(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.

Explanation

(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 with angular 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)`