To find the angular momentum of a disc

[PSN03]

Homework Statement

A block of mass m is attached to one end of a light string which is wrapped on a disc of mass 2m and radius R. The total length of the slack portion of the string is l. The block is released from rest. The angular velocity of the disc just after the string becomes taut is:

Relevant Equations

Conservation of angular momentum
Initial angular momentum=final angular momentum
Moment of intertia about disc about centre is given by MR²/2

I was first wondering wether we can solve this question by applying conservation or energy or not but after googling it I found that we can't apply conservation of energy since there will be some energy lost in this case. I don't know how this energy is getting lost.
My second doubt was if we apply conservation of angular momentum about centre of the disc then won't there be a torque due to the string? So it won't be valid to use conservation of angular momentum.
Can anyone please clarify these doubts.

 

[anuttarasammyak]

In the figure attached is the distance between the center of pulley and the ( center of mass of ) block is R or more ? That may affect the answer.

 

[Frigus]

I was first wondering wether we can solve this question by applying conservation or energy or not but after googling it I found that we can't apply conservation of energy since there will be some energy lost in this case. I don't know how this energy is getting lost.

It depends upon question,
if it given that no energy is lost during the process then you can apply Law of conservation of energy but in case it is not going given then we assume it to be ideal.

My second doubt was if we apply conservation of angular momentum about centre of the disc then won't there be a torque due to the string? So it won't be valid to use conservation of angular momentum.

Yes you are right.
I think more relevant equation should be
$ω_f$ = $ω_i$ + $\frac {τ}{I}$ $t$.
Try solving it using this equation.

 

[PSN03]

In the figure attached is the distance between the center of pulley and the ( center of mass of ) block is R or more ?

How does it even matter. The potential energy is what we are concerned about. I don't think it's a useful information.

 

[PSN03]

It depends upon question,
if it given that no energy is lost during the process then you can apply Law of conservation of energy but in case it is not going given then we assume it to be ideal.

Yes you are right.
I think more relevant equation should be
$ω_f$ = $ω_i$ + $\frac {τ}{I}$ $t$.
Try solving it using this equation.

According to the solution I have we can't apply conservation of energy. If I do so then I will get a different answer. A similar question was asked on another website and the reason given by a user was
'The thing is that you can't use conservation of energy law in this case. At the moment when the string became taut some kind of inelastic impact would happen and some portion of energy would lost. Imagine, that the string is made of elastic rubber so that no energy would be lost. But in this case there will be some oscillations. In case the string is rigid these oscillations will fade out quickly, but some energy will be lost in the process.'
Though I didn't get much of this explanation as to where will the ineleastic impact happen and between which two bodies.

 

[PSN03]

It depends upon question,
if it given that no energy is lost during the process then you can apply Law of conservation of energy but in case it is not going given then we assume it to be ideal.

Yes you are right.
I think more relevant equation should be
$ω_f$ = $ω_i$ + $\frac {τ}{I}$ $t$.
Try solving it using this equation.

Yeah but the answer will be different if I use this. According to them there is no torque. I guess they are just making some assumptions to get their desired answer.

 

[PSN03]

It depends upon question,
if it given that no energy is lost during the process then you can apply Law of conservation of energy but in case it is not going given then we assume it to be ideal.

Yes you are right.
I think more relevant equation should be
$ω_f$ = $ω_i$ + $\frac {τ}{I}$ $t$.
Try solving it using this equation.

Initial angular velocity=0
w=0+(mg*R/2m*R²/2)*t
t=distance/velocity
=l/√2gl
=√l/2g

Now hen I plug these values I will get final angular momentum which is different from the actual answer

 

[Frigus]

Initial angular velocity=0
w=0+(mg*R/2m*R²/2)*t
t=distance/velocity
=l/√2gl
=√l/2g

Now hen I plug these values I will get final angular momentum which is different from the actual answer

Torque was acting only when string was taut.

 

[PSN03]

Torque was acting only when string was taut.

Yes ...totally agree..before that it was just a freely falling body

 

[Frigus]

Yes ...totally agree..before that it was just a freely falling body

So when you should take t=0?

 

[PSN03]

So when you should take t=0?

Ohh sorry...t=0 should be the time when the block has reached the bottomost point and the string is about to be taut

 

[Frigus]

Ohh sorry...t=0 should be the time when the block has reached the bottomost point and the string is about to be taut

So what is answer now?

