Find the tension in the string of this two mass system

[PSN03]

Homework Statement

Two small masses each of mass m are tied through a light inextensible string of lenght l and placed on smooth horizontal surface. At t=0 a sudden impulse delP is given to one of the two masses. Find the tension in the string at t=0 and acceleration of the other mass.

Relevant Equations

delP=delmv (Impulse=change in momentum)
Tension=mv²/r for circular motion

I know that it will probably execute something similar to circular motion. I thought of conserving momentum but I think there is an external force being applied due to the impulse which will prevent me from doing so.
I know that once I find the tension it would become very easy to find the acceleration.
Any help or hint would be appreciated.

 

[etotheipi]

What's the motion of the centre of mass after the impulse has been applied?

 

[PSN03]

What's the motion of the centre of mass after the impulse has been applied?

How do I do that?
I know that the centre of mass is at L/2 distance but what to do after that

 

[etotheipi]

You apply the impulse, and let's say that the mass it's applied to gains a speed $v$ almost instantaneously. (Imagine, for convenience, that the string is horizontal and the impulse applied to one of the masses in the vertical direction.)

The total momentum of a system equals the total mass times the velocity of the centre of mass, i.e. $\vec{P} = M\vec{V}_{CM}$.

What is the total momentum in the $y$ direction? Then, what is the $y$ velocity of the centre of mass?

 

[PSN03]

You apply the impulse, and let's say that the mass it's applied to gains a speed $v$ almost instantaneously. (Imagine, for convenience, that the string is horizontal and the impulse applied to one of the masses in the vertical direction.)

The total momentum of a system equals the total mass times the velocity of the centre of mass, i.e. $\vec{P} = M\vec{V}_{CM}$.

What is the total momentum in the $y$ direction? Then, what is the $y$ velocity of the centre of mass?

Ohk...I got this. By the way the string is horizontal and impulse is in vertical direction.
By the way you said that the total momentum is total mass times the velocity of centre of mass so shouldn't the total mass be 2M instead of M?

 

[etotheipi]

Ohk...I got this. By the way the string is horizontal and impulse is in vertical direction.
By the way u said that the total momentum is total mass times the velocity of centre of mass so shouldn't the total mass be 2M instead of M?

That's correct; I used the variable $M = 2m$. So what do you get?

 

[PSN03]

That's correct; I used the variable $M = 2m$. So what do you get?

I will get P=2m*V=mv
(V=m*v+m*0/m+m=v/2)

 

[etotheipi]

I will get P=2m*V=mv
(V=m*v+m*0/m+m=v/2)

That's right. Since no external forces are acting after the initial impulse, the velocity of the centre of mass of constant. This means that the centre of mass frame is inertial.

What are the velocities of both masses in the centre of mass frame / relative to the centre of mass?

 

[PSN03]

That's right. Since no external forces are acting after the initial impulse, the velocity of the centre of mass of constant. This means that the centre of mass frame is inertial.

What are the velocities of both masses in the centre of mass frame / relative to the centre of mass?

Velocity of Vcm=v/2
Velocity of block 1 with respect to centre of mass would be=v-v/2=v/2
Velocity of block 2 with respect to centre of mass would be=0-v/2=-v/2

 

[etotheipi]

Velocity of Vcm=v/2
Velocity of block 1 with respect to centre of mass would be=v-v/2=v/2
Velocity of block 2 with respect to centre of mass would be=0-v/2=-v/2

Okay, you should be in a position to finish solving it now. The centre of mass, the centre of the string, is fixed in this frame. What type of motion do you get?

 

[PSN03]

Okay, you should be in a position to finish solving it now. The centre of mass, the centre of the string, is fixed in this frame. What type of motion do you get?

I guess it should b something like a circular motion with one mass moving up and the other one down.
Now I guess I can use T=mv²/r=mv²/4l/2=mv²/2l

 

[etotheipi]

That should do the trick!

 

[PSN03]

That should do the trick!

Yes!
You are absolutely right. Thanks a lot!
So to summarise evrything :
1. No external force is acting after the application of impulse therefore conservation of linear momentum can be applied.
2.calculating the velocity of COM
3.finding relative velocity.
4.finding tension

By the way I have a doubt. Why did we chose the centre of mass frame in this problem?

 

[etotheipi]

By the way I have a doubt. Why did we chose the centre of mass frame in this problem?

Mainly because I think it's easier to visualise.

You could alternatively transform into the frame of reference of, say, the ball on the left, about which the ball on the right will perform circular motion. All that is required is to compute the relative acceleration of the ball on the right w.r.t. that on the left in terms of $T$ and $m$, and set this equal to $\frac{v^2}{r}$ where $v$ is the relative velocity of the ball on the right (initially, just $v$).

And for some context, if you choose to do the second method the ball on the right feels an inertial "pseudo-force" of T to the left in addition to the tension. But that's not actually necessary to know to solve the problem!

 

[kuruman]

A more direct way would be to conserve angular momentum about the center of mass.
1. Find the angular impulse about the CM delivered to the system.
2. Set that equal to $I\omega$ and solve for $\omega$.
3. $T=m\omega^2 r$ (what is $r$?).

 

[etotheipi]

A more direct way would be to conserve angular momentum about the center of mass.
1. Find the angular impulse about the CM delivered to the system.
2. Set that equal to $I\omega$ and solve for $\omega$.
3. $T=m\omega^2 r$ (what is $r$?).

