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Derive the expression of Kinetic energy in rotation. |
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Answer» Solution :(i) Let us consider a rigid body rotating with angular VELOCITY `omega` about an axis as shown in Figure. (ii) Every particle of the body will have the same angular velocity `omega` and different tangential velocities v based on its positions from the axis of rotation.  (iii) Let us choose a particle of mass `m_(i)` situated at distance `r_(i)` from the axis of rotation. If 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_(t)=1/2m_(i)v_(i)^(2)` (iv) Writing the expression with the angular velocity, `KE_(t)=1/2m_(i)(r_(i)omega)^(2)=1/2 (m_(i)r_(i)^(2))omega^(2)` (v) For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as, `KE=1/2(summ_(i)r_(i)^(2))` (vi) where, the term `summ_(i)r_(i)^(2)` is the moment of inertia I of the whole body. `I=summ_(i)r_(i)^(2)` (vii) Hence, the expression for KE of the rigid body in ROTATIONAL motion is, `KE=1/2Iomega^(2)` This is analogous to the expression for kinetic energy in TRANSLATIONAL motion. `KE=1/2Mv^(2)` (VIII) Relational between rotational kinetic energy and angular momentum Let a rigid body of moment of inertia I rotate with angular velocity `omega`. The angular momentum of rigid body is, `L=Iomega` (ix) The rotational kinetic energy of the rigid body is, `KE=1/2Iomega^(2)` (x) By multiplying the numerator and DENOMINATOR of the above equation with I, we get a relation between Land KE as, `KE=1/2(I^(2)omega^(2))/I=1/2((Iomega^(2)))/I` `KE=L^(2)/(2I)` |
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