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Ahh, so I found out the reason for the crash from yesterday. We had a power failure which killed the server. I've been meaning to get myself some sort of UPS. I guess this is a good time to pick one up.
Badminton was okay. I didn't have that much energy to play though. I was definitely not pushing the birdie to the back, so there was much running and ducking involved.
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Hey, it seems as if those university kids (kids, ha!) are finishing up their exams now. I've been getting e-mail from a whole bunch of gals who I haven't heard from in many, many moons. *sigh* Just as I was clearning out my inbox too.
This morning I was supposed to show up early to watch over the office, but I got here and found out that I was locked out! Argh! How come every time I get to work early, nobody is ever in?!? This is the reason why I never get in on time! *shakes fist*
I am so full. Ugh. I just ate two leftover meals for some unknown reason. I wasn't even that hungry! *clenches bulging stomach*
Hopefully I'll be burning this off soon.
Anyhoo, I was at work a lot later than I had expected to be, and ended up being late for badminton. I've just finished dinner and I'm downloading some security patches for Win2k. There seem to be a lot of new critical updates that came out in the past couple of months. Scary no?
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Okay, so we have to determine the work done on this system. Consider a point-sized ball of mass m, initial velocity of vo, and initial radius ro. The ball is being pulled towards the centre of the circle at constant velocity vr to a final radial distance of rf and velocity vf. Because the force being applied on the ball is in the same direction as r, the torque on the system is zero, and the angular momentum in the system should not change. (I'd love to throw up a diagram, but hopefully you can still use whatever imagination you've got that hasn't been destroyed by popular entertainment.)
Anyway, the question asks what amount of work has been done, and as we all know, work W is equal to the integral of the force F acting on the system over the distance on which this force is applied. In our case, this force is a function of the radial distance r:
(1) W = ∫ F(r) dr.
But the force being applied in this system is effectively the tension on the string that's attached to the point mass. This tensile force is equal to the centripetal force on the ball. Therefore, the force is:
(2) F(r) = m ar = m v(r)2 r-1.
We know the mass m and the radius r, but v is still unknown. Since angular momentum L is conserved, we can use the definition of angular momentum
(3) L = I ω
to determine a formula for v(r). For such a simple system, the moment of inertia I is
(4) I = m r2
while the definition for angular momentum ω is
(5) ω = v r-1.
Inserting these definitions into equation (3), this gives the following relation:
(6) L = m r2 v r-1 = m r v => v(r) = L m-1 r-1.
Putting equations (1), (2), and (6) together gives
(7) W = ∫ L2 m-1 r-3 dr = L2 m-1 ∫ r-3 dr = 0.25 L2 m-1 (rf-2 - ro-2)
We can use equation (6) to determine a value for L in terms of known values
(8) L = m ro vo
which then can be used in equation (7) to determine the work done on the system:
(9) W = 0.5 m ro2 vo2 (rf-2 - ro-2) = 24 J.
So my original answer was off. (Probably because I integrated r-3 to r-4. Oops.)
Another point I want to make, the other two parts are really easy using conservation of angular momentum, although you'll have to also consider the velocity that the ball is approaching the centre as well, so the actual velocity of the ball v can be derived from
(10) v2 = vc2 + vt2
where vc is the velocity of the ball towards the centre (in this case it's 0.5 m/s) and vt is the tangential velocity of the ball (which can be determined by using the conservation of angular momentum.
Can someone actually check my work? Without something to compare to I'm not sure if my answers are exact.
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