hoop of mass M=2 kg and radius R=0.4 m rolls without slipping down a hill. The l

Question

hoop of mass M=2 kg and radius R=0.4 m rolls without slipping down a hill. The lack of slipping means that when the center of mas of the hoop has speed V, the tangential speed of the hoop relative to the center of mass is equal to Vcm,since in that case the instantaneous speed is zero for the part of the hoop that is in contact with the ground (v-v=0). The initial speed of the hoop is Vinitial =2m/s, and the hill has a height of H=4.1 m. Q. Replace the hoop with a bicycle wheel wheel whose rim has mass M=2 kg and radius r= 0.4 m, and whose hub has a mass m= 1.1 kg. The spokes have negligible mass. What would the bicycle wheel's speed be at the bottom of the hill

Explanation / Answer

Well, you can structure the equation as: .5(moment of inertia)(angular speed final)^2 + .5(mass)(velocity final)^2 = (mass)(acceleration due to gravity)(height) + .5(moment of inertia)(angular speed initial)^2 + .5(mass)(velocity initial)^2 Kinetic(rotational/final) + Kinetic(translational/final) = Potential(gravity) + Kinetic(rotational/initial) + Kinetic(translational/initial) Multiply by 2, factor out the mass, substitute variables (I = MR^2, w = V/R), cancel out everything, you end up with (velocity final)^2 = (acceleration due to gravity)(height) + (velocity initial)^2 Strange, right? Mass, radius? Yep, it all factors out. As for the rest of the problem, I'm still working it out. EDIT: Okay, person from my physics class figured out how to do it: For the second problem, the only thing you do differently is that you use the combined mass for the potential energy of gravity and for the translational kinetic energies on both sides of the equation. Since the new mass is in the middle, it doesn't affect the rotational speed, so you're only concerned with the mass of the rim for the rotational kinetic energy. Then it's just algebra.

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