(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

Medium

$A$ 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

$A$ uniform disc of radius $R$ and mass $M$ can rotate on a smooth axis passing through its centre and perpendicular to its plane. $A$ force $F$ is applied on its rim as shown in the figure. What is the tangential acceleration of the rim?

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$A$ flywheel having mass $3 \ kg$ and radius $5 \ m$ is free to rotate about a horizontal axis. $A$ string having negligible mass is wound around the wheel and the loose end of the string is connected to a $3 \ kg$ mass. The mass is kept at rest initially and released. The kinetic energy of the flywheel when the mass descends by $3 \ m$ is . . . . . . $J$. $(g = 10 \ m/s^{2})$

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How can one verify that an object is undergoing rotation about a fixed axis?

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$A$ uniform solid cylinder of mass $M$ and radius $R$ rotates about a frictionless horizontal axle. Two similar masses $m$ are suspended with the help of two ropes wrapped around the cylinder. If the masses fall down through a distance $h$,the angular velocity of the cylinder will be

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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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$A$ uniform disc of radius $R$ and mass $M$ can rotate on a smooth axis passing through its centre and perpendicular to its plane. $A$ force $F$ is applied on its rim as shown in the figure. What is the tangential acceleration of the rim?

How can one verify that an object is undergoing rotation about a fixed axis?

$A$ uniform solid cylinder of mass $M$ and radius $R$ rotates about a frictionless horizontal axle. Two similar masses $m$ are suspended with the help of two ropes wrapped around the cylinder. If the masses fall down through a distance $h$,the angular velocity of the cylinder will be

$A$ 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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