Relative Motion

Relative Motion

Relative motion means motion relative to another object. However, is it anything different from what is called “motion”? We noted that motion is always with respect to a reference frame, and it’s fruitless to imagine something like absolute state of rest. Hence, at the first sight there doesn’t seem to be any difference between relative motion and the concept of motion we have studied so far. Strictly speaking, there is no such difference. However, there is one subtle difference that we note from the examples of motion we studied in the previous sections – the frame of reference in the examples of motion we studied so far was “ground” which is by assumption always in a state of rest. Thus, when we study ordinary motion it is with respect to an observer standing on ground. In relative motion, we will see examples where in addition to ground, there is one additional frame of reference which moves relative to the ground.

The figure above shows the setting. We select the coordinate axis and imagine an observer at the origin O. The object under motion is denoted by A which is shown in the red and it’s instantaneous position vector relative to the ground is . In addition to ground, there is another frame of reference F whose instantaneous position vector relative to ground is  . Instantaneous position vector of A relative to Frame of reference F is denoted by , which by law of parallelogram addition of vectors is equal  to  . The velocity of A relative to Frame of Reference F is therefore equal to . Similarly, acceleration of A relative to F is equal to

The relative velocity of A with respect to F is equal to which is the velocity of A as seen from F.

We can now find equations of relative motion using the above mentioned relative vector quantities. Let us study examples of relative motion.

Example(Relative Motion in one dimension): Two trains moving in parallel tracks are moving at speeds ‘u’ m/s and ‘v’ m/s respectively with respect to ground. The distance between the two trains at time t=0 is ‘d’ meters. Calculate the time T at which the trains cross each other in the following two cases:

(i)                   The trains are moving in the opposite direction,

(ii)                 The trains are moving in the same direction. Assume that u > v in this case.

Needless to repeat, since the motion in 1D, all the vectors can be represented by real numbers instead of unit vectors. Denote the train moving at speed ‘u’ by A and train moving at speed ‘v’ by B. Further, assume without loss of generality that the train A is moving in the positive direction.

(i)                   If the trains are moving in opposite direction, then the velocity of train A(relative to ground) is simply u whereas velocity of train B(relative to ground) is –v. An observer sitting in train B will see train A moving at velocity . The initial relative distance is d and so it takes T=d / (u+v) amount of time for B to cover this distance.

      (ii)         If the trains are however moving in opposite direction, then the velocity of train A (relative to ground) is u and the velocity of train B(relative to ground) is v. Thus, an observer sitting in train B will see train A moving at velocity   > 0. Now there are two cases to consider here :

               (a) If train A is ahead of train B, then the initial position vector  = d > 0 and since  > 0, the relative distance between A and B will always increase from d>0 with time and so will never become 0. So the trains do never meet in this case.

          
         (b) If train B is ahead of A, then the the initial position vector = -d < 0 and since > 0, the relative distance between A and B will always increase from –d with time and so will become 0 at some time T = d/(u-v).

 Example: Two projectiles are fired simultaneously at time t=0. The first one is fired from the base of the building at angle θ(0<θ<π/2) at initial speed u whereas the second one is fired from the top of the building horizontally (parallel to the ground) at with initial speed v. Under what necessary and sufficient condition do the two particles collide, and assuming those conditions, calculate the time instant of collision, the coordinates of collision (assuming the base of the building to be the origin and the building to be the positive y-axis).

 So if the particles collide at time t=T at coordinates (x,y), then since the horizontal velocity for both the particles remains constant, we have:

x = v.T and x = u cosθ.T => v = u.cosθ.

The above condition is a necessary condition, but is it a sufficient condition. It does not appear obvious, but yes it is a sufficient condition. The reason is if v=ucosθ, then at any instant t, the x-coordinate of the first projectile is ucosθ.t and of the second particle it is v.t = ucosθ.t. Thus, the x-coordinate of both the projectiles at any time instant is equal. Hence, if we observe the motion of first particle (which is fired from the base of the building) relative to the second one, it is seen coming up in straight line i.e. the relative motion is in 1D. The initial relative distance is h, the final relative distance is 0, initial relative velocity is usinθ in the vertical direction, and note that the relative acceleration is 0(=-g – (-g)) at any time. Hence, the first particle is seen coming up uniformly at a speed of usinθ in straight line. Thus, the two particles do meet at some point and time. Infact, it takes time T= h/usinθ for the two particles to meet. The coordinates where the two projectiles collide are given by:

