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Find the expression of KE of the rigid body in rotational motion.

Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C05_E02_124_S01.png" width="80%"/> <br/> Let us consider a rigid body rotating with angualar velocity `<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>` about an axis as shown in the figure. Every particle of the bdoy will have the same <a href="https://interviewquestions.tuteehub.com/tag/angular-11524" style="font-weight:bold;" target="_blank" title="Click to know more about ANGULAR">ANGULAR</a> velocity `omega` and different <a href="https://interviewquestions.tuteehub.com/tag/tangential-1239000" style="font-weight:bold;" target="_blank" title="Click to know more about TANGENTIAL">TANGENTIAL</a> velocities v <a href="https://interviewquestions.tuteehub.com/tag/based-389387" style="font-weight:bold;" target="_blank" title="Click to know more about BASED">BASED</a> on its positions from the axis of rotation. <br/> Let us choose a particle of mass `m_(i)` situatedat distance `r_(i)` from the axis of rotation. It has a tangential velocity`v_(i)` given by the relation, `v_(i)=r_(i) omega`. The kinetic energy `KE_(i)` of the particle is <br/> `KE_(i)=1/2 m_(i)v_(i)^(2)` <br/> Since `v_(i)=r_(i)omega`<br/> `KE_(i)=1/2m_(i)(r_(i)omega)^(2)=1/2m_(i)(r_(i)^(2))omega^(2)` <br/> For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with simmation as <br/> `KE=1/2(sum m_(i)r_(i)^(2))omega^(2)` <br/> where the term `sum m_(i)r_(i)^(2)` is the <a href="https://interviewquestions.tuteehub.com/tag/moment-25786" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENT">MOMENT</a> of inertia I of the whole body. `I=sum m_(i)r_(i)^(2)` <br/> In rotational motion kinetic energy is <br/> `KE=1/2 I omega^(2)`</body></html>


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