on the incline initial energy Ei = M*g*d
at the bottom , final energy Ef = (1/2)*M*v^2 + (1/2)*I*w^2
w = angular speed = v/R
Ef = (1/2)*M*v^2 + (1/2)*I*(v/R)^2
Ef = (1/2)*M*v^2 + (1/2)*I*v^2/R^2
from energy conservation
total energy is conserved
Ef = Ei
(1/2)*M*v^2 + (1/2)*I*v^2/R^2 = M*g*d
(1/2)*I*v^2/R^2 = M*g*d - (1/2)*M*v^2
(1/2)*I*v^2/R^2 = M(g*d - (1/2)*v^2)
I/MR^2 = 2*g*d/v^2 - 1
I/(MR^2) = 2*9.81*2.01/5.8^2 - 1
I/MR^2 = 0.172
(11 %) Problem 7: An object rolls without slipping down a 2.01 m high incline. Randomized...
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