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Physics 8 Spring 2012 NAME: TA: Quiz 8 - Solutions Make sure your name is on your quiz, and please box your final answer. Because we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish them! Starting from rest, a hollow ball rolls without slipping down a ramp inclined at angle θ to the horizontal. Find an expression for its speed, v, after it’s gone a distance d along the incline. Hint: Apply conservation of energy. Furthermore, the moment of inertia of a hollow ball of mass M and radius R is I = 32 M R2 . ———————————————————————————————————— Solution Consider the ball starting at rest at a height yi . It’s initial energy is entirely potential energy, so Ei = M gyi . As the ball rolls down the incline, as seen in the figure to the right, it picks up speed and loses potential energy. Since the ball is rolling it has both translational and rotational kinetic energy. So, the final energy is Ef = KEtrans + KErot + M gyf = 21 M v 2 + 12 Iω 2 + M gyf , where v is the translational speed, I is the moment of inertia, and ω is the rotational speed. Conservation of energy says that H d θ 1 1 M g (yi − yf ) = M v 2 + Iω 2 . 2 2 But, the change in height H = yi − yf . Furthermore, from the geometry, H = d sin θ. Thus, we find 1 1 M gd sin θ = M v 2 + Iω 2 . 2 2 Now, the condition that the ball roll without slipping means that v = Rω, where R is the radius of the ball. Thus, we have 1 1 v2 M I 2 M gd sin θ = M v + I 2 = 1+ v2. 2 2 R 2 M R2 Now, we are told that the moment of inertia is I = 32 M R2 , and so M 2 5M 2 M gd sin θ = 1+ v2 = v . 2 3 6 Canceling off the mass from both sides and solving for the speed gives r 6 v= gd sin θ. 5 1