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The expressions for rotational kinetic energy and translational kinetic energy are:

1. 1/2(mk²)ω²
2. 1/2(mk²)ω² + 1/2mR²ω²
3. 1/2(mk²)ω² + 1/2mR²ω²(k² + R²)
4. 1/2(mk²)ω² + 1/2mR²ω²(k² + R²)

Correct answer: 1/2(mk²)ω² + 1/2mR²ω²

Solution

The total energy of a rolling object is the sum of its rotational kinetic energy (1/2 Iω²) and translational kinetic energy (1/2 mv²). Substituting I = mk² and v = ωR, the total energy becomes 1/2(mk²)ω² + 1/2mω²R².

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