular velocity $\omega$ is given as

The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

$A$ uniform rod of mass $M$ and length $L$ is pivoted at one end and is free to rotate in a vertical plane. The rod is released from rest in a horizontal position. What is the angular velocity of the rod when it reaches the vertical position?

Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$. Column-$I$ Column-$II$ $(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$ $(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$ $(c)$ $L = I \omega$

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Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$