Question

A uniform sphere of mass 500g rolls without slipping on a plane surface so that its center moves at a speed of $0.02{m}/{s}$. The total kinetic energy of the rolling sphere would be? (in J)
A. $1.4 \times {10}^{-4}J$
A. $0.75 \times {10}^{-3}J$
A. $5.75 \times {10}^{-3}J$
A. $4.9 \times {10}^{-5}J$

Answer 1

Hint: Total kinetic energy is the sum of rotational kinetic energy and translational kinetic energy. So, first find the translational kinetic energy which gives a relationship between energy, mass and velocity of an object. Then, find the rotational kinetic energy which gives the relationship between kinetic energy, mass, velocity, radius of gyration and the radius of an object. Add both these kinetic energies. This will give the total kinetic energy of the rolling sphere.

Formula used:
${E}_{trans}= \dfrac {1}{2} m{v}^{2}$
${E}_{rot}= \dfrac {1}{2} m{v}^{2} \times \dfrac {{K}^{2}}{{R}^{2}}$

Complete step by step answer:
Given: Mass (m)= 500g= 0.5 kg
           Speed (v)= $0.02{m}/{s}$
Kinetic Energy of a sphere= Translational Kinetic Energy + Rotational Kinetic Energy
${E}_{total}= {E}_{trans}+ {E}_{rot}$ …(1)
Translational Kinetic Energy is given by,
${E}_{trans}= \dfrac {1}{2} m{v}^{2}$ …(2)
Where, m is the mass of the sphere
            v is the velocity
Substituting values in above equation we get,
${E}_{trans}= \dfrac {1}{2} \times 0.5 \times {0.02}^{2}$
$\Rightarrow {E}_{trans}= 1 \times {10}^{-4} J$ …(3)
Rotational Kinetic Energy is given by,
${E}_{rot}= \dfrac {1}{2} m{v}^{2} \times \dfrac {{K}^{2}}{{R}^{2}}$ …(4)
Where, K is the radius of gyration
             r is the radius of the sphere
We know, for a solid sphere, ${K}^{2}= \dfrac {2}{5}{R}^{2}$
Substituting values in the equation. (4) we get,
${E}_{rot}= \dfrac {1}{2} \times 0.5 \times {0.02}^{2}\times \dfrac {2}{5}$
$\Rightarrow {E}_{rot}= 4 \times {10}^{-5}J$
$\therefore {E}_{rot}= 0.4 \times {10}^{-4}J$ …(5)
Substituting equation. (3) and (5) in equation. (1) we get,
${E}_{tot}= 1 \times {10}^{-4} + 0.4 \times {10}^{-4} $
$\therefore {E}_{tot}= 1.4 \times {10}^{-4}J$
Hence, the total kinetic energy of the rolling sphere would be $1.4 \times {10}^{-4}J$.

So, the correct answer is “Option A”.

Note:
To solve these types of questions, students must know the relationship between radius of gyration (K) and radius (R) of cylinder, hollow cylinder, sphere and a hollow sphere. For a cylinder, the relationship between K and R is given by, ${K}^{2}= \dfrac {1}{2}{R}^{2}$ while for a hollow cylinder it is given by, ${K}^{2}= {R}^{2}$. For a sphere it is given by, ${K}^{2}= \dfrac {2}{5}{R}^{2}$ while for a hollow sphere it is given by, ${K}^{2}= \dfrac {2}{3}{R}^{2}$.