Performance Trade Studies - A Systems level characterization and tradespace evaluation of a sim

In the computation for Tin equation (7), we found that the exponential was negligi-ble. Consequently, ignoring the corresponding exponential term in equation (5), we see that the parachutist’s velocity at impact is

which is the limiting velocity for her fall with the chute open.

In some situations the resistive drag force on an object is proportional to a power of other than 1. Then when the velocity is positive, Newton’s second law for a falling object gen-eralizes to

(8)

where mand ghave their usual interpretation and b 0 and the exponent rare experimental constants. [More generally, the drag force would be written as signy.] Express the solution to (8) for the case r2.

The (positive) equilibrium solution, with the forces in balance, is Other-wise, we write (8) as a separable equation that we can solve using partial fractions or the integral tables on the inside front cover:

and after some algebra,

Here is determined by the initial conditions. Again we see

that yapproaches the terminal velocity y₀as t→ ∞. ◆ c₃c₂ sign3^A^y0y_B

/

^A^y0y_B4

yy₀c₃e^2y⁰^bt/^m c₃e²^y⁰^bt/^m .

`y₀y

y₀y` c₂e^2y⁰^bt/^m ,

^y²₀^dy^y² ^2y¹0

ln`y₀y y₀y` b

mtc₁ , y¿bAy²₀y²_B

/

^m,

yy₀ 2mg

/

^b.

B b0y0^r A B

A mdy

dt mgby^r ,

0y0

b₂ ^A75B A9.81B

105 7.01 m/sec ,

e^1.4T 114 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Unless otherwise stated, in the following problems we assume that the gravitational force is constant with g 9.81 m /sec²in the MKS system and g 32ft/sec²in the U.S. Customary System.

1. An object of mass 5 kg is released from rest 1000 m above the ground and allowed to fall under the inﬂu-ence of gravity. Assuming the force due to air resis-tance is proportional to the velocity of the object with proportionality constant b50 N-sec/m, deter-mine the equation of motion of the object. When will the object strike the ground?

2. A 400-lb object is released from rest 500 ft above the ground and allowed to fall under the inﬂuence of gravity. Assuming that the force in pounds due to air resistance is 10y, where y is the velocity of the object in ft/sec, determine the equation of motion of the object. When will the object hit the ground?

3. If the object in Problem 1 has a mass of 500 kg instead of 5 kg, when will it strike the ground? [Hint:

Here the exponential term is too large to ignore. Use Newton’s method to approximate the time twhen the object strikes the ground (see Appendix B).]

Example 4

Solution

4. If the object in Problem 2 is released from rest 30 ft above the ground instead of 500 ft, when will it strike the ground? [Hint: Use Newton’s method to solve for t.]

5. An object of mass 5 kg is given an initial downward velocity of 50 m/sec and then allowed to fall under the inﬂuence of gravity. Assume that the force in newtons due to air resistance is 10y, where yis the velocity of the object in m/sec. Determine the equa-tion of moequa-tion of the object. If the object is initially 500 m above the ground, determine when the object will strike the ground.

6. An object of mass 8 kg is given an upward initial velocity of 20 m/sec and then allowed to fall under the inﬂuence of gravity. Assume that the force in newtons due to air resistance is 16y, where yis the velocity of the object in m/sec. Determine the equa-tion of moequa-tion of the object. If the object is initially 100 m above the ground, determine when the object will strike the ground.

7. A parachutist whose mass is 75 kg drops from a heli-copter hovering 2000 m above the ground and falls toward the ground under the inﬂuence of gravity.

Assume that the force due to air resistance is propor-tional to the velocity of the parachutist, with the proportionality constant b₁30 N-sec/m when the chute is closed and b₂90 N-sec/m when the chute is open. If the chute does not open until the velocity of the parachutist reaches 20 m/sec, after how many seconds will she reach the ground?

8. A parachutist whose mass is 100 kg drops from a heli-copter hovering 3000 m above the ground and falls under the inﬂuence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b₃ 20 N-sec/m when the chute is closed and b₄100 N-sec/m when the chute is open. If the chute does not open until 30 sec after the parachutist leaves the helicopter, after how many seconds will he hit the ground? If the chute does not open until 1 min after he leaves the helicopter, after how many seconds will he hit the ground?

