- #1
pious&peevish
- 27
- 0
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]
My answer:
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.
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]
My answer:
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: