A thin hoop of radius r = 0.65 m and mass M = 7.1 kg rolls without slipping acro

Question

A thin hoop of radius r = 0.65 m and mass M = 7.1 kg rolls without slipping across a horizontal floor with a velocity v = 3.5 m/s. It then rolls up an incline with an angle of inclination theta = 27 degree . What is the maximum height h reached by the hoop before rolling back down the incline?

Explanation / Answer

initial energy of hoop = translational K.E. + rotational K.E. => Ei = 0.5*m*v^2 + 0.5*I*w^2 => Ei = 0.5*7.1*3.5^2 + 0.5*(7.1*0.65^2/2)*(3.5/0.65)^2 = 65.23 J final energy of hoop = gravitational P.E. = mgh => Ef = 7.1*9.8*H => Ef = 69.58 *H from energy conservation; Ei = Ef => 65.23 = 69.58*H => H = 0.937 m now, H = distance of the centre of mass of the hoop from the ground => H = h + r => h = 0.937 - 0.65 = 0.287 m

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