 

[haruspex]

At the moment when the string became taut some kind of inelastic impact would happen

It's a bit more complicated than that. We are not told the string is inelastic. (We are not told it is inextensible either, but that's another matter.)
But suppose it is fairly elastic, though maybe with such a high constant that it is almost inextensible. Some of the energy will go into stretching the string, but thence into oscillations. So although some energy is absorbed by the string's inelasticity, some goes into oscillations which do not contribute to the overall subsequent motion.

 

[PSN03]

So what is answer now?

I don't know how but it's coming zero as all variables become zero
Initial angular momentum is zero
Time is also zero
w=0+k*0
=0
Where k is angular acceleration

 

[PSN03]

It's a bit more complicated than that. We are not told the string is inelastic. (We are not told it is inextensible either, but that's another matter.)
But suppose it is fairly elastic, though maybe with such a high constant that it is almost inextensible. Some of the energy will go into stretching the string, but thence into oscillations. So although the energy some is absorbed by the string's inelasticity, some goes into oscillations which do not contribute to the overall subsequent motion.

Ohk...this clears my first doubt...thanks a lot.
But what about the 2nd one, won't there be torque acting ?

 

[haruspex]

Ohk...this clears my first doubt...thanks a lot.
But what about the 2nd one, won't there be torque acting ?

While the block is falling there is no tension in the string so no torque.
What is the velocity just before it becomes taut? What angular momentum does that give the block about the disc's centre?

 

[PSN03]

While the block is falling there is no tension in the string so no torque.
What is the velocity just before it becomes taut? What angular momentum does that give the block about the disc's centre?

I clearly understood what you are trying to say here but my doubt is when the string becomes taut won't there be any kind of force acting and hence producing torque about centre?

 

[Frigus]

I don't know how but it's coming zero as all variables become zero
Initial angular momentum is zero
Time is also zero
w=0+k*0
=0
Where k is angular acceleration

It is the answer of this problem,
Doesn't it matches with answer key?

 

[PSN03]

It is the answer of this problem,
Doesn't it matches with answer key?

No...the final answer is
w=√gl/2R

 

[haruspex]

I clearly understood what you are trying to say here but my doubt is when the string becomes taut won't there be any kind of force acting and hence producing torque about centre?

You want the motion immediately after it becomes taut. There is no time for any continuing torques to affect it.

 

[PSN03]

You want the motion immediately after it becomes taut. There is no time for any continuing torques to affect it.

So you mean to say that just before the tension in the string starts to act and produce some torque we are finding the angular velocity...right?

 

[haruspex]

So you mean to say that just before the tension in the string starts to act and produce some torque we are finding the angular velocity...right?

You need to find the angular momentum of the system about the disc's centre just before tautness (i.e. just that of the falling block, since the disc is not turning yet) and equate that to the angular momentum of the system about that same axis immediately after it becomes taut.

 

[PSN03]

You need to find the angular momentum of the system about the disc's centre just before tautness (i.e. just that of the falling block, since the disc is not turning yet) and equate that to the angular momentum of the system about that same axis immediately after it becomes taut.

Yes I know this part of the problem. The only two things bothering me were cleared by you earlier. Thanks a lot for your help sir, I am truly grateful

 

[PSN03]

You want the motion immediately after it becomes taut. There is no time for any continuing torques to affect it.

What do you mean by continuing torques here?

 

[haruspex]

What do you mean by continuing torques here?

After it has become taut, the weight of the descending block continues to exert tension in the string, so a torque on the disc. But this is beyond the time scope of the question.
I presumed this was what concerned you in post #17, but perhaps I misunderstood.

 

[Frigus]

You need to find the angular momentum of the system about the disc's centre just before tautness (i.e. just that of the falling block, since the disc is not turning yet) and equate that to the angular momentum of the system about that same axis immediately after it becomes taut.

Why can't we use this equation
$ω_f$ = $ω_i$ + $\frac {τ}{I}$ $t$
By this equation answers comes out to be 0.
What is wrong in this reasoning?

 

[PSN03]

After it has become taut, the weight of the descending block continues to exert tension in the string, so a torque on the disc. But this is beyond the time scope of the question.
I presumed this was what concerned you in post #17, but perhaps I misunderstood.

