Trajectory in physics characteristics, types, examples and exercises

The trajectory in physics It is the curve that a mobile describes as it passes through successive points during its movement. Since it can adopt an infinite number of variants, so will the trajectories that the mobile can follow.

To get from one place to another, a person can take different paths and different ways: on foot through the sidewalks in streets and avenues, or arriving by car or motorcycle on a highway. During a walk through the forest, the walker can follow a complicated path that includes turns, going up or down in level and even passing through the same point several times.

If the points through which the mobile is traveling follow a straight line, the trajectory will be rectilinear. This is the simplest path, as it is one-dimensional. Specifying the position requires a single coordinate.

But the mobile can follow a curvilinear path, being able to be closed or open. In these cases, tracking the position requires two or three coordinates. These are movements in the plane and in space respectively. This has to do with links: limiting material conditions of movement. Some examples are:

- The orbits that describe the planets around the sun are closed ellipse-shaped paths. Although, in some cases, they can be approximated to a circular, as in the case of the Earth.

- The ball that the goalkeeper kicks in a goal kick follows a parabolic trajectory.

- A bird in flight describes curvilinear trajectories in space, because in addition to moving on a plane, it can go up or down in level at will.

The trajectory in physics can be expressed mathematically when the position of the mobile is known at any instant of time. Be r the position vector, which in turn has coordinates x, Y Y z in the most general case of a movement in three dimensions. Knowing the function r (t) the trajectory will be completely determined.

Article index

• 1 Types
  • 1.1 Movements in one, two and three dimensions
• 2 Examples
  • 2.1 Path of a mobile in an explicit, implicit and parametric way
• 3 Solved exercises
  • 3.1 Resolved exercise 1
  • 3.2 Exercise solved 2
  • 3.3 Exercise solved 3
• 4 References

Types

In general terms, the trajectory can be a rather complicated curve, especially if you want to express it mathematically. For this reason, it begins with the simplest models, where the mobiles travel on a straight line or on a plane, which can be the floor or any other suitable one:

Movements in one, two and three dimensions

The most studied trajectories are:

- Rectilinear, when traveling on a straight horizontal, vertical or inclined line. A ball thrown vertically upward follows this path or an object sliding down an incline follows. They are one-dimensional movements, a single coordinate being enough to determine their position completely..

- Parabolic, in which the mobile describes an arc of a parabola. It is frequent, since any object thrown obliquely under the action of gravity (a projectile) follows this trajectory. To specify the position of the mobile you have to give two coordinates: x Y Y.

- Circular, occurs when the moving particle follows a circle. It is also common in nature and in daily practice. Many everyday objects follow a circular path such as tires, machinery parts, and orbiting satellites, to name a few..

- Elliptical, the object moves following an ellipse. As said at the beginning, it is the path followed by the planets in orbit around the sun..

- Hyperbolic, Astronomical objects under the action of a central force (gravity), can follow elliptical (closed) or hyperbolic (open) trajectories, these being less frequent than the former.

- Helical, or spiral movement, like that of a bird ascending in a thermal current.

- Swing or pendulum, the mobile describes an arc in back and forth movements.

Examples

The trajectories described in the previous section are very useful to quickly get an idea of ​​how an object is moving. In any case, it is necessary to clarify that the trajectory of a mobile depends on the location of the observer. This means that the same event can be seen in different ways, depending on where each person is..

For example, a girl pedals at a constant speed and throws a ball upwards. She observes that the ball describes a rectilinear path. 

However, for an observer standing on the road who sees it pass, the ball will have a parabolic movement. For him, the ball was initially thrown with an inclined speed, a result of the speed upward by the girl's hand plus the speed of the bicycle..

Path of a mobile in an explicit, implicit and parametric way

- Explicit, directly specifying the curve or locus given by the equation y (x)

- Implicit, in which a curve is expressed as f (x, y, z) = 0

-Parametric, in this form the x, y and z coordinates are given as a function of a parameter that is generally chosen as time t. In this case, the trajectory is made up of the functions: x (t), and (t) Y z (t).

Below, two well-studied kinematics trajectories are detailed: the parabolic trajectory and the circular trajectory.

Tilted Launch Into Vacuum

An object (the projectile) is thrown at an angle a with the horizontal and with initial velocity v_or as the picture shows. Air resistance is not taken into account. The movement can be treated as two independent and simultaneous movements: one horizontal with constant speed and the other vertical under the action of gravity..

x (t) = x_or +v_ox.t

y (t) = y_or +v_hey.t -½g.t^two

These equations are parametric equations projectile launching. As explained above, they have the parameter t, what is time.

The following can be seen in the right triangle in the figure:

v_ox = v_or cos θ_i

v_hey = v_or sen θ_i

Substituting these equations containing the launch angle into the parametric equations results:

x (t) = x_or +v_or cos θ_i.t

y (t) = y_or +v_or. sen θ_i.t -½g.t^two

Equation of the parabolic path

The explicit equation of the path is found by solving t from the equation for x (t) and substituting in the equation for y (t). To facilitate algebraic work, it can be assumed that the origin (0,0) is located at the launch point and thus x_or = and_or = 0.

