If a spherical ball rolls on a table without slipping, the fraction of its total energy associated with rotation is ______?
Answer: 2/7
Explanation
When a spherical ball rolls without slipping:
Total energy (E) = Kinetic energy (K) + Rotational energy (U)
K = (1/2)mv^2
U = (1/2)IĻ^2 = (1/5)mv^2 (for a sphere)
E = K + U = (1/2)mv^2 + (1/5)mv^2 = (7/10)mv^2
Fraction of energy associated with rotation:
U/E = ((1/5)mv^2) / ((7/10)mv^2) = 2/7
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