Absolutely correct...I want to give you 100000 likes...just loved your thinking...your logic...your way of answer and everything else sir. I am truly very thankful.
Good day sir and I wish you stay safe and healthy during these hard times.

 

[PSN03]

Why can't we use this equation
$ω_f$ = $ω_i$ + $\frac {τ}{I}$ $t$
By this equation answers comes out to be 0.
What is wrong in this reasoning?

I think it's not actually t=0 but instead it's t tending to 0
So maybe it changes a bit

 

[haruspex]

Why can't we use this equation
$ω_f$ = $ω_i$ + $\frac {τ}{I}$ $t$
By this equation answers comes out to be 0.
What is wrong in this reasoning?

That equation does not represent conservation of angular momentum. It gives the increase in rotation rate of a rigid body when a torque is applied to it for a period t.
When the string becomes taut there is a rotational impulse. This is not the same as a torque. It is to a torque as a linear impulse (momentum change) is to a force.
You can think of it as a very large and unknowable torque acting, and possibly varying, for a very short and unknowable period. We don't need to care exactly what the profile looks like, all we care about is its integral over the period; that is the total change in angular momentum imparted: angular momentum lost by falling block = angular momentum gained by disc.

 

[Frigus]

That equation does not represent conservation of angular momentum. It gives the increase in rotation rate of a rigid body when a torque is applied to it for a period t.
When the string becomes taut there is a rotational impulse. This is not the same as a torque. It is to a torque as a linear impulse (momentum change) is to a force.
You can think of it as a very large and unknowable torque acting, and possibly varying, for a very short and unknowable period. We don't need to care exactly what the profile looks like, all we care about is its integral over the period; that is the total change in angular momentum imparted: angular momentum lost by falling block = angular momentum gained by disc.

So just after the moment string became taut their is no way for the block to move so all the angular momentum goes into disc.
Am I right?
Thanks for clearing doubt.

 

[haruspex]

So just after the moment string became taut their is no way for the block to move so all the angular momentum goes into disc.
Am I right?
Thanks for clearing doubt.

No, the angular momentum gets shared between them, in such a way that the continued downward motion of the block matches the rotation of the disc, keeping the string taut.
It's just like a coalescence in linear collisions.

 

[Frigus]

No, the angular momentum gets shared between them, in such a way that the continued downward motion of the block matches the rotation of the disc, keeping the string taut.
It's just like a coalescence in linear collisions.

I am unable to understand it but I tried to apply my logic again,
When the string became taut then tension will increase enormously so due to which tension decreases the block angular momentum to 0 and the all angular momentum is transferred to disc and here are my calculations,
For block:-
$\int τ ~dt$ = M$\vec V$Rsin90°
Similar for disk:-
$\int τ ~dt$ = $Iω$
And now equating these,
$Iω$ =M$\vec V$Rsin90°
After some little algebra answer comes out to be $\frac {√2gl}{r}$.
Please tell me if it means same as you have written above or different.
Thanks.

No...the final answer is
w=√gl/2R

Friend,please recheck it's answer.

 

[haruspex]

I am unable to understand it but I tried to apply my logic again,
When the string became taut then tension will increase enormously so due to which tension decreases the block angular momentum to 0 and the all angular momentum is transferred to disc and here are my calculations,
For block:-
$\int τ ~dt$ = M$\vec V$Rsin90°
Similar for disk:-
$\int τ ~dt$ = $Iω$
And now equating these,
$Iω$ =M$\vec V$Rsin90°
After some little algebra answer comes out to be $\frac {√2gl}{r}$.
Please tell me if it means same as you have written above or different.
Thanks.

Friend,please recheck it's answer.

You seem still to be assuming the block comes to a stop. It will continue with a reduced velocity, one which matches the rate at which string will unwind from the rotating disc.

 

[Lnewqban]

...
For block:-
$\int τ ~dt$ = M$\vec V$Rsin90°
...

Could you please explain this?
At the impact instant, should we be considering the falling mass:
1) As it were sticking to the edge of the disc and suddenly changing its linear trajectory to rotation?
2) As continuing falling in a linear trajectory at a new slower velocity impossed by the rotational inertia of the disc that it tangentially pulls via the string?

If #1, both moments of inertia (of disc and falling mass), should be considered?
If #2, should we consider the moment of inertia of the disc and the mass of the fallng body?

