# Is My Approach to Calculating Relative Velocity Correct?

• Aurelius120
Aurelius120
Homework Statement
Two cars are moving at constant speed one on a circular track and other on horizontal path. Radius of circular track is 200m. The magnitude of relative velocity is graphed as follows. Find their speeds
Relevant Equations
$$v_{rel}=v_A-v_B$$

In the question I assumed the velocity of the circular object to be :

$$\vec v_A= v cos( \phi + \omega t) \hat i + v sin(\phi + \omega t) \hat j$$
where $$\omega = \frac{v}{R}$$

Velocity of the other particle is
$$\vec v_B = v \hat i$$

Now magnitude of relative velocity comes out on evaluation to be
$$v_{rel} = \sqrt {2v^2(1-cos(\phi+\omega t)}$$

On further solving, by using
##v_{25}=0## and ##v_{0}=v_{50}##
I couldn't solve further and my whatever little solving I did gave incorrect answers.
So, Is my method correct? If yes then how do I proceed further?

I noticed a second method that simply took relative velocity to be maximum when both were moving in opposite directions and since that value was 40 the value of individual speeds must be 20. I could consider it a correct method if I let the fact that the graph doesn't touch 40 slide. But that leaves another problem ##2 \pi R= 2×200× \pi= 20×50## which is not correct.

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You need two dollar signs to delimit the Latex.

Aurelius120 said:
Now magnitude of relative velocity comes out on evaluation to be
$$v_{rel} = \sqrt {2v^2(1-cos(\phi+\omega t)}$$
This is correct. Technically, that is a transcendental equation, so you'll need some numerical method to solve it, approximately.
Aurelius120 said:
On further solving, by using
##v_{25}=0## and ##v_{0}=v_{50}##
I couldn't solve further and my whatever little solving I did gave incorrect answers.
So, Is my method correct? If yes then how do I proceed further?

I noticed a second method that simply took relative velocity to be maximum when both were moving in opposite directions and since that value was 40 the value of individual speeds must be 20. I could consider it a correct method if I let the fact that the graph doesn't touch 40 slide. But that leaves another problem ##2 \pi R= 2×200× \pi= 20×50## which is not correct.

It's not entirely clear to me how you are supposed to tackle this. It looks like the relative speed is heading for a maximum of ##40 \ m/s##, but the graph is not complete in this respect.

You could simplify things by taking ##t = 0## when the relative velocity is 0.

The graph is fairly crude. I put my best estimate of the numbers into a spreadsheet and checked which value of ##v## gave the closest fit. ##v = 20.1 \ m/s## was the best fit. Round that down to ##20 \ m/s##.

The function ##f(v)## is that function of ##v## you have above. I tried to match that with ##v_r^2##. Trial and error.

 v_r Delta t v_r^2 v f(v) v f(v) v f(v) 10​ 5​ 100​ 20​ 97.93395​ 20.1​ 99.88642​ 20.2​ 101.8676​ 20​ 10​ 400​ 20​ 367.7582​ 20.1​ 374.85​ 20.2​ 382.0391​ 28​ 15​ 784​ 20​ 743.4102​ 20.1​ 756.9094​ 20.2​ 770.5694​ 34​ 20​ 1156​ 20​ 1132.917​ 20.1​ 1151.605​ 20.2​ 1170.461​ 38​ 25​ 1444​ 20​ 1440.915​ 20.1​ 1461.354​ 20.2​ 1481.882​

PS I don't see a better way.

The given solution is based on estimating the relative acceleration when their velocities are equal (t=25s). From the graph, it's about ##2m/s^2##. The result follows from ##a=v^2/r##, and the fact that one car is not accelerating.

PeroK
haruspex said:
The given solution is based on estimating the relative acceleration when their velocities are equal (t=25s). From the graph, it's about ##2m/s^2##. The result follows from ##a=v^2/r##, and the fact that one car is not accelerating.
Is it obvious that equation holds when the relative velocity is zero?

Aurelius120 said:
...
I noticed a second method that simply took relative velocity to be maximum when both were moving in opposite directions and since that value was 40 the value of individual speeds must be 20. I could consider it a correct method if I let the fact that the graph doesn't touch 40 slide.
Using the given value of the speed of each car respect to the ground, and assuming certain symmetry of both movements respect to each other, I have put together a graph myself, including summation of the velocity vectors.

I can see in it that the relative velocity does not reach the 40 m/s value because the car on the circular track does not complete a full rotation in the time lapse of 50 seconds.

At the 25 seconds position both car should be moving in the same direction, in order for the differential velocity to be zero, as shown in your graph.

