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A sphere rolls down an inclined plane without slipping. What fraction of its total energy is rotational ?

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  1. \(\frac{2}{7}\)
  2. \(\frac{3}{7}\)
  3. \(\frac{4}{7}\)
  4. \(\frac{5}{7}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{2}{7}\)
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Detailed Solution

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Calculation:
 A sphere rolling without slipping has both translational and rotational kinetic energy.

Rotational Kinetic Energy (KR) = (1/2) × I × (v2 / r2)

Moment of Inertia of a solid sphere, I = (2/5)mr2

So, KR = (1/2) × (2/5)mr2 × (v2 / r2) = (1/5)mv2

Translational Kinetic Energy (KTr) = (1/2)mv2

Total Energy (KTotal) = KR + KTr = (1/5)mv2 + (1/2)mv2

Convert to common denominator:

KTotal = (2/10 + 5/10)mv2 = (7/10)mv2

Fraction of Rotational Energy = KR / KTotal = (1/5)mv2 / (7/10)mv2

Cancel mv2:

Fraction = (1/5) / (7/10) = (2/7)

Answer: Option 1) 2/7

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