This is the equation of the path in explicit way.

Circular path

A circular path is given by:

(x - x_or)^two + (and - and_or)^two = R^two

Here x_or and and_or represent the center of the circle described by the mobile and R is its radius. P (x, y) is a point on the path. From the shaded right triangle (figure 3) it can be seen that:

x = R. cos θ

y = R. sin θ

The parameter, in this case, is the swept angle θ, called the angular displacement. In the particular case that the angular velocity ω (angle swept per unit time) is constant, it can be stated that:

θ = θ_or + ωt

Where θ_or is the initial angular position of the particle, which if taken as 0, reduces to:

θ = ωt

In such case, time returns to parametric equations as:

x = R.cos ωt

y = R. sin ωt

The unit vectors i Y j are very convenient for writing the position function of an object r (t). They indicate the directions on the axis x and on the axis Y respectively. In its terms, the position of a particle that describes a Uniform Circular Motion is:

r (t) = R.cos ωt i + R. sen ωt j

Solved exercises

Solved exercise 1

A cannon can fire a bullet with a velocity of 200 m / s and an angle of 40º with respect to the horizontal. If the throw is on flat ground and air resistance is neglected, find:

a) The equation of the path y (x) ...

b) Parametric equations x (t) Y and (t).

c) The horizontal range and the time that the projectile lasts in the air.

d) The height at which the projectile is when x = 12,000 m

Solution to)

a) To find the trajectory, substitute the values ​​given in the equation y (x) of the previous section:

y (x) = tg 40º. x - 9.8 / (2 '400^two. cos^two40th) x^two ⇒  y (x) = 0.8391 x - 0.0000522x^two

Solution b)

b) The launch point is chosen at the origin of the coordinate system (0,0):

x (t) = x_or +v_ox.t = 400'cos 40º.t = 306.42. t.

y (t) = y_or +v_hey.t -½g.t^two= 400 'sin 40º.t - 0.5 '9.8't^two= 257.12 t - 4.9.t^two

Solution c)

c) To find the time that the projectile lasts in the air, do y (t) = 0, being the launch is made on flat ground:

0 = 257.12.t - 4.9.t^two

t = 257.12 / 4.9 s = 52.473 s

The maximum horizontal reach is found by substituting this value in x (t):

x_max = 306.42'52.47 m = 16077.7 m

Another way to find x_max directly is by making y = 0 in the equation of the path:

0 = 0.8391 x_max - 0.0000522 x^two_max

x = 0.8391 / 0.0000522 m = 16078.5 m

There is a little difference due to rounding of the decimals.

Solution d)

d) To know the height when x = 12000 m, this value is substituted directly in the equation of the path:

and (12000) = 0.8391'12000 - 0.0000522'12000^two m = 2552.4 m

Exercise solved 2

The position function of an object is given by:

r (t) = 3t i + (4 -5t^two) j m

Find:

a) The equation for the path. What curve is?

b) The initial position and the position when t = 2 s.

c) The displacement made after t = 2 s.

Solution

a) The position function has been given in terms of the unit vectors i Y j, which respectively determine the direction on the axes x Y Y, Thus:

x (t) = 3t

and (t) = 4 -5t^two

The Equation of the Path y (x) is clearing t from x (t) and substituting in y (t):

t = x / 3

y (x) = 4 -5. (x / 3)^two = 4 - 5x^two/ 9 (Parable)

b) The starting position is: r (2) = 4 j m ; the position in t = 2 s it is r (2) = 6 i -16 j m

c) Displacement Dr is the subtraction of the two position vectors:

Δr = r (two) - r (2) = 6 i -16 j- 4 j = 6 i - twenty j m

Exercise solved 3

The Earth has a radius R = 6300 km and it is known that the period of rotation of its movement around its axis is one day. Find:

a) The equation of the trajectory of a point on the earth's surface and its position function.

b) The velocity and acceleration of that point.

Solution to)

a) The position function for any point in circular orbit is:

r (t) = R.cos ωt i + R.sen ωt j

We have the radius of the Earth R, but not the angular velocity ω, however it can be calculated from the period, knowing that for circular motion it is valid to say that:

ω = 2π × frequency = 2π / period

The period of the movement is: 1 day = 24 hours = 1440 minutes = 86 400 seconds, therefore:

ω =  2π / 86400 s = 0.000023148 ​​s^-1

Substituting in the position function:

r (t) = R.cos ωt i + R. sen ωt j = 6300 (cos 0.000023148t i + sen 0.000023148t j) Km

The path in parametric form is:

x (t) = 6300. cos 0.000023148t

y (t) = 6300. sin 0.000023148t

Solution b)

b) For circular motion, the magnitude of the linear velocity v of a point is related to the angular velocity w through:

v = ωR = 0.000023148 ​​s^-1'6300 Km = 0.1458 Km / s = 145.8 m / s

Even being a movement with constant speed of 145.8 m / s, there is an acceleration that points towards the center of the circular orbit, in charge of keeping the point in rotation. It is the centripetal acceleration to_c, given by:

to_c = v^two / R = (145.8 m / s)^two / 6300 × 10³ m = 0.00337 m / s^two.

References

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