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Aurelius120
PeroK said:
Is it obvious that equation holds when the relative velocity is zero?
Maybe not obvious, but it seems right.
Just to check…
As established, (shifting time by a half cycle) the magnitude of the relative velocity is ##y=v\sqrt{2(1-\cos(\omega t))}##. Differentiating, ##\dot y=v\frac{\omega}{\sqrt 2}(1-\cos(\omega t))^{-\frac 12}\sin(\omega t)##. In the limit as t tends to zero, ##\dot y=v\omega=\frac{v^2}R##.

PeroK
haruspex said:
Maybe not obvious, but it seems right.
Just to check…
As established, (shifting time by a half cycle) the magnitude of the relative velocity is ##y=v\sqrt{2(1-\cos(\omega t))}##. Differentiating, ##\dot y=v\frac{\omega}{\sqrt 2}(1-\cos(\omega t))^{-\frac 12}\sin(\omega t)##. In the limit as t tends to zero, ##\dot y=v\omega=\frac{v^2}R##.
In this case, with the specific motion involved, it works out. In general, however, ##\big | \frac{dv_r}{dt} \big | \ne a_r##. Even in the case where instantaneously ##v_r = 0##.

I haven't seen the full solution, but it may be based on that fallacy.

We have, in general:$$v_r\frac{dv_r}{dt} = \vec v_r \cdot \vec a_r$$The critical factor here is that the relative acceleration is collinear to the relative velocity as the latter tends to zero. In that case, we do indeed have:$$\bigg | \frac{dv_r}{dt} \bigg | = a_r$$
@Aurelius120 are you following this?

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nasu
@PeroK Yes but I don't see why

$$\frac{d \vec v_r}{dt} \neq \vec a_r$$
If ##\vec a_{rel} = \vec a_A - \vec a_B## and
##\vec v_{rel} = \vec v_A - \vec v_B##
I think it should follow that
$$\frac {d \vec v_r}{dt}=\frac {d \vec v_A}{dt}- \frac{d \vec v_B}{dt}$$
$$\Rightarrow \vec a_r= \vec a_A-\vec a_B$$

Aurelius120 said:
Yes but I don't see why $$\frac{d \vec v_r}{dt} \neq \vec a_r$$
That's true. But ##v_r = |\vec v_r|##, ##a_r = |\vec a_r|## and ##|\vec a_r| \ne \frac d {dt} |\vec v_r|##. In other words, it matters in which order you take the modulus and the derivative.

For example, in uniform circular motion ##v = |\vec v|## is constant: ##\frac {dv}{dt} = \frac {d|\vec v|}{dt}= 0##. But ##a = \big | \frac{d\vec v}{dt} \big |## is non-zero.

WLOG it helps to visualise B’s path as a tangent to A’s circular path. The relative velocity is zero when A is at the tangential contact point, P.

For compactness, I'll abbreviate ##(\omega t + \phi)## to ##()##.

For car A ##\vec {v_A} = v(\cos() \hat i + \sin() \hat j)##. For car B ##\vec {v_B} = v \hat i##

The x-component of the relative velocity (of A wrt B) is ##v_{x,rel} = vcos() - v##. This is zero whenever ##\cos() = 1##. At those times, ##\sin() = 0## . Let T be such a time.

The y-component of the relative velocity is ##v_{y,rel} = v\sin()##. At time T, ##v_{y,rel}## is equal to zero so ## v_{rel} = 0##.

The x-component of relative acceleration is ##a_{x,rel}= -v\omega sin()## . So at time T, ##a_{x,rel} =0##.

That means at time T, the relative acceleration has only a y-component. Since A is at point P, this y-component is A’s inwards radial acceleration. So, at time T, the relative acceleration and A’s centripetal acceleration (##\frac {v^2}{R}##) are one and the same.

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Steve4Physics said:
That means at time T, the relative acceleration has only a y-component. Since A is at point P, this y-component is A’s inwards radial acceleration. So, at time T, the relative acceleration and A’s centripetal acceleration (##\frac {v^2}{R}##) are one and the same.
The relative acceleration is always A's centripetal acceleration. The question is whether the magnitude of this is equal to the rate of change of relative speed. That only happens when the relative velocity and relative acceleration are in the same direction. That's the criterion that must be satisfied. @haruspex calculation indeed confirms this.

Steve4Physics
PeroK said:
The relative acceleration is always A's centripetal acceleration..
Yes.

PeroK said:
The question is whether the magnitude of this is equal to the rate of change of relative speed. That only happens when the relative velocity and relative acceleration are in the same direction. That's the criterion that must be satisfied. @haruspex calculation indeed confirms this.
Yes. But I was hoping to find a geometrical/intuitive argument.

