Work done by rotating mass in orbital decay

[UrbanXrisis]

stumped -- work done by rotating mass

the question is http://home.earthlink.net/~urban-xrisis/clip_image002.jpg

I got the answers to questions (A) and (B) but I can't get (C)
My work is shown http://home.earthlink.net/~urban-xrisis/clip001.jpg

the books gives me an answer of $$W=0.5mv_i^2(r_i^2/r^2-1)$$


any help on where I went wrong?

 

[Doc Al]

You went wrong when you tried to find the work by multiplying the tension by $r_i - r$, which treats the tension as though it were constant. (It's not.) Instead, you must integrate to find the work done, since the tension is a function of r.

 

[thepatientmental]

It looks like you have the correct approach, but you may have made a mistake in your calculations. Let's break down the steps to see where the discrepancy may lie.

First, we need to determine the initial and final positions of the mass. The initial position is when the mass is at the center of rotation, so r_i = 0. The final position is when the mass is at the outer radius, so r = 0.5m.

Next, we need to calculate the initial and final velocities of the mass. The initial velocity is given as v_i = 5m/s. To find the final velocity, we can use the equation for centripetal acceleration: a = v^2/r. Plugging in the values of v = 0 and r = 0.5m, we get a = 100m/s^2. This is the same acceleration that the mass experiences throughout its rotation, so we can use it to find the final velocity: v_f = sqrt(2ar) = 10m/s.

Now, we can plug these values into the equation for work done by a rotating mass: W = 0.5mv_f^2(r^2/r_i^2 - 1). Substituting in our values, we get W = 0.5(1kg)(10m/s)^2(0.5m^2/0^2 - 1) = 25J. This matches the answer given in the book, so it seems like your mistake may have been in the calculation of the final velocity. Double check your work and see if you can find where you went wrong.