(C) Using the principle of conservation of mechanical energy,the loss in potential energy of the block is equal to the gain in kinetic energy of the s

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(C) Using the principle of conservation of mechanical energy,the loss in potential energy of the block is equal to the gain in kinetic energy of the system (block + disc).Loss in potential energy of the block = $m g h$Gain in kinetic energy of the block = $\frac{1}{2} m v^2$Gain in rotational kinetic energy of the disc = $\frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{2} m r^2) (\frac{v}{r})^2 = \frac{1}{4} m v^2$Equating the energies: $m g h = \frac{1}{2} m v^2 + \frac{1}{4} m v^2$$m g h = \frac{3}{4} m v^2$$v^2 = \frac{4 g h}{3}$$v = \sqrt{\frac{4 g h}{3}} = 2 \sqrt{\frac{g h}{3}}$

Class 11 Physics System of Particles and Rotational Motion Rotation Motion Basic, Motion of Connected Mass

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$A$ disc of mass $m$ and radius $r$ is free to rotate about its centre as shown in the figure. $A$ string is wrapped over its rim and a block of mass $m$ is attached to the free end of the string. The system is released from rest. The speed of the block as it descends through a height $h$ is:

A
$\sqrt{2 g h}$
B
$\sqrt{\frac{2}{3} g h}$
C
$2 \sqrt{\frac{g h}{3}}$
D
$\frac{1}{2} \sqrt{3 g h}$

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System of Particles and Rotational Motion

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Rotation Motion Basic, Motion of Connected Mass

$A$ hemisphere of radius $R$ and mass $4m$ is free to slide with its base on a smooth horizontal table. $A$ particle of mass $m$ is placed on the top of the hemisphere. The angular velocity of the particle relative to the hemisphere at an angular displacement $\theta$ when the velocity of the hemisphere is $v$ is:

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For a rigid body in rotational motion,if a particle at a distance of $5 \, cm$ from the axis of rotation has an angular velocity of $10 \, rad/s$,what is the linear velocity of a particle at a distance of $10 \, cm$ from the axis of rotation (in $, cm/s$)?

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$A$ uniform cylinder of mass $m$ can rotate freely about its own axis,which is horizontal. $A$ particle of mass $m_0$ hangs from the end of a light string wound around the cylinder,which does not slip over it. When the system is allowed to move,the acceleration of the descending mass will be

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Obtain the relation between linear velocity and angular velocity of rotational motion of a body.

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$A$ uniform rod of mass $m$ and length $l$ is suspended by means of two identical inextensible light strings as shown in the figure. The tension in one string immediately after the other string is cut is . . . . . . . ($g$ is the acceleration due to gravity)

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$A$ disc of mass $m$ and radius $r$ is free to rotate about its centre as shown in the figure. $A$ string is wrapped over its rim and a block of mass $m$ is attached to the free end of the string. The system is released from rest. The speed of the block as it descends through a height $h$ is:

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For a rigid body in rotational motion,if a particle at a distance of $5 \, cm$ from the axis of rotation has an angular velocity of $10 \, rad/s$,what is the linear velocity of a particle at a distance of $10 \, cm$ from the axis of rotation (in $, cm/s$)?

$A$ uniform cylinder of mass $m$ can rotate freely about its own axis,which is horizontal. $A$ particle of mass $m_0$ hangs from the end of a light string wound around the cylinder,which does not slip over it. When the system is allowed to move,the acceleration of the descending mass will be

Obtain the relation between linear velocity and angular velocity of rotational motion of a body.

$A$ uniform rod of mass $m$ and length $l$ is suspended by means of two identical inextensible light strings as shown in the figure. The tension in one string immediately after the other string is cut is . . . . . . . ($g$ is the acceleration due to gravity)

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