# Motion of connected objects involving a pulley problem

1. Dec 8, 2015

### Theodore Hodson

1. The problem statement, all variables and given/known data
The diagram shows a particle P lying in contact with a smooth table top 1.5m above the floor. A light inextensible string of length 1m connects P to another particle Q hanging freely over a small smooth pulley at the edge of the table. The mass of each particle is 2 kg, and P is held at a point distant 0.5m from the edge of the table. When the system is released from rest find:

a) the speed of each particle when P reaches the edge of the table
b) the tension in the string

If P slips over the pulley without any change in its speed, find, for the subsequent motion:

c) the acceleration of the system
d) the tension in the string

Give the answers in terms of g.

2. Relevant equations

3. The attempt at a solution

Okay so I managed to figure out the first two parts of the problem. Using $T$ for tension and $a$ for acceleration and then setting up these simultaneous equations
$T-2a=0$
$2g-T=2a$

helped me get $a=\frac{g}{2}$ From here, finding the tension and speed was pretty straightforward (i.e rearranging for $T$ and using $v^2-u^2=2as$ for $s=0.5$ to get the speed)

so answer to part a) is $\frac{\sqrt{g}}{\sqrt{2}}$ and part b) is $g$ N.

Parts c) and d) of the problem are puzzling me though. "P slips over the pulley without any change in its speed" - Basically I'm getting confused about what's going on here visually and which bits of information I need to connect together. So is it saying that P is now falling downwards with the same speed as it had when it was at the edge of the table? But then I'm not sure how this can help me find the new tension/acceleration.

Would be very grateful for any hints here.

2. Dec 8, 2015

### Mister T

What would happen to P if, at the instant it slid over the pulley, Q somehow disappeared (like maybe the string broke or something like that). Once you see that, then think about the actual situation where Q is present.

3. Dec 9, 2015

### haruspex

No change in speed is not enough information. I would interpret it as no (instantaneous) change in velocity.

4. Dec 9, 2015

### Theodore Hodson

If Q somehow disappeared, due to the string breaking, then the string would no longer exert any tension on P. So P would move downwards under the action of its weight alone?? so with an acceleration of g?? I'm not too sure about what's happening where Q is present.

5. Dec 9, 2015

### Theodore Hodson

Okay, am I right in thinking that when P slips over the pulley the string will become slack anyway and hence the tension disappears? Cause going back to the equation for the resultant force acting on Q ($2g-2T=2a$) and plugging in $T=0$ yields an acceleration of g, which is the answer they give in the book. The thing I'm not sure about now is why the question is telling me that P slips over the pulley 'without any change in its speed' and how this piece of information relates to the finding of the acceleration/tension.

6. Dec 9, 2015

### Mister T

Is that the way things move? In the direction of the force?

Think about projectiles, for example. The force is always downward, but they don't always move downwards. That is, the direction of $\vec{F}_{net}$ and the direction of $\vec{v}$ are not necessarily the same.

7. Dec 9, 2015

### Mister T

How about, the pulley exerts no force?

8. Dec 9, 2015

### haruspex

Sure, but the OP didn't put it that way. I was just trying to correct 'speed' to 'velocity'. Maybe they didn't write velocity because there is acceleration, but there's no sudden change to velocity.

9. Dec 9, 2015

### Mister T

I'm looking at it from a teacher's frame of reference, trying to find a better way it could have been worded.

10. Dec 9, 2015

### haruspex

Ok.

11. Dec 9, 2015

### Theodore Hodson

Okay I think I understand. The direction of something's velocity does not always equal the direction of the forces acting on it. Though, I'm still kinda unsure about connecting this clue with finding the acceleration of the system/tension in the string. I managed to find the value for the acceleration the book gives as an answer (g m/s^2) by plugging $T=0$ into one of the original equations, but then I've just assumed that the rope goes slack and exerts no tension when P slips over the edge which I'm not sure is right.

12. Dec 9, 2015

### haruspex

Yes, the acceleration of the two mass system will be g (very simple reason why) but the rope will not be slack.