Conservation of Angular Momentum of Puck Problem

[Twiddler]

New to the forums, and I have no experience with LaTex, so if I've done something wrong, I'll try to edit it in a bit.

Homework Statement

A 100 g puck attached to a string is revolving in a 20-cm-radius circle on a frictionless table. The string passes through a hole in the center of the table and is tied to two 200 g weights.

a) What speed does the puck need to support the two weights?
b) One of the two 200 g weights is cut free. What will be the puck's speed and the new radius of its trajectory after this weight is gone?

Homework Equations

$$T = m_2g$$ (Equating tension force with gravitational force of the two weights)

$$v^2 = \frac{m_2gr}{m_1}$$ (Rearranging for the speed of the puck)

Not sure which others I need...

The Attempt at a Solution

So, I managed to solve for the first part of the problem (this question has answers in the textbook, so I use them to verify my results).
However, I'm not sure how I'm supposed to approach the second part of this question. I know that the forces are no longer in balance, but I'm not sure how I use conservation of momentum in this system.
I tried using $$I_iw_i = I_fw_f$$, but that gave me the wrong answer.

I do know that because of the momentum it already has, the radius will increase and the speed will drop, but I really don't know how to approach this, and my TA is pretty anti-social and hard to track down.

Any help on the right approach or formula would be greatly appreciated. Thank you for your time. :D

 

[collinsmark]

Hello Twiddler,

Welcome to Physics Forums!

So, I managed to solve for the first part of the problem (this question has answers in the textbook, so I use them to verify my results).
However, I'm not sure how I'm supposed to approach the second part of this question. I know that the forces are no longer in balance, but I'm not sure how I use conservation of momentum in this system.
I tried using $$I_iw_i = I_fw_f$$, but that gave me the wrong answer.

I think $$I_iw_i = I_fw_f$$ is the right approach. No external torques are applied to the system, so conservation of momentum should hold. Show us your work and maybe we can help figure out what went wrong.

[Edit, by the way, once the weight is cut, the puck no longer will travel in such a simplistic fashion. I'm guessing the question is asking for the instantaneous speed and instantaneous radius at the moment the weight is cut.]

 

[tiny-tim]

Welcome to PF!

… So, I managed to solve for the first part of the problem (this question has answers in the textbook, so I use them to verify my results).
However, I'm not sure how I'm supposed to approach the second part of this question. I know that the forces are no longer in balance, but I'm not sure how I use conservation of momentum in this system.

Hi Twiddler! Welcome to PF!

(have an omega: ω )

Angular momentum will be conserved.

I don't understand what "I" you're using … the angular momentum of a "point" mass is simply r "cross" the momentum, or rxmv.

What do you get? ​

 

[Twiddler]

Hi Twiddler! Welcome to PF!

(have an omega: ω )

Angular momentum will be conserved.

I don't understand what "I" you're using … the angular momentum of a "point" mass is simply r "cross" the momentum, or rxmv.

What do you get? ​

$$I_iw_i = I_fw_f$$

EDIT: Whoops. I was reading notes from a different problem. D'oh!

$$T = m_2g = (0.4 kg)(9.81m/s^2)$$

$$T = 3.924 N$$

$$F_c = F_g = m_1( \frac{v^2}{r})$$

$$\frac{m_2gr}{m_1} = v^2$$

$$v = 2.8m/s$$

So I've got the speed of the system.

Now, if $$I_iw_i = I_fw_f$$ is the way to solve this problem, then my issue becomes that I don't have the final radius of the new path, or its speed. So I can't do simple isolation for it.

Do I need to recalculate the new equilibrium of forces to derive my new balance of momentum? I only worked that out in the first part because I was given the radius.

Oh, I forgot to mention, I paraphrased the question so no one could just Google the answer.
(I'm not the only one stuck here, and I didn't want them to find it up on Yahoo! Answers or what have you.)

The second weight is actually a bag of sand that gets punctured and uniformly drains out (this is to solve that problem caused by the sudden absence of weight). I don't know if there's some formula I can use that takes this into account, so at this point I'm stuck.

 

[Twiddler]

Okay, so, many edits later (*mumble mumble LaTex syntax mumble*)...

So, I solved for the angular velocity of the old system, which is 14 rad/s.
So, I have:

$$I_iw_i = I_fw_f$$

$$(0.2 kg)(14 rad/s) = mr^2w_f$$

 

[tiny-tim]

Oh, I forgot to mention, I paraphrased the question so no one could just Google the answer.
(I'm not the only one stuck here, and I didn't want them to find it up on Yahoo! Answers or what have you.)

The second weight is actually a bag of sand that gets punctured and uniformly drains out (this is to solve that problem caused by the sudden absence of weight).

ahhh! i was wondering how the motion managed to stay circular!

ok, you have found an equation which is valid for any mass: m₂gr = m₁v²

Since you know angular momentum is conserved, you now need to find how r depends on m₂ if angular momentum stays constant.

 

[Twiddler]

I solved it!

Yeah, this turns into quite a mess, so here's the solution:

$$m_1v_1r_1 = m_2v_2r_2$$ (Another way of showing conservation of angular momentum)

$$v_2 = \frac{v_1r_1}{r_2}$$ (We'll use this after we find the radius)

So, I figured out that $$\frac{mv_2^2}{r_2} = Mg$$ (This is how we find our new equilibrium with the missing weight)

If we rearrange this for $$v_2$$, then we get:

$$v_2 = \sqrt{\frac{Mgr_2}{m}}$$

So, now we equate the two for:

$$\frac{v_1r_1}{r_2} = \sqrt{\frac{Mgr_2}{m}}$$

At this point, I'm feeling pretty confident, so I start messing around with it.

I end up with:

$$r_2 = \left [\frac{v_1^2r_1^2m}{Mg} \right]^1/3$$

At this point, I'm almost positive I'm wrong, 'cause it looks horrendous and evil and such.
But I've been stuck on it for hours, and at this point, being wrong again doesn't really hurt me, so, I plug all my numbers in and solve and I get...

$$r_2 = \left[\frac{(2.8 m/s)^2(0.2m)^2(0.1 kg)}{(0.2 kg)(9.81 m/s^2)}\right]^{1/3}$$

$$= 0.25 m$$

And then I look in the back of the book and...it's right! YAY!

At this point, I plug in the radius into the above equation for $$v_2$$ and get 2.2 m/s, which is also right.

The moral of the story is...you really can bash your head against a wall enough times to bring it down. Just got to swing your head hard enough, and not be afraid of wonky exponents.

Thanks for all your help, guys. :)

Edit: I can't get the end number to stay up at the top of the brackets as an exponent...hmm...

 

[nazerofsun]

agreed. ^^

 

[tiny-tim]

Hi Twiddler!

(just got up :zzz: …)

… The moral of the story is...you really can bash your head against a wall enough times to bring it down. Just got to swing your head hard enough, and not be afraid of wonky exponents.

he he

yes … the "bash your head against a wall enough times to bring it down" method really works! …

but just in case you don't have time for that in the exam (or a seat near enough to the wall), here's a tip …

it's all about rearranging the equation you have so that the bit that is constant stands out …

here, you had mv²/r = Mg, and you knew vr was constant …

so rewrite it as m(vr)²/r³ = Mg …

so if M is multiplied by 0.5, … ? ​