Motion in Two and Three Dimensions
7.3 Force Model—Constant Gravity
According to Newton’s law of gravity, there is a gravitational force between any two objects with gravitational masses. For an object close to the Earth’s surface, the gravitational force on the object can be approximated as:
G= −mgj, (7.7)
wheremis the gravitational mass of the object,gis the acceleration of gravity, and the unit vectorjpoints upwards. Upwards is indeed usually defined based on the direction of the force from gravitation. Thisconstant gravityforce model is valid as long as the object does not move far away from the surface of the Earth, and as long as the object does not move too far along the surface of the Earth, since this would lead to a change in the unit vectorj.
We notice that this force model is particularly simple: The force on an object due to gravity is a constant—both in magnitude and direction. Our discussion of the constant gravity force model may be extended to any constant force, such as the force on a charged particle in a homogeneous electric field.
If you throw a ball from the ground, the only forces acting on the ball after it has left your hand are the force from gravity,G, and air resistance,D. It we neglect the effect of air resistance, the only force acting on the ball is gravity. We can therefore apply Newton’s second law to find the acceleration of the ball:
j
Fj =G=mgj=ma ⇒ a= −gj. (7.8)
Since we have wisely chosen the y-axis to correspond to the direction of gravity, the acceleration of the ball is non-zero only in y-direction. The acceleration in the x- andz-directions are zero, and the velocities in these directions do therefore not change. We call such a motiondecoupledbecause the motion in thex- andy-axes are independent of each other. We will use this when we solve problems with constant forces.
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 class- room. 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 heighthabove the ground at an angleαwith the horizontal with an initial speedv0. Find the velocity and position of the ball at a timet.
Identify and Sketch: In this exercise we address the motion of the ball, described by the positionr(t)as a function of time. At the timet0, the ball was thrown. We place the coordinate system so that gravity is acting in they-direction, and we place the x-axis so that the ball is thrown in the positivex-direction. The origin is placed at the ground, directly below the initial position of the ball att =t0. The initial position vector is thereforer(t0)=r0=hj. The initial velocity is directed at an angleαwith the horizontal, this means that the initial velocity isv(t0)=v0cos(α)i+v0sin(α)j.
The situation is illustrated in Fig.7.5.
Fig. 7.5 Illustration of the motion of the ball (left) and free-body diagram of the ball (right)
y
x
h α
(a) (b)
x y
FD v0
v 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 thex- and they-component of the forces independently. In thex-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 thex-direction,axis also zero.
Newton’s law of motion in they-direction gives:
Fy=G= −mg=may, (7.10)
where we have used that the gravitational force from the Earth ismg, and that it acts in the negativey-direction.
The acceleration of the ball is therefore:
a= d2r
dt2 = −gj, (7.11)
and the initial conditions arer(t0)=hjandv(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 thex-direction, the acceleration is zero, and the velocity in this direction is therefore constant.
vx(t)=vx(t0)=v0cos(α). (7.12) Thex-position is given by direct integration:
vx(t)= d x
dt (7.13)
t
t0
vx(t)dt= t
t0
d x
dtdt (7.14)
t t0
v0cos(α)dt=
x
x(t0)(t)d x (7.15)
v0cos(α)(t−t0)=x(t)−x(t0) (7.16) that is, we have recovered motion with constant velocity:
x(t)=x(t0)=v0cos(α)(t−t0), (7.17)
192 7 Forces in Two and Three Dimensions In they-direction, the acceleration is constant,ay= −g. We can find the velocity by direct integration:
t
t0
dvy
dt dt = t
t0
−gdt (7.18)
vy(t)−vy(t0)= −g(t−t0) (7.19) which gives:
vy(t)=(v0sin(α)−g(t−t0)) (7.20) We find the position by integrating once more, using thatvy(t)=d y/dt, and that v0,y =v0sin(α):
d y dt =
v0,y−g(t−t0)
(7.21) t
t0
d y dtdt=
t
t0
v0,y−g(t−t0)
dt (7.22)
y(t)−y(t0)= t
t0
v0,ydt−g t
t0
(t−t0)dt (7.23) y(t)−y(t0)=v0,y(t−t0)−1
2g(t−t0)2 (7.24) which gives
y(t)=h+v0sin(α)(t−t0)−1
2g(t−t0)2, (7.25) where y(t0)=his the launch height of the projectile.
Analyze: We notice that the motion in thex- and y-directions are independent of each other. The motion in thex-direction is simply a motion with constant velocity.
The motion in they-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 they-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 choice of initial direction gives the maximum length.