To find the angular momentum of a disc

In summary: Ohh sorry...t=0 should be the time when the block has reached the bottomost point and the string is about to be...
  • #36
Hemant said:
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.
 
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  • #37
haruspex said:
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.
:doh: 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?
 
  • #38
Hemant said:
:doh: 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.
 
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  • #39
haruspex said:
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😁
 
  • #40
Hemant said:
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.
 
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  • #41
haruspex said:
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.
 
  • #42
Hemant said:
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.
 
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  • #43
PSN03 said:
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😁.
 
  • #44
Hemant said:
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.
 
  • #45
haruspex said:
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?
 
  • #46
Hemant said:
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".
 
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  • #47
Sorry to say buddy but your answer is wrong.
Hemant said:
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
 
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  • #48
PSN03 said:
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.
PSN03 said:
Everything you have done is correct except [that]
No, there is another error. The disc's mass is 2m, not m.
 
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  • #49
PSN03 said:
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.
haruspex said:
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.
 
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  • #50
Hemant said:
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.
 
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  • #51
haruspex said:
The whole string length is L, so the block will fall a distance L before the string becomes taut.
Got it! :doh:
 

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