 

[Frigus]

1) As it were sticking to the edge of the disc and suddenly changing its linear trajectory to rotation?

I found similarity between this question and this example

Of HC Verma(example-24 of chapter 9).
I used statement written in first picture and applied it to the question which is posted here,
Now if tension increases largely then block velocity should became 0 in infinitesimal time and as angular momentum should be conserved then all of the angular momentum should be transferred into disc.
So yes I agree with this point.

If #2, should we consider the moment of inertia of the disc and the mass of the fallng body?

Aren't we just equating the angular momentum and we calculate angular momentum of block using mass and angular momentum of disk using inertia?

 

[haruspex]

if tension increases largely then block velocity should became 0

No. Look at equation (i). It has the block's downward speed reducing from v to V (momentum change = mv-mV). It turns out that V=v/3, so the block's velocity does not become zero.

 

[Frigus]

No. Look at equation (i). It has the block's downward speed reducing from v to V (momentum change = mv-mV). It turns out that V=v/3, so the block's velocity does not become zero.

sorry for this level of dumbness.
I have one another doubt also i.e why do we use integral sign in this question?
If we are just talking about very small interval and not calculating for some finite interval then what is use of integral?

 

[haruspex]

sorry for this level of dumbness.
I have one another doubt also i.e why do we use integral sign in this question?
If we are just talking about very small interval and not calculating for some finite interval then what is use of integral?

It is an accurate representation of what is going on. The force F(t) on one object varies in an unknown way over time, but we know action and reaction are equal and opposite, so -F(t) acts on the other. Since F(t)=m.a(t), $\int F.dt=m\int a.dt=m\Delta v$, the change in momentum. Hence the two objects undergo equal and opposite changes in momentum.

 

[Frigus]

It is an accurate representation of what is going on. The force F(t) on one object varies in an unknown way over time, but we know action and reaction are equal and opposite, so -F(t) acts on the other. Since F(t)=m.a(t), $\int F.dt=m\int a.dt=m\Delta v$, the change in momentum. Hence the two objects undergo equal and opposite changes in momentum.

Is it also correct to say that these equations are valid only for very small interval of time?
And thanks my doubts are cleared now

 

[haruspex]

Is it also correct to say that these equations are valid only for very small interval of time?
And thanks my doubts are cleared now

No, the integrals are valid over any period.

 

[Frigus]

No, the integrals are valid over any period.

I think now it's time to revise old concepts.
Sorry for bothering you,I will comeback again when I will have doubt after working out whole chapter again.
Thanks for helping.

 

[PSN03]

I think now it's time to revise old concepts.
Sorry for bothering you,I will comeback again when I will have doubt after working out whole chapter again.
Thanks for helping.

The best way to solve my problem is to take the disc and mass as a single system. The thing that you and I were doing was to independently study the motion of just disc which was violating the law of conservation of angular moment.

 

[Frigus]

The best way to solve my problem is to take the disc and mass as a single system. The thing that you and I were doing was to independently study the motion of just disc which was violating the law of conservation of angular moment.

Thanks for replying,I just surrendered against this question but it was itching me so I thought I should try to understand it again and you just supported me,
Now if we assume the fact that just after the collision the rate at which string unwinds is equal to the velocity of block then it is very easy to solve,
Momentum just before the collision
M$\vec V$R,
Finding V using $v=u +at$,
$u$ =0,$a=g$ and $\frac {L}{2}$ = $ut$ + $\frac {1}{2}$$a$$t^2$ => $L$ = $at^2$ => $t$ = $√$$\frac {L}{g}$----(1),
Putting (1) in $v=u +at$,
$v$ = $0$ + $g$ $√$$\frac {L}{g}$ => $v$ = $√Lg$----(2),
Putting (2) in M$\vec V$R,
M($√Lg$)R,
Now equating this to final angular momentum,
=>$Iω$ + $MRωR$ = M($√Lg$)R
=> $\frac {MR^2}{2}ω$ + $MωR^2$ = M($√Lg$)R
=> $\frac {3}{2}$ $MR^2ω$ = MR($√Lg$),
dividing both sides by MR,
$\frac {3}{2}$ $Rω$ = ($√Lg$)
=>$ω$ = $\frac {2√Lg}{3R}$
If this is the correct answer then the only thing I can't understand now is that why rate of unwinding is equal to speed of block
Thanks.

