Acceleration due to Impulse

[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?

 

[haruspex]

Does the particle follow a circular path?

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

 

[WubbaLubba Dubdub]

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

$a + qR$

 

[haruspex]

$a + qR$

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

 

[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.

 

[haruspex]

No forces act on the disk in horizontal direction initially.

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

 

[WubbaLubba Dubdub]

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

 

[zwierz]

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?

 

[haruspex]

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.

 

[zwierz]

Then it is a solution is not it?

 

[haruspex]

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

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

 

[haruspex]

Then it is a solution is not it?

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

 

[zwierz]

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

 

[haruspex]

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.

 

[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

 

[DeliriousMistakes]

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?

 

[haruspex]

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.

 

[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.

 

[zwierz]

$a + qR$

Right,

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

 

[haruspex]

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.

 

[WubbaLubba Dubdub]

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

 

[Vibhor]

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

 

[TSny]

Hi, Vibhor.

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]

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 .

 

[TSny]

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 ?

 

[TSny]

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 .

 

[haruspex]

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 a_att then the acceleration of the mass up the page is v²/r-a_att = T/m.

 

[Vibhor]

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

This is what I was missing .

Thanks a lot .

 

[Vibhor]

Thank you very much TSny .