Multiple choice

A solid cylinder is released from the top of an inclined plane of height $h$. What will be the speed of its centre of mass on reaching the bottom?

  1. $\sqrt{2 \,gh}$
  2. $\sqrt{4 \,gh}$
  3. $\left[\dfrac{4}{3}gh \right]^{1/2}$
  4. $\left[\dfrac{2}{3}gh \right]^{1/2}$
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C Correct answer
Explanation

Using conservation of energy: mgh = (1/2)mv^2 + (1/2)Iw^2. For a solid cylinder, I = (1/2)mR^2 and w = v/R. So mgh = (1/2)mv^2 + (1/2)(1/2)mR^2(v^2/R^2) = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2. v^2 = (4/3)gh. v = sqrt(4gh/3).

AI explanation

By the principle of conservation of mechanical energy, the potential energy at the top of the inclined plane converts entirely into translational and rotational kinetic energy at the bottom. The formula equating these energies is mgh = (1/2)mv^2 + (1/2)Iw^2, where the moment of inertia for a solid cylinder is I = (1/2)mr^2 and the rolling condition is w = v/r. Substituting these values simplifies the equation to mgh = (3/4)mv^2, which yields v = [ (4/3)gh ]^(1/2).

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