Multiple choice

A solid cylinder of mass $0.1\, kg$ and radius $0.025\, metre$ is rolling on a horizontal smooth table with uniform velocity of $0.1\,ms^{-1}$. Its total energy will be :

  1. $7.5 \times 10^{-2}\, Joule$
  2. $7.5 \times 10^{-3}\, Joule$
  3. $7.5 \times 10^{-4}\, Joule$
  4. $0.07 \times 10^{-4}\, Joule$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total energy of a rolling cylinder = Translational KE + Rotational KE = 1/2 * m * v^2 + 1/2 * I * w^2. For a solid cylinder, I = 1/2 * m * r^2 and w = v/r. So, Total Energy = 1/2 * m * v^2 + 1/2 * (1/2 * m * r^2) * (v/r)^2 = 1/2 * m * v^2 + 1/4 * m * v^2 = 3/4 * m * v^2. Energy = 0.75 * 0.1 * (0.1)^2 = 0.75 * 0.1 * 0.01 = 0.00075 = 7.5 * 10^-4 Joule.

AI explanation

The total energy of a rolling solid cylinder is the sum of its translational and rotational kinetic energy, given by the formula E = (1/2)mv^2 + (1/2)Iw^2. Substituting the moment of inertia for a solid cylinder I = (1/2)mr^2 and angular velocity w = v/r gives E = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2. Using the provided values m = 0.1 kg and v = 0.1 m/s, we calculate E = (3/4) * 0.1 * (0.1)^2 = 0.75 * 0.001 = 7.5 * 10^-4 Joule.

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