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I find it extremely difficult to resolve velocities into components to solve certain problems . A few examples are the following :-

enter image description here

Point $F$ is pulled down with velocity $u$ . Point $D$ is constrained to move horizontally . Find the instantaneous velocity of point $D$ , given that the angle made with the horizontal is $\theta$.

Note:-I don’t want to solve this problem using the method of derivatives. I know similar questions have been asked , but all the answers have been mathematical , and haven’t clarified my doubt. I want to find a logical approach , that uses the components of velocities , and the string constraint.

I can think of two ways to approach this problem :-

1) By the string constraint , the velocity of point D along the string is $u$ . The horizontal component of the velocity is hence $u\cos{\theta}$

2) Assign velocity $v$ in the horizontal direction to point $D$ . The component of this velocity in the direction of the string must be $u$ , which therefore means that $v=\frac{u}{\cos{\theta}}$

For some reason , the correct approach is the second one .

Another exceedingly similar class of problems , would be the following , which involves the velocity of the point of intersection of two curves:-

enter image description here

Rod $EF$ moves horizontally (to the right) with a velocity $v$ . Find the instantaneous velocity of the point of intersection with the circle , $v_G$ Given that the acute angle the tangent makes with the horizontal is $\theta$.

Again , there exist two methods to solve this question :-

1) The velocity of point $G$ in the horizontal direction is $v$ , therefore the velocity that moves along the circle is $v\cos{\theta}$

And method (2) , which is correct in this case , gives us $v_G=\frac{v}{\cos{\theta}}$

I think my confusion is apparent . I look for a general approach to solve such problems . Who decides which component of velocity is assigned to which entity ?

Any help would be greatly appreciated.

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  • $\begingroup$ in 2nd part, is the rod attached to center of circle ? $\endgroup$
    – maverick
    Apr 12, 2020 at 7:55
  • $\begingroup$ @maverick It is not . $\endgroup$
    – Aspirant
    Apr 12, 2020 at 7:58
  • $\begingroup$ in 2nd part ,answer most probably looks $v\cos \theta$ ??, because rod is constrained to move in horizontal direction , actual velocity is in horizontal direction(if rod is not attached to circle) $\endgroup$
    – maverick
    Apr 12, 2020 at 8:13
  • $\begingroup$ @maverick Precisely my doubt . The “actual” velocity concept is faulty. The correct answer is $v/cos x$ , and can be found by assigning variables , and differentiating. $\endgroup$
    – Aspirant
    Apr 12, 2020 at 8:17

5 Answers 5

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image

In the above figure, the green vector shows the horizontal velocity $v$ and the red vector shows the velocity $u$.

Correct Approach

If you observe the motion of the point D from the ground frame, it would appear to move with horizontal velocity $v$ in the right direction. Now let's resolve this horizontal velocity in two rectangular components, where one is along the string and the other is perpendicular to the string. This way the velocity along the string comes out to be $v\cos \theta$. And thus it follows that $u=v\cos \theta$.

Fallacy in the incorrect approach

You are right in saying that the velocity of D along the string is $u$, however, D also has a velocity along the direction perpendicular to the string. Thus, instead of the horizontal velocity being the component of $u$, it is $u$ that is the component of the horizontal velocity. Also if you use this approach, you wouldn't be able to justify what happened to the component of $u$ which is perpendicular to the string.

General Approach

In such problems, always find the "real/actual velocity" which is almost always the velocity in the ground frame. This velocity is the final velocity with which the object will move under the given constraints. After finding this velocity, break it into its components along the preferred direction and apply the constraints to find the relation between the kinematic parameters (displacement, velocity, acceleration, etc.).

2nd question

In this question, the point G is moving along the circumference of the circle and not along the horizontal direction. Thus its final velocity in this case is the velocity along the circumference and therefore you should take the components of that velocity instead of the horizontal velocity. Again, in this case the fallacy in method (1) is similar to the fallacy in the method (1) of the first question.

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  • $\begingroup$ Thank you , this is really helpful . (1) Is the “Actual Velocity” Anshuman describes in his answer arrived at by studying the motion in the ground frame ? ( As you’ve described in the first sentence under “Correct Approach “ (2) - Can you help me out with the second case? (3) Can you outline a general method for solving similar problems ? (Egs:- rotating rods , and point of contact) $\endgroup$
    – Aspirant
    Apr 12, 2020 at 9:10
  • $\begingroup$ (1) Yes, it seems so. (2) The second case can be derived from the same principles like the first one. (3) I am updating my answer. $\endgroup$
    – user258881
    Apr 12, 2020 at 9:19
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enter image description here

$u$ is the velocity at point F

I see it like this:

you have just one generalized coordinate $q$

with :

$$\dot{q}=u(t)\,\cos(\varphi)\quad , q=\int\dot{q}\,dt$$

and $$\varphi=\varphi(q)$$

so the problem is geometric how to obtain $\varphi(q)$

**Edit **

you can calculate the angle $\varphi(q)$ like this:

enter image description here

So your problem is now solved?

