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A **solid sphere** rolling without slipping enters a vertical loop of radius $R$. The **minimum height** $h$ from which it must start (measuring $h$ from the bottom of the loop) so that it just completes the loop is:
A$2R$
B$2.5R$
C$2.7R$
D$3.5R$
Answer & Solution
Correct answer: C. $2.7R$
At top of loop, minimum condition: $v^2/R = g$ ⇒ $v^2 = gR$. Energy conservation: $mgh = mg(2R) + \tfrac{1}{2}mv^2(1 + 2/5) = 2mgR + (7/10)mgR$ ⇒ $h = 2R + 0.7R = 2.7R$. (Compare: point-particle answer is $2.5R$; the extra $0.2R$ is rotational KE the sphere also needs at top.)
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