It's a good problem because it forces you to justify how the choice of reference frame influences your calculations. The CM frame is non-inertial whilst the impulse is applied and there is a downward impulse acting through the CM due to the resulting inertial force, so that the total momentum in the CM frame is still zero after the impulse. But this inertial force produces no torque about the centre of mass, so we can indeed find the final angular momentum about the CM frame.

 

[PSN03]

It's a good problem because it forces you to justify how the choice of reference frame influences your calculations. The CM frame is non-inertial whilst the impulse is applied and there is a downward impulse acting through the CM due to the resulting inertial force, so that the total momentum in the CM frame is still zero after the impulse. But this inertial force produces no torque about the centre of mass, so we can indeed find the final angular momentum about the CM frame.

Why isn't the impulse producing a torque cause according to em it should. As F*delt=delP which will create a force and this force should generate a torque. So how are we conserving linear and angular momentum.
Are u trying to say that we are conserving it after the impact and not before the impact?

 

[jbriggs444]

The CM frame is non-inertial whilst the impulse is applied

The idea of choosing an accelerating frame while an impulse is applied is not wise. One would typically want to use either a before frame or an after frame and express all quantities in terms of coordinates anchored to the chosen frame.

Often one would choose to change from one frame to another at the impulsive event, taking care to transform all relevant quantities from the before-frame to the after-frame.

 

[etotheipi]

The idea of choosing an accelerating frame while an impulse is applied is not wise. One would typically want to use either a before frame or an after frame and express all quantities in terms of coordinates anchored to the chosen frame.

Often one would choose to change from one frame to another at the impulsive event, taking care to transform all relevant quantities from the before-frame to the after-frame.

Right, sure. I just did it in this case to demonstrate that linear momentum is only conserved in the CM frame because of an inertial force.

Why isn't the impulse producing a torque cause according to em it should. As F*delt=delP which will create a force and this force should generate a torque. So how are we conserving linear and angular momentum.
Are u trying to say that we are conserving it after the impact and not before the impact?

There is an angular impulse; you're right, angular momentum is not conserved before and after the impulse. But the impulse is the change in angular momentum, so if you know the initial angular momentum is zero, then you know the final angular momentum equals the change. The angular impulse is $\frac{l}{2} \Delta P$ about the CM.

 

[PSN03]

Right, sure. I just did it in this case to demonstrate that linear momentum is only conserved in the CM frame because of an inertial force.
There is an angular impulse; you're right, angular momentum is not conserved before and after the impulse. But the impulse is the change in angular momentum, so if you know the initial angular momentum is zero, then you know the final angular momentum equals the change. The angular impulse is $\frac{l}{2} \Delta P$ about the CM.

Can we say that after the impact since no force is acting hence we can conserve both angular momentum as well as linear momentum?

 

[jbriggs444]

Why isn't the impulse producing a torque cause according to em it should. As F*delt=delP which will create a force and this force should generate a torque. So how are we conserving linear and angular momentum.
Are u trying to say that we are conserving it after the impact and not before the impact?

My preferred approach is to examine the system entirely from the perspective of the center of mass frame after the impulse is already complete. We do not need to know how the objects came to acquire their initial velocities. All we need to know is what those initial velocities are.

Sure, the impulse means that energy, linear momentum and angular momentum are all different from what they were a moment before the event. But we do not care.

The very first thing that I did when examining the question was to realize that the asked-for quantities (tension and acceleration) would be undefined exactly at the instant of the impulse. The tension would be discontinuously changing. I then re-interpreted the question to be asking about the tension a very short time after the impulse. It is then abundantly clear what frame should be used to do the analysis. One should use the post-impulse frame.

[One can always do the analysis in any frame, of course. But why make one's life difficult with a sub-optimal choice?]

 

[etotheipi]

Can we say that after the impact since no force is acting hence we can conserve both angular momentum as well as linear momentum?

Sure, so long as your frame is inertial.

 

[PSN03]

Sure, so long as your frame is inertial.

I also hav another doubt about the tension that we found out. The tension here is obtained by considering the relative velocity of the body with respect to COm, so will this tension be equal to the actual tension? And will the tension remain same after some time? Cause according to me it should be same as there isn't anything affecting it to change.

 

[etotheipi]

Real forces are invariant under changes of reference frame. So yes.

(I believe this is the case, if not then I'm sure @jbriggs444 will correct me )

 

[jbriggs444]

I also hav another doubt about the tension that we found out. The tension here is obtained by considering the relative velocity of the body with respect to COm, so will this tension be equal to the actual tension? And will the tension remain same after some time? Cause according to me it should be same as there isn't anything affecting it to change.

The tension is a real property of the string. It does not depend on frame of reference or coordinate system or anything else. If you can argue correctly that tension is constant based on considerations using one frame of reference then the tension must be constant regardless of what other frame might be used to re-analyze the problem.

Edit: Beaten to exactly the same argument by @etotheipi

 

[PSN03]

Real forces are invariant under changes of reference frame. So yes.

(I believe this is the case, if not then I'm sure @jbriggs444 will correct me )

Ohk thanks a lot. Also will the tension change after a while?

 

[jbriggs444]

Ohk thanks a lot. Also will the tension change after a while?

Two objects rotating on a frictionless surface -- sounds like a long-term stable situation to me.

 

[PSN03]

Two objects rotating on a frictionless surface -- sounds like a long-term stable situation to me.

Thanks a lot to everyone for helping me out. I don't have any doubts now and I hope I won't have any in future. You people are great :). Thank you once again and I hope you guys stay safe.