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(i) A uniform sphere of mass 200 g rotates on a horizontal surface without shipping. If centre of the sphere moves with a velocity 2.00 cm/s then its kinetic energy is? (ii) Derive the expression for kinetic energy in rotating object and also derive the relation between rotational kinetic energy and angular momentum. |
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Answer» Solution :(i) As the SPHERE rolls without slipping on the plane surface, it.s angular speed about the centre is `OMEGA = (v_(CM) )/(r )` Kinetic ENERGY `k = 1/2 I_(CM) omega^2 + 1/2 M_(CM)^2` `= 1/2 (2/5 Mr^2) omega^2+ 1/2 Mv_(CM)^2` ` = 1/5 M_(CM)^2 + 1/2 Mv_(CM)^2` `= 7/10 Mv_(CM)^2` ` 7/10 (0.200) (0.02)^2` `k = 5.6 xx 10^(-5) J` (ii)  Let us consider a rigid body rotating with angular velocity `omega ` about an axis as shown in figure. 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. Let us CHOOSE a particle of MASS `m_i`situated at 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` Writing the expression with the angular velocity ` KE = 1/2 m_i (r_i omega)^2= 1/2 (m_i v_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 summation as, `KE = 1/2 (sum m_i r_i^2) omega^2` where , the term `(sum m_i r_i^2) omega^2` is the moment of inertia 1 of the whole body . `sum m_i r_i^2` . Hence, the expression for KE of the rigid body in rotational motion is, `1/2 I omega^2` Relation between rotational kinetic energy and angular momentum Let a rigid body of moment of inertia I rotate withangular velocity `omega` The angular momentum of a rigid body is, `L = I omega ` By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as ` KE = 1/2 (I^2 omega^2)/(I) = 1/2 ( (I omega)^2)/(I)` `KE = (L^2)/(2I)` |
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