To find the angular momentum of a disc

[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.

 

[Frigus]

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

Got it!