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Unformatted text preview: Physics 105 Solutions to Midterm 1 The motion of a moving wheel can be decomposed into the linear motion of the axel and the rotational motion about the axel. Since the horizontal direction isn’t of interest to us, we’ll ignore the translational motion except to note that the require ment of rolling without slipping enforces the relationship v = ωa between the linear speed v and the angular speed ω of the wheel. We will denote points on the wheel by an angle θ with the top of the wheel being θ = 0. Using a little trigonometry and some standard results from circular motion, we conclude that a mud spec still attached to the tire at an angle θ has vertical position a (1 + cos θ ) and vertical velocity aω sin θ = v sin θ . The height of a mass under the influence of gravity satisfies: y ( t ) = y + v y, t 1 2 gt 2 This is a quadratic fucntion which has a unique maximum at: y max ( θ ) = y + v 2 y, 2 g = a (1 + cos θ ) + v 2 sin 2 θ 2 g = a (1 + cos θ ) + v 2 2 g (1 cos 2 θ ) = v 2 2 g cos 2 θ + a cos θ + v 2 2 g + a Again, we have a quadratic function, this time in cos θ . Using the same technique, the maximum of this quadratic is y ¨ubermax = a 2 g 2 v 2 + v 2 2 g + a . This maximum is obtained when cos θ = ag v 2 which fortunately is less than one. What if ag v 2 > 1? Well, quadratic functions are very simple in form. If we aren’t allowed to reach the unique extremum of a quadratic, then we simply get as close to it as we can. Setting cos θ = 1 gives y max = 2 a . This makes sense. After all, in the limit that the wheel is motionless, a free mud spec just falls straight down. 1 Alternatively, we could perform a more detailed maximization of y max with respect to θ , which produces the same result. 2 (a) We could use the result from the homework that this system is equivalent to a system with the same mass on a spring with the same spring constant but a different natural length and no gravity. However, let’s first tackle this problem using conservation of energy. Initially, the mass is at rest and the spring is at its natural length. If we also choose the initial position of the mass as the zero for gravitational potential energy, then the total energy is zero.potential energy, then the total energy is zero....
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This note was uploaded on 04/01/2008 for the course PHYSICS 105 taught by Professor Edgarknobloch during the Fall '07 term at University of California, Berkeley.
 Fall '07
 EdgarKnobloch
 Physics

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