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6 - Circular Motion and Other Applications of Newton's Laws

Diagram of the sphere as it falls c speed time graph

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Unformatted text preview: R (6.2) bv where v is the speed of the object and b is a constant whose value depends on the properties of the medium and on the shape and dimensions of the object. If the object is a sphere of radius r, then b is proportional to r. Consider a small sphere of mass m released from rest in a liquid, as in Figure 6.15a. Assuming that the only forces acting on the sphere are the resistive force bv and the force of gravity Fg , let us describe its motion.1 Applying Newton’s second law to the vertical motion, choosing the downward direction to be positive, and noting that Fy mg bv, we obtain mg bv ma m dv dt (6.3) where the acceleration dv/dt is downward. Solving this expression for the acceleration gives dv dt Terminal speed g b v m (6.4) This equation is called a differential equation, and the methods of solving it may not be familiar to you as yet. However, note that initially, when v 0, the resistive force bv is also zero and the acceleration dv/dt is simply g. As t increases, the resistive force increases and the acceleration decreases. Eventually, the acceleration becomes zero when the magnitude of the resistive force equals the sphere’s weight. At this point, the sphere reaches its terminal speed vt , and from then on 1 There is also a buoyant force acting on the submerged object. This force is constant, and its magnitude is equal to the weight of the displaced liquid. This force changes the apparent weight of the sphere by a constant factor, so we will ignore the force here. We discuss buoyant forces in Chapter 15. 6.4 165 Motion in the Presence of Resistive Forces it continues to move at this speed with zero acceleration, as shown in Figure 6.15b. We can obtain the terminal speed from Equation 6.3 by setting a dv/dt 0. This gives mg bvt 0 or vt mg/b The expression for v that satisfies Equation 6.4 with v v mg (1 b e bt/m) vt (1 e t/ 0 at t 0 is (6.5) ) This function is plotted in Figure 6.15c. The time constant m/b (Greek letter tau) is the time it takes the sphere to reach 63.2% ( 1 1/e) of its terminal speed. This can be seen by noting that when t , Equation 6.5 yields v 0.632vt . We can check that Equation 6.5 is a solution to Equation 6.4 by direct differentiation: mg d mg bt/m dv d mg e e bt/m ge bt/m dt dt b b b dt Aerodynamic car. A streamlined body reduces air drag and increases fuel efficiency. (See Appendix Table B.4 for the derivative of e raised to some power.) Substituting into Equation 6.4 both this expression for dv/dt and the expression for v given by Equation 6.5 shows that our solution satisfies the differential equation. EXAMPLE 6.11 Sphere Falling in Oil A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil, where it experiences a resistive force proportional to its speed. The sphere reaches a terminal speed of 5.00 cm/s. Determine the time constant and the time it takes the sphere to reach 90% of its terminal speed. Solution vt Because the terminal mg/b, the coefficient b is b mg v...
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