alleryA solid cylinder (I = MR²/2) of mass 2 kg, radius 0.1 m, rolls without slipping at 6 m/s. Total kinetic energy:
A48 J
B27 J
C54 J
D36 J
Answer & Solution
Correct answer: C. 54 J
We have KE = (1/2)Mv² + (1/2)I omega².
So KE = (1/2)(2)(36) + (1/2)(0.5 × 2 × 0.01)(60²).
With v = omega R, omega = 6/0.1 = 60. KE_rot = (1/2)(0.01)(3600) = 18. KE_trans = 36. Total = 54 J. (Or use formula: KE_total = (1/2)Mv²(1 + I/MR²) = (1/2)(2)(36)(1 + 1/2) = 54 J.)
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