**Kinetic Energy of a rolling body**

Consider a wheel of mass m and radius R rolling along a smooth surface in a straight line on a horizontal plane surface without slipping as shown in figure. When the body rolls, it rotates about the horizontal axis through the center of mass and undergoes displacement in the forward direction. So, the body possesses both horizontal and translational motion.

Let v be the velocity of the center of mass and T is the period of rotation of the body. During this time, It describes and angle of 2π radians about the axis. If the body covers a linear distance x in one revolution, the distance covered in one revolution is,

x = 2πR

The angular velocity of the body is

Ω = 2π/ T

And the velocity of translation of its center of mass is

v = 2πR/ T

Comparing these two equations, we get

v = ωR

.: Kinetic energy of rotation, E

_{rot}= ½ Iω^{2}And kinetic energy of translation, E

_{trans}= ½ mv^{2}The total K.E. of the rolling body is given by

E = E

_{rot}+ E_{trans}= ½ Iω

^{2}+ ½ mv^{2}……. (i)= ½ m K

^{2}ω^{2}+ ½ mv^{2}[Since, I = mK^{2}]Total K.E. = ½ m v

^{2}[(K^{2}/ R^{2}) + 1] …..... (ii) which is the total kinetic energy for rolling body
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