MHT CET · Physics · Rotational Motion

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

A 3 J
B 4 J
C 5 J
D 6 J

Verified Solution

Answer & Solution

Correct Answer

(A) 3 J

Step-by-step Solution

Detailed explanation

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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