Thetime of motion
"t=\\sqrt{\\frac{2L}{g\\sin\\theta}\\left(1+\\frac{I}{mr^2}\\right)}"The moment of inertia
"I_{hoop}=mr^2"
"I_{disk}=\\frac{1}{2}mr^2"
"I_{hoop}=\\frac{2}{5}mr^2"
Hence, the time of motion is smallest for the sphere.
Answer: A. The sphere reaches the bottom first.
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