9. An object of mass 100 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of times the weight of the object is pushing the object up (weight mg). If we assume that water resistance exerts a force on the object that is proportional to the velocity of the object, with

/

⁴⁰

Section 3.4 Newtonian Mechanics 115

proportionality constant 10 N-sec/m, ﬁnd the equa-tion of moequa-tion of the object. After how many seconds will the velocity of the object be 70 m/sec?

10. An object of mass 2 kg is released from rest from a platform 30 m above the water and allowed to fall under the inﬂuence of gravity. After the object strikes the water, it begins to sink with gravity pulling down and a buoyancy force pushing up.

Assume that the force of gravity is constant, no change in momentum occurs on impact with the water, the buoyancy force is the weight (weight mg), and the force due to air resistance or water resistance is proportional to the velocity, with proportionality constant b₁10 N-sec/m in the air and b₂ 100 N-sec/m in the water. Find the equa-tion of moequa-tion of the object. What is the velocity of the object 1 min after it is released?

11. In Example 1, we solved for the velocity of the object as a function of time (equation (5)). In some cases, it is useful to have an expression, independent of t, that relates y and x. Find this relation for the motion in Example 1. [Hint:Letting ,

then .]

12. A shell of mass 2 kg is shot upward with an initial velocity of 200 m/sec. The magnitude of the force on the shell due to air resistance is When will the shell reach its maximum height above the ground? What is the maximum height?

13. When the velocity yof an object is very large, the mag-nitude of the force due to air resistance is proportional to y²with the force acting in opposition to the motion of the object. A shell of mass 3 kg is shot upward from the ground with an initial velocity of 500 m/sec. If the magnitude of the force due to air resistance is , when will the shell reach its maximum height above the ground? What is the maximum height?

14. An object of mass mis released from rest and falls under the influence of gravity. If the magnitude of the force due to air resistance is byⁿ, where band n are positive constants, find the limiting velocity of the object (assuming this limit exists). [Hint:Argue that the existence of a (finite) limiting velocity

implies that .]

15. A rotating flywheel is being turned by a motor that exerts a constant torque T(see Figure 3.10 on page 116). A retarding torque due to friction is pro-portional to the angular velocity If the moment of inertia of the ﬂywheel is Iand its initial angular veloc-ity is v₀,ﬁnd the equation for the angular velocity v

/

^dt^S^{0 as t}^S^q

A0.1By² 0y0

/

^20.

/

^dt ^A^dV

/

^dx^B^V

yAtB VAxAtBB 1

/

116 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

19. An object of mass 60 kg starts from rest at the top of a 45º inclined plane. Assume that the coeffi-cient of kinetic friction is 0.05 (see Problem 18).

If the force due to air resistance is proportional to the velocity of the object, say,3y, ﬁnd the equa-tion of moequa-tion of the object. How long will it take the object to reach the bottom of the inclined plane if the incline is 10 m long?

20. An object at rest on an inclined plane will not slide until the component of the gravitational force down the incline is sufﬁcient to overcome the force due to static friction. Static friction is governed by an experimental law somewhat like that of kinetic friction (Problem 18); it has a magnitude of at mostmN, where mis the coefﬁcient of static fric-tion and Nis, again, the magnitude of the normal force exerted by the surface on the object. If the plane is inclined at an angle determine the criti-cal value for which the object will slide if

but will not move for

21. A sailboat has been running (on a straight course) under a light wind at 1 m/sec. Suddenly the wind picks up, blowing hard enough to apply a constant force of 600 N to the sailboat. The only other force acting on the boat is water resistance that is proportional to the velocity of the boat. If the pro-portionality constant for water resistance is b 100 N-sec/m and the mass of the sailboat is 50 kg, ﬁnd the equation of motion of the sailboat. What is the limiting velocity of the sailboat under this wind?

22. In Problem 21 it is observed that when the velocity of the sailboat reaches 5 m/sec, the boat begins to rise out of the water and “plane.” When this hap-pens, the proportionality constant for the water resistance drops to b060 N-sec/m. Now ﬁnd the equation of motion of the sailboat. What is the lim-iting velocity of the sailboat under this wind as it is planing?

