A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 4 J then total kinetic energy of the disc is

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TestHub · Accepted Answer

Total K.E of the rolling disc or ring is given by.
$K . E = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
For ring and disk, translational kinetic energy $\frac{1}{2} m v^{2}$ is constant.
Rolling K.E. of disc is $\frac{1}{4} m R^{2} ω^{2}$
Rolling K.E. of ring is $\frac{1}{2} m R^{2} ω^{2}$
As for ring, $4 J = \frac{1}{2} m v^{2} + \frac{1}{2} m R^{2} ω^{2}$
$∴ m R^{2} ω^{2} = 4 J$
For disc $\frac{1}{2} m R^{2} ω^{2} + \frac{1}{4} m R^{2} ω^{2} = (\frac{4}{2} + \frac{4}{4}) J = (2 + 1) J = 3 J$

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