(A) Given: Mass of the wheel $M = 10 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Force $F = 20 \ N$,Displacement $d = 1 \ m$.Since the wheel is supported by

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(A) Given: Mass of the wheel $M = 10 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Force $F = 20 \ N$,Displacement $d = 1 \ m$.Since the wheel is supported by spokes of negligible mass,its moment of inertia is $I = MR^2 = 10 \times (0.1)^2 = 0.1 \ kg \ m^2$.The work done by the force $F$ is $W = F \times d = 20 \times 1 = 20 \ J$.According to the work-energy theorem,the work done is equal to the change in rotational kinetic energy: $W = \Delta KE = \frac{1}{2} I \omega^2$.Substituting the values: $20 = \frac{1}{2} \times 0.1 \times \omega^2$.$20 = 0.05 \times \omega^2$.$\omega^2 = \frac{20}{0.05} = 400$.$\omega = \sqrt{400} = 20 \ rad/s$.

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

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$A$ cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of the wheel is $10 \ kg$ and the radius is $10 \ cm$. It can rotate freely without any friction. Initially,the wheel is at rest. If a steady pull of $20 \ N$ is applied on the cord,the angular velocity (in $rad/s$) of the wheel,after the cord is unwound by $1 \ m$,would be:

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A
$20$
B
$30$
C
$10$
D
$0$

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Class 11

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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 of mass $1 \ kg$ and length $1 \ m$ is hinged at one end to the ground. It originally stands vertically and is allowed to fall to the ground. If the rod hits the ground with angular speed $\omega$, then the correct statement is (Assume $g = 10 \ m \ s^{-2}$):

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