Velocity - String(Pulley) constraints

[phoenixXL]

Homework Statement

A bead C can move freely on a horizontal rod. The bead is connected by blocks B and D by a string as shown in the figure. If the velocity of B is v. Find the velocity of block D.

Homework Equations

As the string is inextensible the velocity of the string along the length is const.

The Attempt at a Solution

The doubt I have is that, the following should be true
$$v_c\ =\ v_b.cos53°\ \ \ \ ...(1)$$

But in the solution from the book I get that
$$v_b\ =\ v_c.cos53°\ \ \ \ ...(2)$$

I used relation 1 and got wrong results, I'm just confused how do we get relation 2.
Kindly help me out.
Thanks for your time

 

[Tanya Sharma]

[
The doubt I have is that, the following should be true
$$v_c\ =\ v_b.cos53°\ \ \ \ ...(1)$$

But in the solution from the book I get that
$$v_b\ =\ v_c.cos53°\ \ \ \ ...(2)$$

I used relation 1 and got wrong results, I'm just confused how do we get relation 2.
Kindly help me out.
Thanks for your time

How is the tip of the string connected to bead C moving ? What is its velocity ?

 

[phoenixXL]

Its velocity is $v_b$,along the length of the string, the same as that of block B.

 

[Tanya Sharma]

Its velocity is $v_b$,along the length of the string, the same as that of block B.

That means the tip of the string and bead C have different velocities . Is that so ?

 

[phoenixXL]

I think that so (may be my misconception). The bead is confined to the horizontal bar, and can only have velocity along the horizontal bar.

Furthermore, is the velocity of TIP and the whole of the string, not the same.

 

[Tanya Sharma]

The red spot in the picture depicts the tip of the string .Forget about all the pulleys and blocks for a moment.Just focus on the red spot .How does it move ?

 

[phoenixXL]

It should have the velocity same as that of the bead $v_c$.

 

[haruspex]

It hasn't been specified anywhere that I can see, but I presume the bead is tied to the string, so the string cannot slide through it.

phoenixXL, if the bead moves to the right at velocity v_b, what is the component of that towards the top-right pulley?

 

[phoenixXL]

haruspex,
As you are pointing, the following would be the diagram
( assuming $v_b$ to be the velocity of the block( or string) and $v_c$ the velocity of the bead. )

But, I got confused and tried to solve the problem using the following diagram,

So, I will be grateful to know what misconception I do have, and how can I further prevent from getting errors.

 

[haruspex]

So, I will be grateful to know what misconception I do have, and how can I further prevent from getting errors.

The question of which vector equals a component of the other comes up in a few guises. In the present case, you just have to remember that it's the string length that's constant, so the rate at which the string passes over the pulley equals the rate at which the bead gets closer to that pulley.
I find it can also help to imagine the process happening. If you pull down steadily on the string hanging from the pulley, do you expect the bead to get faster or slower?

 

[phoenixXL]

so the rate at which the string passes over the pulley equals the rate at which the bead gets closer to that pulley

How can we conclude this?

I find it can also help to imagine the process happening. If you pull down steadily on the string hanging from the pulley, do you expect the bead to get faster or slower?

Faster, of course.

 

[dauto]

haruspex,
As you are pointing, the following would be the diagram
( assuming $v_b$ to be the velocity of the block( or string) and $v_c$ the velocity of the bead. )

But, I got confused and tried to solve the problem using the following diagram,

So, I will be grateful to know what misconception I do have, and how can I further prevent from getting errors.

Your misconception is the belief that the opposite ends of the string must have the same velocities. They don't. Only the component of the motion parallel to the string is constrained. The ends of the string are completely free to move in the direction perpendicular to the string.

 

[haruspex]

How can we conclude this?

The distance from bead to pulley is the length of string joining them. The rate at which the string passes over the pulley is the rate at which that length decreases.

 

[phoenixXL]

The distance from bead to pulley is the length of string joining them. The rate at which the string passes over the pulley is the rate at which that length decreases.

Got it.

For anyone with similar problem

Let the distance of the bead from the pulley be r. Then $$-\frac{dr}{dt}\ =\ v_b\\<br /> \implies\ -\frac{d\sqrt{x^2\ +\ y^2}}{dt}\ =\ v_b\\<br /> \implies\ -\frac{1}{2\sqrt{x^2\ +\ y^2}}.\frac{dx^2}{dt}\ =\ v_b\\<br /> \implies\ -\frac{1}{2\sqrt{x^2\ +\ y^2}}.2x.v_c\ =\ v_b\\<br /> \implies\ -v_c.cosθ\ =\ v_b\\$$

Thank you so much when I derive it myself, meager confidence builds up.
Thanks you so much