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Find the expression of KE of the rigid body in rotational motion. |
Answer» Solution : Let us consider a rigid body rotating with angualar velocity `OMEGA` about an axis as shown in the figure. Every particle of the bdoy will have the same ANGULAR velocity `omega` and different TANGENTIAL velocities v BASED on its positions from the axis of rotation. 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 `KE_(i)=1/2 m_(i)v_(i)^(2)` Since `v_(i)=r_(i)omega` `KE_(i)=1/2m_(i)(r_(i)omega)^(2)=1/2m_(i)(r_(i)^(2))omega^(2)` 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 `KE=1/2(sum m_(i)r_(i)^(2))omega^(2)` where the term `sum m_(i)r_(i)^(2)` is the MOMENT of inertia I of the whole body. `I=sum m_(i)r_(i)^(2)` In rotational motion kinetic energy is `KE=1/2 I omega^(2)` |
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