Ask Question
11 August, 17:52

Consider a uniform solid sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, translational or rotational?

+2
Answers (1)
  1. 11 August, 20:26
    0
    Translational kinetic energy is equal to one half the product of mass andthe square of the linear velocity. Rotational kinetic energy, on the other hand, is one half the product of the mass of the object in circular motion and the square of the angular velocity where angular velocity is equal linear velocity divided by the radius of the circular motion. To compare these energies, we do as follows:

    KE (t) = 0.5mv^2

    KE (r) = 0.5mw^2 = 0.5mv^2/r^2

    KE (r) = KE (t) / r^2

    From the relation, rotational kinetic energy is 1/r^2 times than the translational kinetic energy. Therefore, translational kinetic energy would be bigger than the rotational kinetic energy.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Consider a uniform solid sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, translational ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers