Collision involving blocks and pulley

[Tanya Sharma]

Homework Statement

Homework Equations

The Attempt at a Solution

When the particle of mass M/2 collides with the pan ,an impulsive tension acts on the (pan+particle) . Let us call this J . Tension acting in the string between the two blocks is T .

Speed of particle before collision = u = √(gl)
Speed of (pan+particle) after collision = v

Applying impulse momentum theorem on (pan+particle), Mu/2 -J = Mv/2

Applying impulse momentum theorem on block A, J = Mv

From the above two equations we get v = √(gl)/3 .

Is my reasoning correct ? How should I proceed ? I would be grateful if somebody could help me with the problem .

 

[andrevdh]

Since A and B is connected by a taught string shouldn't the magnitude of their acceleration be the same?
Problem is the tension in the string between A and B also changes during the collision.
So I am getting the impression that you need to consider the system as a whole.
Setup looks very much like a swing/pendulum on the left.

 

[Tanya Sharma]

Since A and B is connected by a taught string shouldn't the magnitude of their acceleration be the same?

Yes .

Problem is the tension in the string between A and B also changes during the collision.
So I am getting the impression that you need to consider the system as a whole.
Setup looks very much like a swing/pendulum on the left.

So , how should I proceed ?

 

[AlephNumbers]

So , how should I proceed ?

What would you expect to happen if you dropped a mass on the edge of a pan that is connected by a string to the ceiling?

 

[TSny]

Is my reasoning correct ? How should I proceed ? I would be grateful if somebody could help me with the problem .

I think you are correct so far. Draw a free body diagram for each mass M. Apply 2nd law to the vertical direction for each mass.

The vertical acceleration of the mass M on the left will consist of a superposition of two parts: one part from the change in length of the left portion of the string and one part due to the fact that the sideways motion makes the mass move in circular motion.

 

[Tanya Sharma]

Hello TSny ,

Thanks for replying .

Draw a free body diagram for each mass M. Apply 2nd law to the vertical direction for each mass.

The vertical acceleration of the mass M on the left will consist of a superposition of two parts: one part from the change in length of the left portion of the string and one part due to the fact that the sideways motion makes the mass move in circular motion.

Assuming tension T exists in the string connecting A and B after the collision and A moves upwards as well as towards left and B downwards .

For A , $T - Mg = M(a+\frac{v^2}{l})$ ,

For B , $Mg-T = Ma$

Does it make sense ?

 

[TSny]

Yes. You are letting "$a$" be the upward acceleration of A due to the changing length (and therefore the downward acceleration of B). Looks good.

 

[TSny]

Do you think that B will move downward?

 

[Tanya Sharma]

But this gives $M(2a+\frac{v^2}{l}) = 0$ ? Does this mean , my assumption that A going upwards was wrong . Instead B moves upwards with acceleration g/18 .

Is that so ?

 

[TSny]

But this gives $M(2a+\frac{v^2}{l}) = 0$ ? Does this mean , my assumption that A going upwards was wrong . Instead B moves upwards with acceleration g/18 .

Is that so ?

I think so. I also get that B accelerates upward at g/18.

 

[TSny]

Here’s another way to look at it. Treat masses A and B as one system. The gravity force on each side cancels. So, the net force accelerating the system is just the “centrifugal” force due to the swinging of A.

So, for the system: $F = ma$ gives $Mv^2/l = (2M)a$. So $a = v^2/(2l)$ where $v^2 = gl/9$.

 

[Tanya Sharma]

I think so. I also get that B accelerates upward at g/18.

The answer given is option b) i.e g/9

 

[TSny]

The answer given is option b) i.e g/9

Good. We must still have something to learn! I don't see a mistake, so we can hope that someone will set us straight.

 

[Tanya Sharma]

Here’s another way to look at it. Treat masses A and B as one system. The gravity force on each side cancels. So, the net force accelerating the system is just the “centrifugal” force due to the swinging of A.

So, for the system: $F = ma$ gives $Mv^2/l = (2M)a$. So $a = v^2/(2l)$ where $v^2 = gl/9$.

How does gravity cancels if we treat A and B as one system ? It is acting downwards on both the blocks .

 

[TSny]

Weight of A tries to make the system move counterclockwise while weight of B acts clockwise.

 

[Tanya Sharma]

We get g/9 as answer if we consider only the centripetal acceleration of block A .

For block A , $T - Mg = M\frac{v^2}{l}$ . So, T = (10/9)Mg

Now for block B , acceleration = net force /mass . So acc = (T - Mg) / M = g/9 .

But the problem I see is that if this is the case , both the blocks then accelerate upwards with g/9 . Doesn't look convincing .

I saw the solution given , and this is how the option b) is obtained .

 

[TSny]

Doesn't look convincing .

I agree. This solution appears to neglect the downward acceleration of A due to the changing length of the string on the left.

 

[Tanya Sharma]

This solution appears to neglect the downward acceleration of A due to the changing length of the string on the left.

How does leftwards impulse on block A gives it a downwards acceleration ? I understand that it makes A move in circular path about the top pulley , but how does it provide downward acceleration to A ?

Downwards acceleration of block A means that string length of left string increases in vertical direction and that of right string decreases in vertical direction. Isn't that so ?

I would like to understand your reasoning .

 

[TSny]

Downwards acceleration of block A means that string length of left string increases in vertical direction and that of right string decreases in vertical direction. Isn't that so ?

The string on the left side is accelerating downward, but mass A has a total acceleration upward due to the upward centripetal acceleration. For mass A, the net upward acceleration is

$a_{net, A} = v^2_A/l - a$

where $a$ is the downward contribution to the acceleration of A due to the lengthening of the string on the left. Note, $a = a_{B}$ where $a_B$ is the upward acceleration of B.

2nd law for mass A: $T-Mg = M(v_A^2/l-a_B)$

For mass B: $T-Mg = Ma_B$.

These are the essentially the same as you already set up.

 

[Tanya Sharma]

Sorry...but some confusion has crept in .

So,basically both A and B have a net acceleration g/18 upwards . Right ?

The string on the left side is accelerating downward

How ? I can't visualize it .

I still do not see how leftwards movement of A causes vertically lengthening of the left string . I can see horizontal lengthening of left string causing an overall lengthening of left string .Please help me understand how A accelerates downwards i.e how string length of left string increases in vertical direction.

 

[TSny]

Sorry...but some confusion has crept in .

So,basically both A and B have a net acceleration g/18 upwards . Right ?

Yes. Just after the collision A and B experience exactly the same vertical forces: Mg acts down on each mass while T acts up on both masses. So, A and B must have the same vertical component of acceleration at that time.

How ? I can't visualize it .

I still do not see how leftwards movement of A causes vertically lengthening of the left string . I can see horizontal lengthening of left string causing an overall lengthening of left string .Please help me understand how A accelerates downwards i.e how string length of left string increases in vertical direction.

That was poor wording on my part. If we let $r$ be the length of the string on the left side, then we agree that $\ddot{r} > 0$. We are interested in the time just after the collision when the string on the left is still vertical. The vertical component of acceleration of A, with upward taken as positive, is given by $a_y = v_A^2/r - \ddot{r}$. Thus, $\ddot{r}$ is seen to contribute in the downward direction to the acceleration of A. That’s all I should have said. I did not mean to imply that the "length of left string increases in vertical direction". That would have more to do with y-component of velocity of A rather than acceleration.

 

[Tanya Sharma]

The vertical component of acceleration of A, with upward taken as positive, is given by $a_y = v_A^2/r - \ddot{r}$.

I think i am messing up with the basics . Why can't $a_y = v_A^2/r$ ? How did you write the expression $a_y = v_A^2/r - \ddot{r}$ ? Can you relate this expression to some other simple kinematic problem ?

 

[TSny]

If $r$ is the length of the left string and $\theta$ is the angle of the left string from vertical, then $r$ and $\theta$ are just polar coordinates for A with the origin at the point where the string meets the pulley. The radial component of acceleration in polar coordinates is $a_r = \ddot{r} - r\dot{\theta}^2$. When the string is vertical, the radial direction is downward. So, the upward component of acceleration is $a_y = - a_r = r\dot{\theta}^2 - \ddot{r} = v_A^2/r - \ddot{r}$.

