Hoop Rolling up Hill Must answer both parts A thin hoop of radius r = 0.71 m and

Question

Hoop Rolling up Hill

Must answer both parts

A thin hoop of radius r = 0.71 m and mass M = 9.7 kg rolls without slipping across a horizontal floor with a velocity v = 1.1 m/s. It then rolls up an incline with an angle of inclination ? = 20°. What is the maximum height h reached by the hoop before rolling back down the incline?

Part B: Now, suppose a uniform solid sphere is used instead of a hoop. Use the same values of r, M, and v as stated in the previous problem, and calculate the maximum height reached in this case.

Explanation / Answer

You can use conservation of energy for this. The moment of inertia of a hoop is Mr2 so the initial KE of thehoop is:

KE = (1/2) Mv2 + (1/2) I 2 = (1/2)Mv2   + (1/2) Mr2 (v/r) 2

or   KE = (1/2) Mv2 + (1/2) Mv2   = Mv2    

The initial KE equals the final PE so

       Mv2   =  Mgh

   h = v2 / g =1.12 / 9.8 = 0.123 meters

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b)

Now for sphere
At the bottom of the ramp, the hoop has rotational KE and translational KE. At the max height, the hoop is all PE. Conservation of energy says KE initial = PE final
(1/2)*m*v2 + (1/2)*I*w2 = mgh
The Moment of Inertia(I) for a sphere is (2/5)*m*r2
w = v/r
(1/2)*m*v2 + (1/2)*((2/5)*m*r2)*(v/r)2 = mgh
(1/2)*m*v2 + (1/5)*m*r2*v2/r2 = mgh
(1/2)*m*v2 + (1/5)*m*v2 = mgh
(7/10)*m*v2 = mgh
(7/10)*v2 = gh
h = (7/10)*v2 /g
h = 0.0864 m

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