Pulleys + Convex Mirror Reflection

[zorro]

Homework Statement

In the given arrangement pulley P1 and P2 are moving with constant speed vo downward and the centre of the pulley P lies on the principal axis of a convex mirror having radius of curvature R. Find the speed of image of pulley P when it is at a distance x from the surface of convex mirror in terms of vo, R, x and θ

The Attempt at a Solution

This question is pretty easy but I got stuck at one point.

Using constraint relations,
I got 2v_o - vcosθ + (L-x_p)sinθ =0
where v is the velocity of Pulley P and L is the distance of the mirror from the upper fixed support.

i.e. v= (2v_o + (L-x_p)sinθ)/cosθ

This is not the final answer but as per the solution given v = 2vo/cosθ (which is written directly)

Can somebody explain me my mistake?

Thanks

 

[tiny-tim]

Hi Abdul!

I got 2vo - vcosθ + (L-x)sinθ =0

I don't understand where (L-x)sinθ came from.

What is your constraint equation for the lengths (not speeds) of the vertical and diagonal bits of string? ​

 

[zorro]

Hi tiny-tim!

This is my equation in terms of length of the string-

2x_p1 + 2x_pcosΦ + 2x_p2 = Total length of the string,

where Φ = angle at any instant

Now if we consider the change in lengths and differentiate w.r.t time,
The equation reduces to

v_p1 - v_pcosΦ + x_psinΦ + v_p2 = 0
On substituting the values given in the question,
2v_o - vcosθ + x_psinθ = 0

I considered all distances from the fixed support. Let the distance of the mirror from the fixed support be L. So distance of the pulley P from the mirror is L - x_p (sorry I missed s.s. 'p' earlier).

The above equation reduces to
2v_o - vcosθ + (L-x_p)sinθ = 0

 

[tiny-tim]

It'll be easier if you use the fact that the horizontal distance is constant.

(btw, x₁ etc is a lot clearer and easier to write than x_p1 )

 

[zorro]

Where shall I use that fact?

2x_p1 + 2x_pcosΦ + 2x_p2 = Total length of the string,

Here in x_pcosΦ both x_p and cosΦ are variables.

That fact would have been useful if
2v_o - vcosθ + x_psinθ = 0
was differentiated again so that x_psinθ vanishes

 

[tiny-tim]

x₀/H = … ?

 

[zorro]

what is x_o and H ?

 

[tiny-tim]

uhh? …

dx₀/dt = v₀, and what do you think H is?

 

[zorro]

I am getting confused
H might be the distance of the mirror from the support ( I used L for it ).
I still don't understand how x_o/H will help :|

 

[tiny-tim]

Abdul, in the original diagram, L and x_p1 were not marked, and you had to write them in.

Sometimes questions are like that.

They don't always spoon-feed you with the information you require.

You need to look at the diagram, find something else you haven't used yet, call it H, and use it.