Tape unwinding from drum

1. Nov 12, 2013

Pranav-Arora

1. The problem statement, all variables and given/known data
A disk of mass M and radius R unwinds from a tape wrapped around it. The tape passes over a frictionless pulley, and a mass m is suspended from the other end. Assume that the disk drops vertically.

a. Relate the accelerations of m and the disk, a and A, respectively to the angular acceleration of disk.

b. Find a, A and $\alpha$.

2. Relevant equations

3. The attempt at a solution
Applying Newton's second law on disk: $Mg-T=MA$ where T is the tension in tape.

For m: $mg-T=ma$. I still need one more equation.

I am not sure but do I have to equate the acceleration along string? For instance, the point in contact with disk has acceleration $A-\alpha R$. Does this mean, $a=A-\alpha R$?

Any help is appreciated. Thanks!

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2. Nov 12, 2013

haruspex

That's about right, but be careful with the signs. Which way are you defining as positive for each variable?

3. Nov 12, 2013

Pranav-Arora

I am defining downward to be positive (the direction of g). But I still don't see why there is a sign error in my equation.

4. Nov 12, 2013

haruspex

There's also the question of which direction of rotation is positive. Taking that to be clockwise in the diagram, I have no problem with the A-αR part. But remember that a is upwards on this side of the pulley. E.g. suppose A=0. A positive a for the descending mass should produce a positive α, but in your equation it gives negative α.

5. Nov 13, 2013

Pranav-Arora

So that means, $a=-A+\alpha R$? This also looks incorrect to me. The clue for the answer states that if A=2a, then $\alpha=3A/R$ but if I substitute a=A/2, I get $\alpha=3A/(2R)$.

6. Nov 13, 2013

haruspex

That's not what I get with that substitution. You might be mixing up your A and a.

7. Nov 13, 2013

Pranav-Arora

I don't see what I did wrong. Given clue: $A=2a \Rightarrow a=A/2$. Plugging this in $a=-A+\alpha R$, $A/2=-A+\alpha R \Rightarrow 3A/2=\alpha R$. Is something wrong with my expression? I have tried to make the equations according to the diagram. Please tell me if something is unclear.

8. Nov 13, 2013

Enigman

Try to use the length (or the change in length) of the tape for another constraint.

9. Nov 13, 2013

haruspex

Oh, sorry - it's me that's getting my A and a confused. Anyway, I insist that A+a = αR. Consider two special case: if m is held still then a = 0 and A = αR; likewise if the centre of M is held fixed than A=0 and a = αR.

10. Nov 13, 2013

Pranav-Arora

Umm...what do I have to do after considering these two cases? I still don't get the right $\alpha$ after substituting a=A/2. :uhh:

11. Nov 13, 2013

Enigman

Let length be L(t)
$\frac{d^2 L}{dt^2}=?$
How does that ^ relate to α? What about a and A?
:zzz:need sleepzzz...

12. Nov 13, 2013

Tanya Sharma

Hi Pranav...

Please write down all the equations you have formed.Somehow I am not able to follow.

13. Nov 13, 2013

Pranav-Arora

Hi Enigman!

I am honestly lost at finding out the expression for L(t).

Let $L_0$ be the initial length of tape. If the block moves down x and the disk moves down y, the new length of string is:

$$L(t)=L_0+x+y$$

Differentiating wrt time twice,

$$\frac{d^2L}{dt^2}=a+A-\alpha R$$.

What should I do with this?

Hi Tanya!

Here are the equations I wrote before this post:
For disk: $Mg-T=MA$
For block: $mg-T=ma$

Relating the accelerations: $a=-A+\alpha R$.

The above equations are written with reference to the attached diagram in #1.

14. Nov 13, 2013

Tanya Sharma

Good...These are correct equations .Now what is the problem ?

15. Nov 13, 2013

Pranav-Arora

The problem is that these equations do not satisfy the given clue for part a. I quote the clue from the book:

What I have is $a=-A+\alpha R$. Substituting a=A/2, $\alpha = 3A/(2R)$ which is incorrect. But if I substitute A=2a, I get $\alpha =3a/R$. Error in the book I guess?

16. Nov 13, 2013

Tanya Sharma

$\alpha =3a/R$ is correct

17. Nov 13, 2013

Pranav-Arora

Thanks haruspex, Tanya and Engiman! :)

@Tanya: Can you please confirm if my expression for d^2L/dt^2 is correct? Also, what am I supposed to do with that? Equate it to zero?

18. Nov 13, 2013

Tanya Sharma

Why are you doing all this ? This isn't required .

19. Nov 13, 2013

Pranav-Arora

Since I have written it down, I would like to know if it's valid or not.

20. Nov 13, 2013

Tanya Sharma

No...that's wrong .

What does the term d2L/dt2 represent ?