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Work, Energy And Power For Rotating Body MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Work, Energy and Power for Rotating Body is considered one of the most asked concept.

  • 11 Questions around this concept.

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QuestionMEDIUM

Two rotating bodies A and B of masses m and 2m with moments of inertia IA and I_B \left( {I_B > I_A } \right) have equal kinetic energy of rotation. If LA and LB are their angular momenta respectively, then

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A body of radius \mathrm{R} and mass \mathrm{M} is initally rolling on a level surface with speed \mathrm{\omega}. It then rolls up an inclined to a maximum height \mathrm{h}. If \mathrm{h=\frac{u^2}{g}}. The geometrical shape of the body is.
 

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A solid sphere is rolling on a surface its speed is \mathrm{(V)} and mass \mathrm{(M)}. The its kinetic energy is
 

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A dog of mass 10kg runs a speed of 5m/sec at surface of a circular drum of radius 1m. The drum can rotate about its own axis AB freely as shown in the figure. The motion of the dog causes the rotation of drum. The speed of rotation is such that relative position of the dog is unaltered. The kinetic energy of rotation of dog is

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An object of mass '2kg' is rotating with an angular velocity of '1rad/s' about an axis that is '1m' away from the centre of mass. If the radius of gyration of this object is '2m'. Then rotational kinetic energy of the body is given by:

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Concepts Covered - 1

Work, Energy and Power for Rotating Body

  1. Work-

For translation motion   W=\int Fds

           So for rotational motion  W=\int \tau d\theta

  1. Rotational kinetic energy-

The energy of a body has by virtue of its rotational motion is called its rotational kinetic energy.

 

    

 

Rotational kinetic energy

Translatory kinetic energy

1

K_R=\frac{1}{2}I\omega ^2

  

K_T=\frac{1}{2}mV ^2

 

2

K_R=\frac{1}{2}I\omega

  

K_T=\frac{1}{2}PV

 

3

K_R= \frac{L^2}{2I}

 

K_T= \frac{P^2}{2m}

 

 

  1. Power =Rate of change of kinetic energy  

 

For translation motion  P=\vec{F}.\vec{V}

So for rotational motion  

 P=\frac{d(K_R)}{dt}=\frac{d(\frac{1}{2}I\omega ^2)}{dt}=I\omega \frac{d\omega }{dt}=I\alpha \omega =\tau .\omega

             Or   P= \vec{\tau }.\vec{\omega }

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Work, Energy and Power for Rotating Body

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