Portion AB of the wedge shown in the figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. If AB = BC, then ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches point C is

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answer is B.

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Detailed Solution

For a cylinder $\frac{K_{T}}{K_{R}} = 2$ in case of pure rolling.

Upto B:

$K_{T} = \frac{2}{3} mgh and K_{R} = \frac{1}{3} mgh$

After .B: rotational kinetic energy is constant; however, translational kinetic energy increases.

$At C : K_{T} = \frac{2}{3} mgh + mgh = \frac{5}{3}$

$While K_{R} = \frac{1}{3} mgh, \frac{K_{T}}{K_{R}} = 5$

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