Example: A helicopter is ascending vertically upwards at a constant speed of ‘v’ m/s relative to the ground. When it reaches height ‘h’, a ball is thrown from the helicopter vertically upwards in the air at an initial speed ‘u’ m/s relative to the helicopter. Calculate the time at which the helicopter and the ball cross each other, and also the time when the ball hits the ground.Let the helicopter and ball cross each other at time t=T. If we observe the ball from inside the helicopter, the initial relative distance is 0, the final relative distance is 0, initial relative velocity is +u, relative acceleration is always –g at any point and time, and thus equations of motion say

The river is flowing in the horizontal direction with velocity  . The particle is moving with velocity  whose magnitude is always constant and equal to u. The velocity of boat relative to river is  = – or + =. Because  and are perpendicular, we have

Where we used the fact that since and are perpendicular, .= 0. The time taken to cross the river is therefore

Example : Consider the setting of example 12 with the difference that the boat now chooses to move at an angle θ so as to reach the other end in the shortest possible time. The speed of current and the speed of boat relative to ground are v and u which are to be considered as fixed. Calculate angle θ.

 Under the same notation of example 12, we have + = , except that  and are not perpendicular but at an angle θ.

 The observer on ground sees the boat moving with velocity whose magnitude ‘u’ is fixed. If θ is the angle between  and, then we have

because is the resultant velocity(i.e. the velocity of boat as seen from the ground) of  and . The time taken to cross the river is t = d / u.sinα and since d and u are fixed, t is minimized when sin α = 1and so α=π/2. To find θ when α=π/2, we note that when α=π/2 the above equations become

 Example: Consider a modification of examples 12 and 13 where the boat moves with speed  λv (0<λ<1) relative to water moving at speed v. The boat chooses the angle θ with an aim to minimize the horizontal drift (i.e. the horizontal distance it moves away from the opposite end of the bank) when it reaches the other bank. Find angle θ.

This example is a bit different from the previous ones in that the magnitude of velocity of boat relative to river is given to be fixed. In the previous examples, we assumed magnitude of velocity relative to ground to be fixed.

The figure below demonstrates the setting.

Time taken by the boat to reach other end is given by

The horizontal drift is therefore horizontal velocity multiplied by T which is

which is minimized when

.

Case Study: Cycloid Motion

Let us now apply the theory of motion and relative motion to study cycloidal motion, i.e. motion of a particle along a cycloid.

Imagine a wheel of radius R rolling forward. The center of the wheel as seen from the ground moves in a straight line with velocity v. With respect to center of the wheel, any point on the wheel executes a uniform circular motion with angular velocity ω. Assume that v = ωR and we shall see later when we study Rolling that this is a necessary condition for “Pure Rolling” where the wheel rolls smoothly without slipping.

Consider point O shown in the above figure. Relative to the center C of the wheel, it executes uniform circular motion. Let us find it’s equations of motion and it’s trajectory for an observer on the ground.

To find equations of motion, we select point O to be at origin at t=0. At time t, it is at some point O. The center of the wheel moves from C to C . The position vector of particle at time t=t is therefore =   which is given by

Now, from the figure we can see that

We now need to calculate  , which is the position vector of O relative to the center of the wheel. From the center of the circle, the particle will be seen executing uniform circular motion at any time. Hence,  

as calculated in the following figure.

and so,

Therefore,

which is an equation of a cycloid. It is a parametric curve where x and y are expressed as parameters. We note one thing from the equations, the x-coordinate is a strictly increasing function of t (because it’s derivative with respect to t is non-negative). Thus, a point never revisits the same x-coordinate in future and thus the curve is indeed a function. The y-coordinate is a periodic function of t and becomes 0 at every time instant of the form 2nπ/ω for n=0,1,2,3,….The y-coordinate achieves maximum value of R at time instants (2n+1)π/2ω for n=0,1,2,3,….

There are many cusps in the trajectory. From the equations and also from the figure above, we see that the point O traverses a distance 2πR horizontally, which is equal to the full circumference of the circle before it comes back to the ground. The total distance it travels before it again touches the ground is however the length of a single cusp. The total distance traveled from t=0 to t=2π/ω is given by

Now,

And so,