You should not worry about this homework too much, until the midterm2 is over.
You should go over the practice exam first, and do all other preps for the midterm2 first.
So, come back to this posting later.
10.40: (a) 30 J, (b) 12 J
11.12: 61 rad/sec (You should figure the direction out! Discuss with your friends before submitting your solution!)
11.38: (a) 14 J s, (b) 57 N m
NOTE: L = I omega for almost all problems in this course, as in many problems of this homework set. Next week, we will see an example, where we have to use L = r x p (the more fundamental formula), but such an example is more an exception than a rule.
Keep up your good work in doing the homework!!!
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12 comments:
Hi Sam,
I have a question about part K of the potential energy graphs and motion from the homework. Part K said Consider points A, E, and G. Of these three points, which one corresponds to the greatest magnitude of acceleration of the particle?
I thought that the force is greater where the curve is steeper, and therefore the acceleration would be big too. Wouldn't the answer be A and E, because ther are both at points where there is a steep hill. But the answer is only A. So why wouldn't it be A and E ?
- Gurpreet
Yes, Gurpreet, you need to consider which one is steeper.
Just to give you guys a heads up, for 10.37, the answer they want is the actual fraction, not the percentage (which is what the answer in the back of the book says)...
Can anyone help me on 10.49? I know that an additional torque is needed to keep the pulley stationary.. and that the torque of one mass is rm(1)g and the torque of the second mass is rm(2)g. I also know that these two torques are in opposite directions. Can anyone else elaborate?
Tara, if you subtract those two torques (forget about their opposite directions for the moment or just think about their magnitudes) then you should get the torque that must be added to the system to keep it from moving.
I'm having trouble with 10.40 a. I got 30 joules as well but when I try to enter it I get a message that I might have rounded wrong or my significant figures may be off. I tried 3.0*10 with no luck either.
-Kevin
Oh, and one other question. How are you supposed to figure out what the rotational kinetic energy is without the radius?
Hey Kevin, I'm not sure what you're rounding issue is, but I don't understand how you got 30J for 10.40a. I know that we have different numbers, but I got ~twice that for a, and I did got 30J for part b. Are you sure that you are solving for translational Kinetic energy (K = 1/2mv^2)?
Anyway, for part b, you don't need to know the radius. If you go through all of the variables r will cancel out [Krot = 1/2Iw^2, w=v/a, and I=2/5mr^2]
My mass is 1.6kg and 6.0m/s. Energy should be 1/2*1.6*6^2 or 28.8 J, right?
Yes, you're right. Does it let you enter the right answer for rotational energy?
Oh, I screwed something up. The answer is 29, not 30.
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