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  • $\begingroup$ I appreciate the answer , but I was looking for a more “physical” approach to solve the problem ( more on the lines of a high school student). $\endgroup$
    – Aspirant
    Apr 12, 2020 at 8:10
  • $\begingroup$ I don't think this is physical problem, this is geometric (mathematical) problem as i wrote $\endgroup$
    – Eli
    Apr 12, 2020 at 8:38
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We take the components of the actual velocity of any point , not the reverse way. So in such questions, general approach is to assume the velocity of body/particle/point and to apply constraints on them. Edit: I think I should explain further why (1) is incorrect approach for first problem. It is correct that velocity of point D towards the string is u. But that isn't its actual velocity as it's velocity must be horizontal (by constraint). And as stated above, we take components of actual velocity to find the point's velocity in some direction, but not he other way around.

Edit: Actual velocity of any particle can be defined as net instantaneous displacement of particle/ time. Components of actual velocity is only and not the other way around.

Complete explanation(skip if you understood):

Let dr â (position vector) be the actual displacement of body in time dt. In order to see by how much a body is displaced along, let's say û direction, we take component of dr â along û, i.e. dr(â.û)û.

On the other hand if we know that a body is displaced along û by dx(let) but its actual displacement is in direction â, you can't take the component of the component of some original vector to find that vector.

Analogously, treat the original vector(dr â) as a set, then the component(along û) will be its subset, and taking the component's component along â will give the sub-subset, NOT the original set. I hope you understood now.

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  • $\begingroup$ I’m sorry , but I find this unclear. In the second example I provided , the “actual” velocity of point $G$ is $v$ , the velocity of the rod . Your answer doesn’t describe any way of determining “the actual velocity”. Who decides what “the actual velocity” is? If I didn’t have any trouble determining what the “actual velocity” was , I wouldn’t have asked this question. $\endgroup$
    – Aspirant
    Apr 12, 2020 at 7:46
  • $\begingroup$ Assume actual velocity to be Vx and Vy in perpendicular directions wherever you want to define your X and Y axes $\endgroup$
    – ba-13
    Apr 12, 2020 at 7:57
  • $\begingroup$ How are we to decide what the “actual velocity” is? That is precisely the doubt . Also , it is left unexplained why we take the components of this actual velocity , and not the other way around . $\endgroup$
    – Aspirant
    Apr 12, 2020 at 7:59
  • $\begingroup$ This is the most general approach, but for the second question, I will add the solution in my answer $\endgroup$
    – ba-13
    Apr 12, 2020 at 8:00
  • $\begingroup$ I have added 2 pictures as I couldn't type all that $\endgroup$
    – ba-13
    Apr 12, 2020 at 8:15
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@FakeMod gave an excellent answer, and I would like to summarize the key point. Consider your first example. The general principle is -

The velocity of both $F,D$ ALONG the string must be the same. The string wouldn't be taut otherwise. Then, from your figure,$$v(D)_{string}=v(F)_{string}=u$$ $$\implies v(D)cos\theta=u$$ as the LHS tells you the component of $v(D)$ along the string. This is the result.

To reiterate, the velocity along the constraining surface must be the same, in order to remain under that constraint(e.g. the string being taut here).

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There is a formula relevant in such problems $$\sum T\cdot v = 0$$ where $T$ and $v$ are the tension and velocity vectors associated with a point on the string. I don't know the exact derivation, but I suppose from the constraint of inextensibility of the string and using energy conservation you can arrive at the result.

So, for the first question, we get, $$T v \cos\theta + Tu\cos(180^{\circ})=0$$

And we get the correct result: $$v=\frac{u}{cos(\theta)}$$

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  • $\begingroup$ Good answer, but it could be even better if you also include the explicit derivation of the relevant formula. $\endgroup$
    – user258881
    Apr 12, 2020 at 16:45
  • $\begingroup$ I haven't read the derivation, nor is this formula needed for me at this level. I just remember this as a "trick" to solve competitive exam problems faster. I would've surely given the derivation if I had a concrete mathematical proof. $\endgroup$ Apr 12, 2020 at 18:42

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