Question

A disc rolls without slipping with constant velocity. What fraction of total kinetic energy of the disc is in the form of rotational kinetic energy?

Answer 1

Answer: 2/7

Explanation:

Answer 2

Answer:

fraction of total kinetic energy of the disc is in the form of rotational kinetic energy = $\dfrac{1}{3}$

Explanation:

Since, the disc is rolling without slipping,

hence, v-ω.R = 0

        => ω = $\dfrac{v}{R}$

where,   ω = angular velocity of disc

              v = Translational velocity of disc

Now, the total kinetic energy of disc is given by

                   K=translational kinetic energy + rotational kinetic energy

                      =$\dfrac{1}{2}.m.v^{2} +\dfrac{1}{2}.I.\omega^{2}$

      where, m = mass of disc

                   I = moment of inertia of disc

as moment of inertia of disc, I =$\dfrac{m.R^2}{2}$

Hence, the total kinetic energy, K =$\dfrac{1}{2}.m.v^{2} +\dfrac{1}{2}.I.\omega^{2}$

                                              =$\dfrac{1}{2}.m.v^{2} +\dfrac{1}{2}.\dfrac{m.R^{2}}{2}.\dfrac{v^{2}}{R^{2}}\\\\=\dfrac{1}{2}.m.v^{2}+\dfrac{1}{4}.m.v^{2}\\\\=\dfrac{3}{4}.m.v^{2}$

Since, the rotational kinetic energy is $\dfrac{1}{4}.m.v^{2}$ and total kinetic energy is $\dfrac{3}{4}.m.v^{2}$. So, the ratio of rotational kinetic energy to total kinetic energy will be $\dfrac{1}{3}$