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Unformatted text preview: PHYS 230 Fall 2007 Assignment 10 1. French 7.21, slightly extended. 5 points. A trick cyclist rides his bike around a “wall of death” in the form of a vertical cylinder (see the figure). The maximum frictional force parallel to the surface of the cylinder is equal to a fraction μ of the normal force exerted on the bike by the wall. (a) At what (minimum) speed must the cyclist go to avoid slipping down? There are three, and only three, forces acting on the cyclist. The vertical forces are mg and the force of friction on his wheel(s) at the wall. These must be equal and opposite: mg = F f The horizontal force is the normal force of the wall, which provides the centripetal acceleration: N = mv 2 /R Since F f = μN , we obtain mg = μmv 2 /R or v 2 = gR/μ (b) Show that he must also be inclined at an angle φ = arctan μ to the horizontal. In principle, the cyclist would seem to rotate about his Centre of Mass – he will “topple over”. However, if we choose to take the torques about the Centre of Mass, we see that they will balance if the resultant of F f and N passes through the CofM. @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ ? 6 mg F f N φ 1 Thus F f cos φ must balance N sin φ , or μ = tan φ . (c) To answer b), in what frame of reference did you work? There is another frame in which you could work. Redo the solution in that frame. We worked in the usual frame of the Earth, but we could equally well have worked in the rotating frame. In fact, in the rotating frame, we no longer have to choose to balance torques only about the CofM. The centrifugal force, which acts on the CofM, will provide a torque around the contact point to balance the torque due to gravity. We obtain mv 2 /R sin φ = mg cos φ However, since, in the rotating frame, the cyclist is not moving, we have...
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 Fall '07
 Harris
 Force, Friction, Light, 2m, 100 m, coriolis force, 5000 radians

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