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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
dt Terminal speed g b
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.
mg bvt 0
The expression for v that satisﬁes Equation 6.4 with v
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:
e bt/m ge bt/m
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
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This document was uploaded on 09/19/2013.
- Fall '13
- Circular Motion