What is the Acceleration of C in This Constrained Motion Problem?

[Saitama]

Homework Statement

If acceleration of A is 2m/s² to the left and acceleration of B is 1m/s² to left, then acceleration of C is (see attachment 1)
A)1m/s² upwards
B)1m/s² downwards
C)2m/s² downwards
D)2m/s² upwards

Ans: A

Homework Equations

The Attempt at a Solution

(see attachment 2, I hope the labels are clear)
Let the radius of smaller pulleys be r and that of bigger pulleys be R. I have marked the lengths of different portions of string. L is the length of total string.
L=x_1+2 \times \pi r+x_2+x_3+2\times \pi R/4+2x_4
Differentiating twice w.r.t time
0=\ddot{x_1}+\ddot{x_2}+\ddot{x_3}+2\ddot{x_4}
Now I am confused as to what should be the value of $\ddot{x_1}$.

Any help is appreciated. Thanks!

 

[verty]

I suggest expressing $x_1$ as a function of time, then differentiating twice.

Without calculus, it is not as easy. I can't think of a useful hint to give in this case except to say, use your intuition.

 

[tiny-tim]

Hi Pranav-Arora!

Isn't x₁ a function of x₂ and x₃ ? ​

 

[Saitama]

Isn't x₁ a function of x₂ and x₃ ? ​

Yes but then how do I write $x_1$ in terms of $x_2$ and $x_3$? Something like this: $x_2+x_3+\text{some constant}$?

 

[tiny-tim]

$x_2+x_3+\text{some constant}$?

yes!

(what's wrong with that?? )

 

[Saitama]

Okay so I get,
0=2\ddot{x_2}+2\ddot{x_3}+2\ddot{x_4}
\ddot{x_4}=-(\ddot{x_2}+\ddot{x_3})
$\because \ddot{x_2}=2 m/s^2$ and $x_2=\ddot{x_2}=1 m/s^2$
\ddot{x_4}=-3 m/s^2

This is definitely wrong.

 

[tiny-tim]

acceleration of A is 2m/s² to the left and acceleration of B is 1m/s² to left

 

[Saitama]

Can you please explain a bit more? I am still not able to understand what to do with that info. :(

 

[tiny-tim]

acceleration of A is 2m/s² to the left and acceleration of B is 1m/s² to left

so x₂' = 2, and x₃' = … ?

 

[haruspex]

$x_2=\ddot{x_2}=1 m/s^2$

I assume you meant $\ddot{x_3}=1 m/s^2$, but that's still wrong. Which way is B moving?

 

[PeterO]

Can you please explain a bit more? I am still not able to understand what to do with that info. :(

If A moves left while B is stationary, C would move up.

If B moved left while A was stationary, C would move down.

So the effects of leftward motion of A & B have opposite effects on C.

 

[Saitama]

so x₂' = 2, and x₃' = … ?

Is that a typo or are you actually asking me to calculate the first derivative?

 

[tiny-tim]

sorry, i meant '' not '

 

[Saitama]

sorry, i meant '' not '

But that still doesn't help. Both the accelerations are in the same direction. Should I work on the signs of first derivatives?

 

[tiny-tim]

But that still doesn't help. Both the accelerations are in the same direction.

But x₂ is measured to the left, and x₃ is measured to the right.

 

[Saitama]

But x₂ is measured to the left, and x₃ is measured to the right.

So $\ddot{x_2}=2 m/s^2$ and $\ddot{x_3}=-1 m/s^2$?

But is it wrong to think about the signs of first derivative? I never specified any directions, like you said that x₂ is measured to the right.

 

[tiny-tim]

But is it wrong to think about the signs of first derivative?

i don't see how it helps

I never specified any directions …

yes you did! …

it's clear from your diagram that x₂ and x₃ are always positive, and so must be measured from the fixed end, which is in the middle

 

[Saitama]

it's clear from your diagram that x₂ and x₃ are always positive, and so must be measured from the fixed end, which is in the middle

Okay, so in general, I have to specify the directions and measure distances from a fixed end. I guess I have to do some more practice on these problems. -.-'

Anyways, I get $\ddot{x_4}=-1 m/s^2$. Do I have to put my reasoning for the negative sign like this: Since $x_4$ is positive, the negative sign indicates that the block C moves upwards. Would that be enough?

Thanks a lot tiny-tim!

 

[tiny-tim]

Do I have to put my reasoning for the negative sign like this: Since $x_4$ is positive, the negative sign indicates that the block C moves upwards. Would that be enough?

It probably doesn't need any reasoning.

If you want to provide a reason, just say that x₄ is decreasing, and x₄ is the depth below the main level

 

[Saitama]

If you want to provide a reason, just say that x₄ is decreasing, and x₄ is the depth below the main level

Okay. Thanks again! :)

 

[Tanya Sharma]

Here is my proposed solution :

Instead of five string lengths x1,x2,x3 and two x4's. I will take two string lengths y and z .

y = x1+x2+x4 and z = x3+x4

Now,I will work in the frame of top left block (A) ,the one moving to the left with acceleration 2 m/².In this frame the top right block(B) appears to move with an acceleration 1 m/² towards right .

When the block B moves a distance x towards right ,it produces a change Δy in string y and a change Δz in string z such that 2(x) = Δy + Δz.

Now we see that the middle block (C) is attached from two strings (y and z).This allows us to infer that the displacement of the block would cause an equal displacement in the two strings (y and z) i.e,Δy = Δz.

Now,2(x) = Δy + Δz ,so,2(x) = 2(Δy) ,or x=Δy=Δz.

Movement of block B towards right would cause block C moving upwards.

From this we can infer that the acceleration of block B towards right would produce an equal acceleration in block C in upwards direction ,1m².

If instead of block A ,we work from the frame of block B,we again end up with the same result.

It would be nice if the mentors could confirm my way of approaching the problem

 

[tiny-tim]

Hi Tanya!

How is that quicker than …

0=2\ddot{x_2}+2\ddot{x_3}+2\ddot{x_4}
\ddot{x_4}=-(\ddot{x_2}+\ddot{x_3})

?​

 

[Tanya Sharma]

Hi Tanya!
How is that quicker than …

Hi tiny-tim

I didnt proposed it as a better solution.I just put forward my way of thinking about the problem.The solution looks lengthy as I had to explain the details.All this was done mentally.

The approach outlined by you is very nice and applies in all situations.

Nevertheless,you didnt tell if my approach is okay.

 

[tiny-tim]

you can certainly choose the reference frame of either block

your actual calculations look rather complicated (and i honestly haven't checked them) …

you can simly say that there are two strings to the right block, and two strings to the lowerr block, so the lengthening of one must equal the shortening of the other

(you could also do that in the lab frame, but the fact that there are three pairs of strings can cause confusion as to which are the same sign! )

 

[Tanya Sharma]

Thanks