A loop and a disc roll without slipping with same linear velocity v. The mass of the loop and the disc is same. If the total kinetic energy of the loop is 8J, find the kinetic energy of the disc (in J).

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answer is 6.

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Detailed Solution

Total kinetic energy of loop $\frac{1}{2} {Iω}^{2} + \frac{1}{2} {Mv}^{2}$

$= \frac{1}{2} ({MR}^{2} ω^{2}) + \frac{1}{2} {Mv}^{2} = {Mv}^{2} = 8 J (given)$

Total kinetic energy of $disc = \frac{1}{2} {Iω}^{2} + \frac{1}{2} {Mv}^{2}$

$= \frac{1}{2} (\frac{{MR}^{2}}{2} ω^{2}) + \frac{1}{2} {Mv}^{2} = \frac{3}{4} {Mv}^{2} = \frac{3}{4} \times 8 = 6 J$

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