posted by Morgan on .
15. [1pt] A thin hoop of radius r = 0.59 m and mass M = 9.2 kg rolls without slipping across a horizontal floor with a velocity v = 3.3 m/s. It then rolls up an incline with an angle of inclination θ = 31°. What is the maximum height h reached by the hoop before rolling back down the incline?
The initial kinetic energy before it starts climbing is
KEmax = (1/2)MV^2 + (1/2)I*w^2
where I = M R^2 and w = V/R.
I is the moment of intertia, R is the radius, V is the initial velocity and and w is the initial angular velocity.
KEmax = M V^2.
It stops rolling when
KEmax = M g h,
the potential energy increase.
h = V^2/g. The ramp angle, hoop mass and radius do not matter.