# Homework Help: Acceleration due to Impulse

Tags:
1. May 1, 2017

### WubbaLubba Dubdub

1. The problem statement, all variables and given/known data
A uniform circular disc of mass 2m and radius R placed freely on a horizontal smooth surface as shown in the figure. A particle of mass m is connected to the circumference of the disc with a mass less string. Now an impulse J is applied on the particle in the directions shown in the figure. The acceleration of center of mass of the disc just after application of impulse is

2. Relevant equations

$$F = ma$$

3. The attempt at a solution
assuming tension T in the string
Force acting on particle = T
Force acting on disc, $T = 2ma$ , a is the acceleration of center of mass of disc
Torque on disc $TR = Iq$ , q is the angular acceleration
hence $TR = \frac {2mR^2} {2}q$
thus $T = mRq$
also since impulse is change in momentum, velocity of particle, $v= \frac{J}{m}$

I cant seem to find a relation between velocity and tension. Does the particle follow a circular path? If so, do i take tension to be equal to the centripetal force?

2. May 1, 2017

### haruspex

What is the acceleration of the point on the disc where the string attaches?

3. May 1, 2017

### WubbaLubba Dubdub

$a + qR$

4. May 1, 2017

### haruspex

Right, so how does that modify the equation of motion of the mass m?

5. May 1, 2017

### zwierz

The problem looks strange. Initially no forces act on the disk in horizontal direction . Acceleration of the disk's center is equal to zero at initial moment of time.

6. May 1, 2017

### haruspex

Not in the left-to-right direction, no. (It is all horizontal.)

7. May 1, 2017

### WubbaLubba Dubdub

$T - m(a + qR) = \frac{mv^2}{2R}$

8. May 1, 2017

### zwierz

Pardon, I do not understand little bit.Do you agree that in the left-to-right direction there are no forces at initial moment?

9. May 1, 2017

### haruspex

Yes, I already agreed with that. No acceleration, initially, in the left-to-right direction.

10. May 1, 2017

### zwierz

Then it is a solution is not it?

11. May 1, 2017

### haruspex

Not quite. That would imply the tension is even greater than if the disc were fixed.

12. May 1, 2017

### haruspex

No, what about the other horizontal direction, up and down the page?

13. May 1, 2017

### zwierz

As I understand the statement of the problem the disk can not move up or down

14. May 1, 2017

### haruspex

Not up or down in space, up or down in the page. It is a plan view.

15. May 1, 2017

### zwierz

O now I see, the picture is a view from above on a table where the disk lies. Then it is a good problem

16. May 1, 2017

### DeliriousMistakes

But if i were to add the forces instead of subtracting them wouldn't it mean that the string would go slack?

17. May 1, 2017

### haruspex

Only if the acceleration of the point of attachment exceeds v2/2R, which presumably it won't.

18. May 1, 2017

### WubbaLubba Dubdub

I got confused...shouldn't $ma + mqR$ be because of the tension in the string and so, the net force acting on the particle is T which would provide centripetal acceleration.

19. May 1, 2017

### zwierz

Actually this needs proof. It is a sole nontrivial point I guess.

20. May 1, 2017

### haruspex

But how much acceleration is needed? The standard formula v2/r assumes a fixed centre of rotation. In this case, it is accelerating towards the mass. The fixed quantity is the length of the string. The acceleration the mass needs to achieve that is reduced to the extent that the other end of the string is accelerating towards it.

21. May 1, 2017

### WubbaLubba Dubdub

Finally got it! Thank you so much!
$a = \frac{J^2}{10m^2R}$

22. May 2, 2017

### Vibhor

Hi haruspex ,

So , if the string length is constant , does this mean the particle is accelerating downwards (in the direction along the length of string away from disk ) with same acceleration as that of the point where string is attached to disk I.e $a+qr$ ?

I tried approaching this problem mathematically using polar coordinates .

Let the origin be a fixed point on the table just below the point where string leaves the disk and use polar coordinates $r$ and $\theta$ for the position of the particle. In polar coordinates the radial component of acceleration is $a_r = \ddot{r} – r\dot{\theta}^2$

If I use $\ddot{r} = a+qR$ , the equation of motion of particle would be

$-T = m[(a+qR) - \frac{v^2}{2R}]$ .

This does give correct result

Is my choice of origin correct ?

Have I used correct value of $\ddot{r}$ considering the string constraint ? I am especially quite doubtful in this .Is particle actually accelerating downwards instantaneously with magnitude ( a+qR ) ?

In case haruspex has signed off for the day , would @TSny look at my approach .

Thanks

23. May 2, 2017

### TSny

Hi, Vibhor.
Did you mean to write $a+qR$ here?

The particle is not accelerating downward. The particle is accelerating upward (note the direction of the force on the particle).

However, in your polar coordinate system, you can show that $\ddot{r} = a+qR$ just after the impulse. That is, $\ddot{r}$ does equal the magnitude of the acceleration of the point of attachment of the string to the disk.

But $\ddot{r}$ does not represent the downward acceleration of the particle. The downward acceleration of the particle is $\ddot{r} - r\dot{\theta}^2$.

24. May 2, 2017

### Vibhor

Could you please explain this .

25. May 2, 2017

### TSny

Let $b$ denote the point of attachment of the string to the disk and let $p$ refer to the particle. Let $\hat{r}$ be the radial unit vector in your polar coordinate system. $\hat{r}$ points downward at the time of interest.

Relative acceleration formula: $\vec{a}_p = \vec{a}_{p/b} + \vec{a}_b$

At the time of interest
$\vec{a}_p = (\ddot{r} - r\dot{\theta}^2) \hat{r} \,\,\,$ (1)
$\vec{a}_{p/b} = - r\dot{\theta}^2 \hat{r} \,\,\,\;\;\;\;$ (2)

To obtain(2), use the fact that the string can't stretch.

Substitute (1) and (2) into the relative acceleration formula.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted