Question

Show that the rotational kinetic energy of a ball rolling without slipping over a horizontal plane is 2/7th of its total kinetic energ

Answer 1

Rolling KE =  KEr = 1/2 I (V / R)^2   where (V / R) = angular velocity

KEr = 1/2 * (2/5 M R^2) * (V / R)^2 = 1/5 M V^2

Translational KE = KEt = 1/2 M V^2

Total KE = KEt + KEr = (1/2 + 1/5) * M V^2 = 7/10 M V^2

1/5 M V^2 / (7/10 M V^2) = (1/5 ) / (7/10) = 10 / 35 = 2 / 7

Answer 2

Explanation:

The Rotational Kinetic energy $(KE_{r} )$ of an object is the energy possessed by an object with respect to the angular velocity of the object. It is given,

$KE_{r} =\frac{1}{2}Iw^{2}$

Where, $I$ is the moment of inertia and ω is the angular velocity of the object.

The linear or Translational Kinetic Energy $(KE_{t})$ of the object is the energy possessed by the object by virtue of its velocity.

$KE_{t}=\frac{1}{2}mv^{2}$

The total kinetic energy for a body rolling without slipping is the sum of Rotational as well as Translational Kinetic energy.

$KE=KE_{r} +KE_{t}$

$KE=\frac{1}{2} Iw^{2} +\frac{1}{2}mv^{2}$

Now,

According to the question, we have the object as ball,sphere, and, we know the moment of Inertia of a sphere of mass $m$ and radius $r$ is given as

$I=\frac{2}{5}mR^{2}$

and from the relation $v=R$ω, we get

ω $=\frac{v}{R}$

Hence, we get the Rotational Kinetic energy of the ball as

$KE_{r} =\frac{1}{2}(\frac{2}{5}mR^{2} )(\frac{v}{R} )^{2}$

$KE_{r}=\frac{1}{2} (\frac{2}{5}mv^{2} )$          $(i)$ $or$

$KE_{r}=\frac{1}{5}mv^{2}$              $(ii)$

Therefore,

The Total Kinetic Energy of the ball from the equation above will be

$KE=\frac{1}{2}(mv^{2} +\frac{2}{5} mv^{2} )$

$KE=\frac{1}{2}mv^{2} (1+\frac{2}{5} )$

$KE=\frac{1}{2}mv^{2} \frac{7}{5}$

$KE=\frac{7}{10}mv^{2}$

If we multiply both sides by $\frac{2}{7}$, we get

$\frac{2}{7}KE=\frac{2}{7}(\frac{7}{10}mv^{2} )$

$\frac{2}{7}KE=\frac{1}{5}mv^{2}$

We know,   from $(ii)$,  $\frac{1}{5}mv^{2}=KE_{r}$

$KE_{r}=\frac{2}{7}KE$

Hence proved.