# How Does Impulse Affect Acceleration and Tension in a Disk-Particle System?

• WubbaLubba Dubdub
In summary: Yes, the origin is correct.In summary, the particle is accelerated downward by the same amount as the point where the string is attached to the disk.
WubbaLubba Dubdub

## Homework Statement

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

## Homework Equations

$$F = ma$$
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## 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}##
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I can't 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?

WubbaLubba Dubdub said:
Does the particle follow a circular path?
What is the acceleration of the point on the disc where the string attaches?

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

##a + qR##

WubbaLubba Dubdub said:
##a + qR##
Right, so how does that modify the equation of motion of the mass m?

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.

zwierz said:
No forces act on the disk in horizontal direction initially.
Not in the left-to-right direction, no. (It is all horizontal.)

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

haruspex said:
Not in the left-to-right direction, no. (It is all horizontal.)
Pardon, I do not understand little bit.Do you agree that in the left-to-right direction there are no forces at initial moment?

zwierz said:
Do you agree that in the left-to-right direction there are no forces at initial moment?
Yes, I already agreed with that. No acceleration, initially, in the left-to-right direction.

Then it is a solution is not it?

WubbaLubba Dubdub said:
##T - m(a + qR) = \frac{mv^2}{2R}##
Not quite. That would imply the tension is even greater than if the disc were fixed.

zwierz said:
Then it is a solution is not it?
No, what about the other horizontal direction, up and down the page?

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

zwierz said:
As I understand the statement of the problem the disk can not move up or down
Not up or down in space, up or down in the page. It is a plan view.

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

haruspex said:
Not quite. That would imply the tension is even greater than if the disc were fixed.
But if i were to add the forces instead of subtracting them wouldn't it mean that the string would go slack?

DeliriousMistakes said:
But if i were to add the forces instead of subtracting them wouldn't it mean that the string would go slack?
Only if the acceleration of the point of attachment exceeds v2/2R, which presumably it won't.

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.

WubbaLubba Dubdub said:
##a + qR##

haruspex said:
Right,
Actually this needs proof. It is a sole nontrivial point I guess.

WubbaLubba Dubdub said:
provide centripetal acceleration.
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.

DeliriousMistakes and WubbaLubba Dubdub
Finally got it! Thank you so much!
##a = \frac{J^2}{10m^2R}##

haruspex said:
In this case, it is accelerating towards the mass. The fixed quantity is the length of the string.

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

Hi, Vibhor.
Vibhor said:
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## ?
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##.

Vibhor
TSny said:
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.

Could you please explain this .

Vibhor said:
Could you please explain this .

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.

Vibhor
Ok .

Could we choose the point on the disk where string attaches , as the origin of polar coordinate system ?

I am asking this since it is an accelerating point and ##\ddot{r} =0## in this case . How would we work in this system ?

I don't see any advantage of working in a frame accelerating with the point of attachment, ##b##. But, maybe I'm not seeing something.

Vibhor
Fair enough .

Is there a straightforward way to think about this problem rather than using polar coordinates ? I haven't been able to properly understand haruspex's explanation in post#20 .

Last edited:
Vibhor said:
haven't been able to properly understand haruspex's explanation in post#20 .
Then let me reword it...

v2/r is the acceleration necessary for a particle moving at speed v to stay at distance r from a fixed point.
In this problem, it does not need to stay at a fixed distance from a fixed point. The point it must stay at a constant distance from is accelerating towards it anyway. Its acceleration relative to the attachment point needs to be v2/r (where r=2R). If the attachment point is accelerating down the page at aatt then the acceleration of the mass up the page is v2/r-aatt = T/m.

Vibhor
haruspex said:
Its acceleration relative to the attachment point needs to be v2/r (where r=2R).

This is what I was missing .

Thanks a lot .

Thank you very much TSny .

Last edited:

## 1. What is acceleration due to impulse?

Acceleration due to impulse is a measure of how quickly an object's velocity changes when it experiences a sudden force, known as an impulse. It is calculated by dividing the magnitude of the impulse by the mass of the object.

## 2. How is acceleration due to impulse related to Newton's Second Law?

Acceleration due to impulse is directly related to Newton's Second Law, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In the case of impulse, the force is applied over a short period of time, resulting in a change in velocity and acceleration.

## 3. What is the formula for calculating acceleration due to impulse?

The formula for acceleration due to impulse is a = F∆t/m, where a is the acceleration, F is the force applied, ∆t is the time over which the force is applied, and m is the mass of the object.

## 4. How does the duration of the impulse affect the acceleration due to impulse?

The longer the duration of the impulse, the smaller the acceleration due to impulse will be. This is because a longer duration means the force is applied over a longer period of time, resulting in a smaller change in velocity and therefore a smaller acceleration.

## 5. Can acceleration due to impulse be negative?

Yes, acceleration due to impulse can be negative. This would occur if the force applied causes the object to decelerate, resulting in a negative change in velocity and acceleration.

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