Motion in 2D - Uniform Circular Motion and Projectiles

In this article, we study examples of motion in both 1D and 2D. The simplest example of motion in 2D is a particle moving along the circumference of a circle with uniform speed. Such a motion is called uniform circular motion which we study now.

Uniform Circular Motion

The figure below shows example of a particle moving along a circle at constant speed v. Therefore the magnitude of velocity is constant but velocity is not constant because there is a constant change in direction – infact the velocity vector is always tangential to the circle which is intuitive and we shall also prove the same fact below:

So without wasting any more time, let us calculate the velocity and acceleration of point mass undergoing uniform circular motion. We can write the position vector as where ? is the angle it subtends at origin. Of course, as particle moves along circle over time, the angle ? also changes with time.

Thus, it’s velocity at time t is given by

 Because the speed is uniform across the circle, we have

and so we see that the rate of change of angle d?/dt = v/r is a constant since v and r are constants. We denote this quantity by ?. ? is called the angular velocity of the particle exhibiting angular motion.

To show that the velocity vector is always along the tangential direction, we take it’s dot product with position vector and show that it is equal to 0. Let us do this,

And hence, at any time t, the velocity vector is always perpendicular to the radial vector and hence is in the tangential direction.

 The acceleration vector is given by,

Thus, the magnitude of acceleration is equal to –?²r and it’s direction is radially inwards toward the center of the circle.

Hence, in uniform circular motion, the magnitude of the position vector, the velocity vector, and the acceleration vector remain constant. The only thing that changes is their direction.

 Acceleration Due to Gravity and Projectile Motion

 The force of gravitation is something whose discovery dates back to times of Newton who discovered it. An object dropped from a height always moves downward under a constant acceleration given by g = 9.8m/s². Why it moves downwards and more than that why does it have a constant acceleration is something explained by Newton’s Law of Gravitation. For the purposes of Kinematics, we assume that a free falling object (under absence of all other forces) has acceleration g = 9.8m/s² in downward direction.

 So in this section, we assume that a free falling object has acceleration in the downward direction with magnitude g.

Example: An object is dropped from a building of height ‘h’ meters. Assume that it’s initial velocity is 0, calculate the time it takes to hit the ground and it’s velocity just before it hits the ground.

The above is actually example of 1D motion along the vertical line, and we know that when we work in 1D, all the vector quantities can be represented by means of real numbers. So, the object has initial velocity 0, moves down with acceleration g, hits the ground at time t=T after traversing distance h. Hence,

h = 0.T + ½ gT² => T = ?(2g / h)

and it’s velocity at time t=T is therefore v = 0 + gT=?(2gh)

Example : Consider the previous example with a modification that the object has initial velocity of ‘u’ m/s in the horizontal direction. Calculate the total time it takes to hit the ground and the total horizontal distance it travels. Also find equations of motion at any time ‘t’.

 Now this is an example of motion in 2D, so need to use the vector notation to safely solve the equations of motion without errors. So using Cartesian coordinates, the initial velocity of the particle is uî , the particle undergoes constant acceleration -g , the initial position vector is given by h, so let us calculate the position vector at any given time ‘t’.

The figure above derives some equations of motion. So assume at time t, the position vector is given by which in 2D can be written as x î + y . The velocity vector  is therefore (dx/dt) î + (dy/dt) =

 v_x î + v_y where v_x = dx/dt is the rate of change of x-coordinate and so is called the horizontal component of velocity vector . Similarly, v_y = dy/dt is the rate of change of y-coordinate and hence is called the vertical component of velocity vector . The acceleration is constant and given by -g at any given time, and so

 From the above equation, we get

 What do the above equations mean? The first one says that the horizontal component of velocity is independent of time and hence is constant. It’s initial value was ‘u’ and thus v_x = u at any time t.

The second one says that the vertical component of velocity varies linearly with time. It’s initial value was 0, and so integrating the second equation from 0 to t, we get v_y = -gt at any time t. Thus, the velocity vector at any time ‘t’ is given by

Now when we have got the velocity vector, let us also calculate the position vector:

And the results we derived are intuitive. The horizontal distance traveled at time t is simply u.t because there is no acceleration in the horizontal direction. For the vertical direction, there is a constant acceleration of g in the downward direction and the equation of motion is the same as the equation of motion of particle moving downwards under gravity. To find the time T at which the particle hits the ground, we must have y=0 at t=T and so

Projectiles:

Let us now consider the general case of projectile motion where a particle with initial speed ‘u’ at an angle ? (0 < ? < ?)  relative to the ground as shown in the figure below:

 The equations of motion are more or less same we derived in example above – the horizontal velocity remains constant. The initial horizontal velocity is u_x = u.cos? and so v_x = u.cos? at any given time ‘t’.

 The initial vertical velocity is u_y = u.sin?. The vertical velocity changes at a constant rate of –g with time, and so at any time ‘t’, v_y = u_y – gt = u.sin? – gt.

Hence, the vertical velocity as a function of t is given by:

 at time t=0, v_y = u_y = usin? > 0. At time T = u.sin?/g, v_y = 0 and when t > T, v_y < 0. And when t < T, v_y > 0. This means that the particle moves in the vertical direction till time t=T, then changes it’s direction after time T and moves in the downward direction. This is along with horizontal motion with velocity of u.cos? at any time t. The particle is bound to hit the ground again at some point and time since the vertical velocity reverses the direction after time T.

 Let us now start calculating the height h at which it reverses it’s direction, the total range R which is the horizontal distance it travels till it hits the ground, the total time it takes to hit the ground. At time t=0, both x and y are 0. So we have,

To find the total time of journey, note that if the particle hits the ground at time t=T₁ , then y = 0 at t=T₁ . That is,

The equation shows that the particle is at the ground at time t= 0 or at t= 2u.sin?/g = 2T. So, the particle hits the ground again at t=2T.

The Range R of the projectile is nothing but the horizontal distance it travels from t=0 to t=2T, and since the horizontal velocity is constant, the range is equal to

The maximum height achieved by the projectile is given by the value of y-coordinate at time t=T which is therefore equal to

 Let us now find the equation of the trajectory which the particle follows from t=0 to t=2T. It can be found by eliminating the common variable ‘t’ from x and y,

which is an equation of a parabola assuming ???/2, because then cos ?=0 and so we can not divide by cos? in any of the above steps. When ?=?/2, we have x=0 and y = ut – ½ gt² which therefore turns out to be pure rectilinear motion in 1D. So, when ever we refer to projectile motion, unless stated otherwise we assume the initial angle to be ???/2.

And because , which means that  and so from trigonometry we know that the above trigonometric equation has a solution. Moreover there are always two solutions of an equation of the form sin2? = r as the following graph show