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190

7 Forces in Two and Three Dimensions

Fig. 7.4 The motion of a ball thrown across the lecture room

7.3.1 Example: Motion of a Ball with Gravity

Throughout this chapter we will follow a particular problem as we gradually increase the complexity of the physical model: The motion of a ball thrown across the classroom. The experiment is illustrated in Fig. 7.4, which illustrates the observed path of a ball in an experiment. How can we develop a realistic model for the motion of this ball? We start from the simplest description: The motion when affected by gravity alone.

Problem: A ball is thrown from a height h above the ground at an angle α with the horizontal with an initial speed v0. Find the velocity and position of the ball at a time t .

Identify and Sketch: In this exercise we address the motion of the ball, described by the position r(t ) as a function of time. At the time t0, the ball was thrown. We place the coordinate system so that gravity is acting in the y-direction, and we place the x -axis so that the ball is thrown in the positive x -direction. The origin is placed at the ground, directly below the initial position of the ball at t = t0. The initial position vector is therefore r(t0) = r0 = h j. The initial velocity is directed at an angle α with the horizontal, this means that the initial velocity is v(t0) = v0 cos(α) i + v0 sin(α) j. The situation is illustrated in Fig. 7.5.

Fig. 7.5 a Illustration of the motion of the ball.

b Free-body diagram of the ball

(A)

y v0

α

h

x

(B)

FD

y

 

 

v

x G

7.3 Force Model—Constant Gravity

191

Model: The motion of the ball is determined by the forces acting on it. The only contact force acting on the ball is air resistance, FD , but we will here assume that this force is negligible. The only long-distance force acting on the ball is gravity, G, as illustrated in the free-body diagram in Fig. 7.5.

Newton’s second law is applied to both the x - and the y-component of the forces independently. In the x -direction Newton’s second law gives:

Fx = max = 0 .

(7.9)

There are no horizontal forces. The sum of the forces in the horizontal, x -direction is therefore zero. Consequently, the acceleration in the x -direction, ax is also zero.

Newton’s law of motion in the y-direction gives:

Fy = G = −mg = may , (7.10)

where we have used that the gravitational force from the Earth is mg, and that it acts in the negative y-direction.

The acceleration of the ball is therefore:

 

d2r

 

 

a = dt 2

= −g j ,

(7.11)

and the initial conditions are r(t0) = h j and v(t0) = v0.

Solve: We find the motion of the ball by solving the differential equation in (7.11). Since the acceleration is constant, we can solve it by direct integration for each of the components.

In the x -direction, the acceleration is zero, and the velocity in this direction is therefore constant.

vx (t ) = vx (t0) = v0 cos(α) .

(7.12)

The x -position is given by direct integration:

 

 

vx (t ) =

d x

 

 

 

(7.13)

 

 

dt

 

 

 

 

 

 

 

 

 

 

 

t

 

t

d x

 

 

vx (t )dt =

 

 

 

dt

(7.14)

 

 

 

dt

t

 

t0

 

 

t0

 

 

 

v0 cos(α)dt =

x

(t )d x

(7.15)

t0

 

 

 

 

x (t0)

 

v0 cos(α)(t t0) = x (t ) − x (t0)

(7.16)

that is, we have recovered motion with constant velocity:

x (t ) = x (t0) = v0 cos(α)(t t0) ,

(7.17)

192

7 Forces in Two and Three Dimensions

In the y-direction, the acceleration is constant, ay = −g. We can find the velocity by direct integration:

t

t0

vy (t ) −

dv

 

t

 

 

 

y

dt =

gdt

(7.18)

dt

 

 

 

t0

 

vy (t0) = −g(t t0)

(7.19)

which gives:

vy (t ) = (v0 sin(α) − g(t t0))

(7.20)

We find the position by integrating once more, using that vy (t ) = d y/dt , and that v0,y = v0 sin(α):

d y = dt

t d y

dt =

t0 dt

v0,y g(t t0)

 

(7.21)

t

v0,y g(t t0)

dt

(7.22)

t0

 

 

 

y(t ) − y(t0) = t

v0,y dt g t (t t0)dt

(7.23)

t0

 

 

 

t0

 

y(t ) − y(t0) = v0,y (t t0) −

1

g(t t0)2

(7.24)

 

 

 

2

 

 

 

which gives

 

 

 

 

 

 

y(t ) = h + v0 sin(α)(t t0) −

1

g(t t0)2 ,

(7.25)

 

 

2

 

 

where y(t0) = h is the launch height of the projectile.

Analyze: We notice that the motion in the x - and y-directions are independent of each other. The motion in the x -direction is simply a motion with constant velocity. The motion in the y-direction is the same as for the one-dimensional problem. If the ground is flat, it is the motion in the y-direction that determines how long time it takes to reach the ground. We can therefore answer questions about flight time, and maximum height by just studying the one-dimensional motion along the y-direction.

We have now found the complete solution for the motion of a ball subject only to gravity. From this solution, we can answer any complicated question, such as how far the projectile travels or what choise of initial direction gives the maximum length.

7.4 Force Model—Viscous Force

For an object moving through a fluid, such as a projectile flying through the air, a meteor entering the Earth’s atmosphere, or a tiny microrobot navigating through your bloodstream, there is a contact force on the object due to the motion of the

7.4 Force Model—Viscous Force

193

object relative to the fluid. The fluid has to flow around the object when the object moves, as a result the fluid exerts a force on the object. This force is distributed: It acts everywhere on the surface of the object, and it may also vary in magnitude and direction along the surface of the object. Usually, we will simplify the effect of the this distribution of forces into a single force acting in a single point on the object, and we will call this force the drag force or the “fluid resistance” acting on the object. For most purposes this is a sufficiently precise description of the interaction with the fluid.

The form of the drag force depends on the velocity of the object relative to the fluid. We discern between a behavior at low velocities, where the drag force is proportional to the velocity, and high velocities, where the drag force depends on the square of the velocity:

The drag force on an object moving at a velocity v relative to a fluid moving with a velocity w is:

FD

kv (v w) at small velocities

(7.26)

D |v w| (v w) at high velocities

The constant kv depends on the object’s size, shape and surface, as well as on the (dynamic) viscosity of the fluid. For a sphere Stokes found that

kv = 6π R η

(7.27)

where R is the radius of the sphere, and η is the viscosity of the fluid. The viscosity of air is η = 1.82 × 10−5 Nsm−2 and for water it is η = 1.00 × 10−3 Nsm−2, both at room temperature.

Experimental data indicates an approximative value for D for a spherical object:

D 12.0 ρ R2 .

(7.28)

where v = |v|, ρ is the density of the fluid, and R is the radius of the sphere.

Versatility of the viscous force model: The viscous force model

FD = −kvv ,

(7.29)

is much more versatile than suggested by its application to fluid drag forces. It is often used as a general damping term—a term reducing relative motion that also introduces dissipation and heat generation. For example, you will find that the viscous force model used to model damping of vibrations in solid object, to model the damping of vibrations in macroscopic objects and macroscopic springs, and to model surface forces in nano-scale surface contact. The viscous force model is a first order model to study any velocity-dependent force that tends to reduce velocity differences.