See equation (4) here:
http://ocw.mit.edu/courses/aeronaut...fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf

 

[Tanya Sharma]

My sincere apologies for being so careless . I understood the expression but just as I was about to edit my previous post , saw your above response .

 

[Tanya Sharma]

Doesn't $\ddot{r}>0$ imply that left string length increases in vertical direction at time just after the collision ?

 

[TSny]

Just after the collision, $\dot{r} = 0$ and $\ddot{r} > 0$. This means that $r$ will increase. So, the string on the left will get longer. But it doesn't necessarily mean that the string length increases "in vertical direction". That is, block A does not necessarily move downward as it moves to the left.

 

[Tanya Sharma]

Okay .

How did you determine that $\ddot{r} > 0$ ? I am asking this because earlier I was incorrectly believing that the string length changes is the reason behind $\ddot{r} > 0$ .

 

[TSny]

How did you determine that $\ddot{r} > 0$ ? I am asking this because earlier I was incorrectly believing that the string length changes is the reason behind $\ddot{r} > 0$ .

Applying Newton's second law to each mass and solving for $a_B$ gave us a positive answer for $a_B$. And we have $\ddot{r} = a_B$.

 

[Tanya Sharma]

I regret my inability to understand trivial issues . Thanks for your time and energy in putting up the explanations .The question is a bit tricky . Even good teachers and students I know are fumbling on this question . Everyone is going by the solution I mentioned previously.I am trying hard to understand this question .

Applying Newton's second law to each mass and solving for $a_B$ gave us a positive answer for $a_B$.

You mean by solving $T-Mg = M(v_A^2/l-a_B)$ and $T-Mg = Ma_B$ , we get $a_B = v_A^2/2l$ , a positive number. Right ?

Does that mean that the component of radial acceleration $r\dot{\theta}^2$ i.e $v_A^2/l$ necessitates the presence of $\ddot{r}$ component ,else the two force equations would not be consistent ?

I mean , just as in the solution given they have assumed $a=0$ , but then it made the two force equations inconsistent.

Is that so ?

And we have $\ddot{r} = a_B$.

Sign issues .

1) If we take downward positive for both A and B , would we have written $\ddot{r} = -a_B$ ?

2) If we take downward positive for both A and B , should the force equations be $Mg-T = M(a_B-v_A^2/l)$ and $Mg-T = -M(a_B)$ ?

But then we discussed in a previous thread that we consider signs with forces , not with 'a' in ΣF = Ma . Why should we put a negative sign in front of $a_B$ ?

3) Since in polar coordinates with origin at pulley, increasing r is downwards , sign of $a_B$ should be negative irrespective of whether we choose upwards or downwards as positive or negative ?

4) How do we determine sign of $\ddot{r}$ term ?

Again, thanks for your patience . I hope it is not getting irksome for you .

 

[TSny]

Let $r$ be the length of the string on the left and $s$ be the length of the string on the right. Let’s take downward as the positive y-direction for both A and B. The 2nd law is

$\sum{F_{A,y}} = M a_{A,y}$ and $\sum{F_{B,y}} = M a_{B,y}$

Then $a_{A,y} = \ddot{r} - \frac{v_A^2}{r^2}$. No assumption is being made about whether or not $\ddot{r}$ is a positive or negative quantity. Solving the equations will determine the value of $\ddot{r}$.

$a_{B,y} = \ddot{s}$. No assumption is being made about whether or not $\ddot{s}$ is positive or negative. But $a_{B,y}$ and $\ddot{s}$ must have the same sign if positive y is downward.

We have the constraint that $r + s$ = const. So, $\ddot{r} =-\ddot{s}$. Or, $\ddot{r} =-a_{B,y}$.

So, the 2nd law equations are

$Mg-T = M \left (-a_{B,y} - \frac{v_A^2}{r^2} \right )$ and $Mg-T = Ma_{B,y}$

Solving gives $a_{B,y} = -\frac{v_A^2}{2r^2} = -\frac{g}{18}$. So, B accelerates upward at g/18.