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Unformatted text preview: PHY 7A  Review for Final Exam (Solutions) Dimitri DounasFrazer Spring 2008 Strategy for Harmonic Motion Problems 1. Draw a picture. Choose an origin and a define a set of coordinate axes. Make sure to draw the system slightly displaced from equilibrium. If necessary, choose a pivot point as well. 2. Draw an extended free body diagram (FBD). Decompose all forces along the axes. 3. Apply ∑ F i = m a and/or ∑ τ i = I α . Make approximations if necessary (e.g., sin θ ≈ θ ). 4. Solve for the equation of motion of the system: d 2 /dt 2 = ω 2 . Identify the angular frequency ω in terms of the parameters of the problem. 5. Solve. Check your answer (units, limits, etc.). Problem 1 1. Draw a picture. Choose an origin and coordinate axes. There are three forces acting on the ball: gravity F , friction f , and the normal force N . Because the ball rolls back and forth along the tunnel, a convenient set of axes is parallel ( x ) and perpendicular ( y ) to the tunnel. With this convention, f points in the xdirection and N points in the positive ydirection. The gravitational force F points from the center the ball to the center of the earth. Choose the origin O at the earth’s center of mass and the pivot point P at the ball’s center of mass. 2. Draw an extended FBD. Decompose forces along axes. Because f and N already point along the axes, we only need to decompose F . The magnitude F of F is given by F = GM m r 2 = GMmr R 3 = mg R r, (11) where r is the distance from the center of the earth to the center of the ball. Here m is the mass of the ball, M = M ( r 3 /R 3 ) is the mass of sphere “underneath” the ball, and M and R are the mass and radius of the earth, respectively. (See pages 142143 of Giancoli.) I have used g = GM/R 2 = 9 . 8 m / s 2 to simplify this expression. Let x and y denote the horizontal and vertical distances from the origin to the center of the ball. After drawing a right triangle, you can see that F x = mg R x and F y = mg R y, (12) where F x and F y are the components of F along the x and y axes, that is, F = F x ˆ i + F x ˆ j . 1 3. Apply Newton’s second law. Since there is no motion in the ydirection, we ignore it com pletely. In the xdirection, we have X F x,i = F x + f = ma, (13) where a is the acceleration of the ball’s center of mass. Furthermore, the torque equation is X τ i = r obj f...
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 Fall '08
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 Light, Moment Of Inertia, Giancoli, compass needle

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