23. Sailboats A and B each have a mass of 60 kg and cross the starting line at the same time on the ﬁrst leg of a race. Each has an initial velocity of 2 m/sec. The wind applies a constant force of 650 N to each boat, and the force due to water resistance is proportional to the velocity of the boat. For sail-boat A the proportionality constants are b1 80 N-sec/m before planing when the velocity is less

a 6 a₀. a 7 a₀

a₀

a, as a function of time. [Hint:Use Newton’s second law

for rotational motion, that is, moment of inertia angular acceleration sum of the torques.]

16. Find the equation for the angular velocity in Problem 15, assuming that the retarding torque is proportional to

17. In Problem 16, let I 50 kg-m²and the retarding torque be N-m. If the motor is turned off with the angular velocity at 225 rad/sec, determine how long it will take for the ﬂywheel to come to rest.

18. When an object slides on a surface, it encounters a resistance force called friction. This force has a mag-nitude of mN, where mis the coefﬁcient of kinetic frictionand Nis the magnitude of the normal force that the surface applies to the object. Suppose an object of mass 30 kg is released from the top of an inclined plane that is inclined 30º to the horizontal (see Figure 3.11). Assume the gravitational force is con-stant, air resistance is negligible, and the coefﬁcient of kinetic friction m 0.2. Determine the equation of motion for the object as it slides down the plane. If the top surface of the plane is 5 m long, what is the veloc-ity of the object when it reaches the bottom?

51v 1v.

v Motor

T, Torque from motor

Retarding torque I = d /dt

Figure 3.10Motor-driven ﬂywheel

x (0) = 0 x(t)

− N mg sin 30°

30° 30°

mg cos 30° mg

Figure 3.11Forces on an object on an inclined plane

than 5 m/sec and b₂60 N-sec/m when the velocity is above 5 m/sec. For sailboat B the proportionality constants are b₃ 100 N-sec/m before planing when the velocity is less than 6 m/sec and b₄50 N-sec/m when the velocity is above 6 m/sec. If the ﬁrst leg of the race is 500 m long, which sailboat will be leading at the end of the ﬁrst leg?

24. Rocket Flight. A model rocket having initial mass m₀ kg is launched vertically from the ground. The rocket expels gas at a constant rate of kg/sec and at a constant velocity of m/sec relative to the rocket.

Assume that the magnitude of the gravitational force is proportional to the mass with proportionality con-stant g.Because the mass is not constant, Newton’s second law leads to the equation

where is the velocity of the rocket,xis its height above the ground, and is the mass of the rocket at tsec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for 25. Escape Velocity. According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the dis-tances between them. That is,

where M₁and M₂are the masses of the objects,ris the distance between them (center to center),F_gis the attractive force, and Gis the constant of propor-tionality. Consider a projectile of constant mass m being fired vertically from Earth (see Figure 3.12).

Let trepresent time and ythe velocity of the pro-jectile.

F_gGM₁M₂

/

^r²^,

0t 6 m₀

/

^a.

m₀at ydx

/

^dt

Am₀atBdy

dt ab gAm₀atB , b

Section 3.5 Electrical Circuits 117

r R

m Earth

Figure 3.12 Projectile escaping from Earth

(a) Show that the motion of the projectile, under Earth’s gravitational force, is governed by the equation

where r is the distance between the projectile and the center of Earth,Ris the radius of Earth, Mis the mass of Earth, and

(b) Use the fact that to obtain

(c) If the projectile leaves Earth’s surface with velocity y₀, show that

(d) Use the result of part (c) to show that the veloc-ity of the projectile remains positive if and only

if The velocity is

called the escape velocityof Earth.

(e) If g9.81 m/sec²and R6370 km for Earth, what is Earth’s escape velocity?

(f ) If the acceleration due to gravity for the Moon is and the radius of the Moon is R_m 1738 km, what is the escape velocity of the Moon?

g_mg

/

⁶

y_e 22gR y²₀2gR 7 0.

y²2gR²

r y²₀2gR . ydy

gR² r² .

/

^dt^y

gGM

/

^R²^.

dy dt

gR² r² ,

In document A Systems level characterization and tradespace evaluation of a simulated airborne fourier transform infrared spectrometer for gas detection (Page 139-141)