2)t = 10 s
a)the moment of inertia of hoop is:
Ih = 2/3 m a^2
Is = 2/5 m a^2
from conservation of energy
m g h = 1/2 m v^2 + 1/2 x 2/3 m a^2 x v^2/a^2 = 1/2 m v^2 + 1/3 m v^2 = 5/6 m v^2
v = sqrt (6 g h/5) [hoop]
similarly for speher we get
v' = sqrt (10 g h/7) [sphere]
v' is more than v, v = d/t => t = d/v
more the velocity less the time, so the hoop will take more time as its moment of inertia is more than sloid sphere.
b)I' = 2/5 (2m) (2a)^2 = 15/6 m a^2
I' > I(solid)
Since the moment of inertia is now bigger, the time taken will be higher.
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