I think the missing step is that, when the relative velocity becomes zero, the magnitude of the y-component of acceleration and the rate of change of relative speed are equal. (Because the x-component of acceleration and the x-component of the relative velocity are both zero; relative motion is effectively limited to the y-direction.)

It then follows that the rate of change of relative speed equals the magnitude of the centripetal acceleration.

PeroK said:
In this case, with the specific motion involved, it works out. In general, however, ##\big | \frac{dv_r}{dt} \big | \ne a_r##. Even in the case where instantaneously ##v_r = 0##.

I haven't seen the full solution, but it may be based on that fallacy.

We have, in general:$$v_r\frac{dv_r}{dt} = \vec v_r \cdot \vec a_r$$The critical factor here is that the relative acceleration is collinear to the relative velocity as the latter tends to zero. In that case, we do indeed have:$$\bigg | \frac{dv_r}{dt} \bigg | = a_r$$
@Aurelius120 are you following this?
If the relative velocity is tending smoothly to zero at t=0 then to a first approximation ##\vec v_r=\vec ct## for some constant vector ##\vec c##. So the relative acceleration tends to ##\vec c##, making it necessarily collinear with ##\vec v_r##, no?

PeroK
Steve4Physics said:
I think the missing step is that, when the relative velocity becomes zero, the magnitude of the y-component of acceleration and the rate of change of relative speed are equal. (Because the x-component of acceleration and the x-component of the relative velocity are both zero; relative motion is effectively limited to the y-direction.)
There is a technical problem. The y-component of relative velocity also goes to zero - and, crucially, more slowly than the x-component. In the limit, the motion tends towards motion in the y-direction. But, it's for this reason, not simply that the x-component goes to zero, that motion is in the y-direction in the limit.

Note that the relative speed is technically non-differentiable at the critical point itself.

If you could construct an example where the y-component tended to zero faster than the x-component, you might be able to find a counterexample to your reasoning.
Steve4Physics said:
It then follows that the rate of change of relative speed equals the magnitude of the centripetal acceleration.
It's true in this case, but whether it's true in general for these criteria is not clear.

haruspex said:
If the relative velocity is tending smoothly to zero at t=0 then to a first approximation ##\vec v_r=\vec ct## for some constant vector ##\vec c##. So the relative acceleration tends to ##\vec c##, making it necessarily collinear with ##\vec v_r##, no?
It follows from your calculation (or using Taylor series): ##1 - \cos(\omega t)## tends to zero faster than ##\sin(\omega t)##.

PeroK said:
It follows from your calculation: ##1 - \cos(\omega t)## tends to zero faster than ##\sin(\omega t)##.
Yes, but you seemed to be saying this is a special case :
PeroK said:
In general, however, ##\big | \frac{dv_r}{dt} \big | \ne a_r##. Even in the case where instantaneously ##v_r = 0##.
According to my post #17, it only requires that the relative velocity tends smoothly to zero.

PeroK
haruspex said:
Yes, but you seemed to be saying this is a special case :

According to my post #17, it only requires that the relative velocity tends smoothly to zero.
Ah! That's why it's not a special case. I missed that.

## 1. What is relative velocity?

Relative velocity is the velocity of one object as observed from the reference frame of another object. It is a vector quantity that represents how fast and in what direction one object is moving relative to the other.

## 2. How do you calculate the relative velocity of two objects when one is on a circular path and the other on a straight path?

To calculate the relative velocity of two objects where one is on a circular path and the other on a straight path, you first need to determine the velocity vectors of both objects in a common coordinate system. Once you have these vectors, you subtract the velocity vector of the object on the straight path from the velocity vector of the object on the circular path.

## 3. How does the centripetal acceleration of the object on the circular path affect the relative velocity?

Centripetal acceleration affects the velocity vector of the object on the circular path by continuously changing its direction. This changing direction must be taken into account when calculating the relative velocity, as it means the velocity vector of the object on the circular path is not constant over time.

## 4. What are the common coordinate systems used for these calculations?

The most common coordinate systems used for these calculations are Cartesian coordinates (x, y, z) and polar coordinates (r, θ). Cartesian coordinates are often used for the object on the straight path, while polar coordinates can be convenient for the object on the circular path. However, both objects' velocities need to be converted to a common coordinate system for the relative velocity calculation.

## 5. Can the relative velocity be zero? If so, under what conditions?

Yes, the relative velocity can be zero if both objects have the same velocity vector in a common reference frame. This means they would have to be moving with the same speed and direction at the same instant in time, even if one is on a circular path and the other on a straight path.

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