Relative velocity of two objects: one on a circular path, one on a horizontal straight path

[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.

Please help. Thank you

 

[PeroK]

You need two dollar signs to delimit the Latex.

 

[Aurelius120]

Sorry about that. Done.

 

[PeroK]

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.

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.

Please help. Thank you

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.

 

[PeroK]

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​

 

[PeroK]

PS I don't see a better way.

 

[haruspex]

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]

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?

 

[Lnewqban]

...
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.

 

[haruspex]

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]

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?

 

[Aurelius120]

@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
$$\frac {d \vec v_r}{dt}=\frac {d \vec v_A}{dt}- \frac{d \vec v_B}{dt}$$
$$\Rightarrow \vec a_r= \vec a_A-\vec a_B$$

 

[PeroK]

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.

 

[Steve4Physics]

I’ve been following the thread with interest. Can I add this...

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.

EDIT. That's not very good. See also posts 15 and 16.

 

[PeroK]

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]

The relative acceleration is always A's centripetal acceleration..

Yes.

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.

 

[haruspex]

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]

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.

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.

 

[PeroK]

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 $\sin(\omega t)$.

 

[haruspex]

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 :

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]

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.