Question

Derive an expression for kinetic energy of a rolling body.​

Answer 1

Derive an expression for kinetic energy, when a rigid body is rolling on a horizontal surface without slipping. Hence find the kinetic energy of a solid sphere.

-Total kinetic energy = translational kinetic energy + rotational kinetic energy.

-Total kinetic energy = 12MV2+12Iω2.

-ω=VR.

-Total kinetic energy = 12MV2+12I(VR)2.

❣️Hope it helps u mate❣️

Answer 2

Consider a body of mass m rolling with a velocity v$\sf _{cm}$ The total kinetic energy of a rolling body is the sum of the kinetic energy of the body due to the motion of the centre of mass mv²/2 and kinetic energy of rotational motion about the centre of mass of the system of particles (K¹)

Hence,

$\sf K = K^1 + \dfrac{1}{2}mv_{cm}^2 \: \: \: \: \: ...(1)$

Here the kinetic energy of the centre of mass that is the K.E of translation of the rolling body is $\sf \dfrac{mv^2_{cm}}{2}$

where, m is the mass of the body and v$\sf _{cm}$ is the velocity of the centre of mass.

since the motion of the rolling body about the centre of mass is rotation, K¹ is the kinetic energy of rotation of the body that is,

$\sf K^1 = \dfrac{I\omega ^2}{2} \: \: \: \: ...(2)$

where, I is the moment of inertia about the appropriate axis, which is the symmetry axis of the rolling body.

Now put eq (2) in eq(1)

Therefore the kinetic energy of a rolling body

$\sf K = \dfrac{1}{2} I\omega ^2+ \dfrac{1}{2}mv_{cm}^2 \: \: \: \: \: ...(3)$

Now substituting I = mK² where K is the corresponding radius of gyration of the body and v$\sf _{cm}$ = ωR in eq (3)

$\sf K = \dfrac{1}{2} I\omega ^2+ \dfrac{1}{2}mv_{cm}^2$

$\sf K = \dfrac{m {K}^{2} {v}^{2}_{cm} }{2} + \dfrac{1}{2}mv_{cm}^2$

$\underline{ \boxed{\sf K = \dfrac{1}{2}mv_{cm}^2 \bigg[ 1 + \dfrac{k^2}{R^2} \bigg]}}$