# Acceleration due to Impulse

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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.