Problems regarding components of velocity

Question

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

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:-

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.

Answer 1

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.).

2^nd 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.

Answer 2

$u$ 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:

So your problem is now solved?

Answer 3

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.

Answer 4

@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).

Answer 5

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: $$v=\frac{u}{cos(\theta)}$$