Homework Help: Some more questions

1. Dec 12, 2012

pious&peevish

Some more urgent questions

Hi! So I know I made a thread earlier today in which I asked for help on a couple of past final problems. For the other ones I'm about to ask here, I think I generally "get" the underlying concepts, but I'm not so sure that my answers are satisfactory. This is just a way of confirming the accuracy of what I have so far.

1.

http://imageshack.us/a/img51/7451/screenshot20121212at938.png [Broken]

Part A: Call the 4.0 kg mass m1, and the 3.0 kg mass m2.

m1 has 2 forces acting on it: m1g and (m1 v^2)/r.

By substituting the values into the variables, I found that the tension came out to be about 53.4 N.

Part B: m2 also has 2 forces acting on it: m2g and (m2 v^2)/r.

The tension in this string was the tension caused by the mass m2 (approx. 42 N), PLUS the tension found in Part A (53.4 N).

Part C: Because of the greater tension in the 4.0 m string, I wrote that the 4.0 m string would break first as both strings had the same strength.

Somehow I think I messed up something here, so I need someone to check my solution.

2.

http://imageshack.us/a/img715/5200/screenshot20121212at937.png [Broken]

I used the apparent weight formula w = mg - m(v^2/r), but currently don't have much beyond that point. I also have no idea in which direction the plumb bob will end up pointing - I tentatively answered "south" but I really don't have a solid justification for that answer.

3.

http://imageshack.us/a/img141/1887/screenshot20121212at921.png [Broken]

The graph drawing was relatively straightforward - the potential would come up from a really negative value and then asymptotically approach 0 as it went to the surface of the other body, and there were two such lines going in both directions since gravitation represents an action-reaction pair. So I wrote that this would give a local maximum. What I don't understand is how to calculate the *exact* position of the extremum, as well as the "calculate the work" portion.

Last but not least...

4.

http://imageshack.us/a/img526/4638/screenshot20121212at951.png [Broken]

Relevant equation: T = 2*pi*sqrt(m/k).

Case 1: The 2 masses had the same period but different amplitudes.
Case 2: The second mass had a period that was *three times* that of the first mass.

I tentatively answered that in Case 1, the particles would be out of phase and would be going in different directions when they did pass. In Case 2, they would be going in the same direction, since mass 2's period was a multiple of that of mass 1. But I don't know if this is right, and I also don't know how I can determine precisely how much time passes (first part of the question). Do I just plug in the values into the equation I mentioned above?

Thanks to you all for your help in advance! :) I know this is a lot, so you don't need to answer all of them if you're pressed for time.

Last edited by a moderator: May 6, 2017
2. Dec 12, 2012

haruspex

Re: Some more urgent questions

It's better to put each problem on a separate thread. Different responders can choose which threads to engage with.
Doesn't seem quite enough. What are you using for v here?
Sounds right to me. If you wish to justify such an answer, for which plane should you draw the free body diagram?
What expression can you write down, and then differentiate, for the potential as a function of distance along that line?
What is the work required to move the particle fro the earth's surface to the local max of potential?
Why? They start in phase and have the same period, so how would they get out of phase?
Write out the equations for their motions. What solution sets do you get for each for x=0?

3. Dec 13, 2012

pious&peevish

Re: Some more urgent questions

Thank you. I think I understand the spring problem and plumb bob problem now.

For #1, I used the same speed for m1 as the one for m2, but this is mostly where I think I messed up because I can't say that for certain...

4. Dec 13, 2012

haruspex

Re: Some more urgent questions

You can say for certain they are not the same. The ratio of the speeds is evident from the diagram.