 

[haruspex]

why rate of unwinding is equal to speed of block

Because the string is not stretching or shrinking. I guess I don't understand how it isn't obvious.

 

[Frigus]

Because the string is not stretching or shrinking.

It makes too much sense if we talk about the case in which block moves smoothly without any initial angular impulse but in this case one can think of many possibilities like the block will fly off or block stops momentarily and meanwhile rope unwinds making the string slack again e.t.c.
How one can know that this will happen?

 

[haruspex]

It makes too much sense if we talk about the case in which block moves smoothly without any initial angular impulse but in this case one can think of many possibilities like the block will fly off or block stops momentarily and meanwhile rope unwinds making the string slack again e.t.c.
How one can know that this will happen?

Ok, I see your point and it is a good one.

In most mechanics problems at this level some things are idealised - no air resistance, no friction, inextensible strings, perfectly elastic or inelastic...
This is ok if more realistic models converge to the answer as the idealisation is progressively applied. E.g. take the string to have spring constant k, solve, then let the constant tend to infinity. If the question is valid, this should produce the intended answer.
(Occasionally I do come across problems that fail this requirement.)

Applying that here, yes, there would in reality be some bounce. But if we take it as only weakly elastic and with a high spring constant then the bounce can be small, small enough that the string does not become slack, though the tension will reach a peak and decline. We could solve to find the velocity at different times: at max and min subsequent tension, say. Then, letting the elasticity tend to zero and k tend to infinity check that these two velocities converge to the same value. If so, the question is valid and we have its solution.

But this does identify a flaw in the problem statement. It fails to state that the string is both "inextensible" and "inelastic".

 

[PSN03]

Sorry to say buddy but your answer is wrong.

Thanks for replying,I just surrendered against this question but it was itching me so I thought I should try to understand it again and you just supported me,
Now if we assume the fact that just after the collision the rate at which string unwinds is equal to the velocity of block then it is very easy to solve,
Momentum just before the collision
M$\vec V$R,
Finding V using $v=u +at$,
$u$ =0,$a=g$ and $\frac {L}{2}$ = $ut$ + $\frac {1}{2}$$a$$t^2$ => $L$ = $at^2$ => $t$ = $√$$\frac {L}{g}$----(1),
Putting (1) in $v=u +at$,
$v$ = $0$ + $g$ $√$$\frac {L}{g}$ => $v$ = $√Lg$----(2),
Putting (2) in M$\vec V$R,
M($√Lg$)R,
Now equating this to final angular momentum,
=>$Iω$ + $MRωR$ = M($√Lg$)R
=> $\frac {MR^2}{2}ω$ + $MωR^2$ = M($√Lg$)R
=> $\frac {3}{2}$ $MR^2ω$ = MR($√Lg$),
dividing both sides by MR,
$\frac {3}{2}$ $Rω$ = ($√Lg$)
=>$ω$ = $\frac {2√Lg}{3R}$
If this is the correct answer then the only thing I can't understand now is that why rate of unwinding is equal to speed of block
Thanks.

And shouldn't the distance be L and not L/2?
Everything you have done is correct except the L/2 part

 

[haruspex]

shouldn't the distance be L and not L/2?

Yes. I would guess @Hemant was misled by the way the diagram shows the string folded in half.

Everything you have done is correct except [that]

No, there is another error. The disc's mass is 2m, not m.

 

[Frigus]

Sorry to say buddy but your answer is wrong.

And shouldn't the distance be L and not L/2?
Everything you have done is correct except the L/2 part

Thanks for figuring it out,
My answer was not matching with your answer so I wrote the whole solution here to minimize discrepancies.

I would guess @Hemant was misled by the way the diagram shows the string folded in half.

I thought $\frac {L}{2}$ portion is right because their was written total slack portion and to gave it more credibility I searched for meaning of slack and it's meaning was loose and whole rope is loose so i thought total length of rope is L and thus heights is $\frac {L}{2}$.
I am now trying to understand post #46 and it will take me some time so after then I will come here again after understanding it.

 

[haruspex]

I thought $\frac {L}{2}$ portion is right because their was written total slack portion and to gave it more credibility I searched for meaning of slack and it's meaning was loose and whole rope is loose so i thought total length of rope is L and thus heights is $\frac {L}{2}$.

The whole string length is L, so the block will fall a distance L before the string becomes taut.