Question

A solid sphere is in rolling motion. In rolling motion, a body possesses translational kinetic energy (${{K}_{t}}$) as well as rotational kinetic energy (${{K}_{r}}$) simultaneously. The ratio ${{K}_{t}}$ :( ${{K}_{t}}+{{K}_{r}}$ ) for the sphere is?
A- 10:7
B- 7:10
C- 2:5
D- 5:7

Answer 1

when the body is in rolling motion on an uneven surface, it has both rotation and translation, there are some cases when body rolls without slipping. The role played by mass in translation is played by moment of inertia in rotation.

Complete step by step answer: Linear kinetic energy of a body of mass, m and moving with linear velocity v, is given by ${{K}_{t}}=\dfrac{m{{v}^{2}}}{2}$
Rotational kinetic energy of a body having moment of inertia I, and angular velocity $\omega$is given by ${{K}_{r}}=\dfrac{I{{\omega }^{2}}}{2}$
Now a body possessing both translational kinetic energy and rotational kinetic energy is given by, ${{K}_{t}}+{{K}_{r}}=\dfrac{m{{v}^{2}}}{2}+\dfrac{I{{\omega }^{2}}}{2}$
Now to find the desired ratio,
${{K}_{t}}$ :( ${{K}_{t}}+{{K}_{r}}$)= $\dfrac{{{K}_{t}}}{{{K}_{t}}+{{K}_{r}}}=\dfrac{\dfrac{m{{v}^{2}}}{2}}{\dfrac{m{{v}^{2}}}{2}+\dfrac{I{{\omega }^{2}}}{2}}$--(1)
Moment of inertia of solid sphere is $\dfrac{2m{{r}^{2}}}{5}$ and angular velocity can be written as $v=r\omega$, putting the value in eq (1) we get,
$\dfrac{{{K}_{t}}}{{{K}_{t}}+{{K}_{r}}}=\dfrac{\dfrac{m{{v}^{2}}}{2}}{\dfrac{m{{v}^{2}}}{2}+\dfrac{m{{v}^{2}}}{5}}=\dfrac{\dfrac{1}{2}}{\dfrac{7}{10}}=\dfrac{5}{7}$
So, the correct option is (D)

Additional information- The moment of inertia of a uniform square plate of mass m and edge an about one of its diagonals comes out to be $\dfrac{m{{a}^{2}}}{12}$

Note: Rotational energy is on account of the motion of the body. Always while finding rotational kinetic energy we use moment of inertia, instead of mass. The moment of inertia plays the same role in rotation as it is played by mass in translation. Also, since we have taken the ratio, the result is dimensionless.