How much work is required to reduce the radius of the circle from r0 to r0/2?

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The discussion centers on calculating the work required to reduce the radius of a block's circular motion from r0 to r0/2 on a frictionless table. Key calculations involve determining angular momentum, kinetic energy, and tension in the string. Initial attempts at solving for work yielded an incorrect answer of 9/16mv02, while a referenced solution suggested -2/3mv02, which was also disputed. The importance of conserving angular momentum in the context of changing radius is highlighted, despite the torque not being constant. Ultimately, integrating the force of tension was recommended as a simpler approach to find the correct answer.
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Homework Statement



A block of mass m sliding on a frictionless table is attached to a string that passes through a narrow hole through the center of the table. The block is sliding with speed v0 in a circle of radius r0. (Use any variable or symbol stated above as necessary.)

(a) Find the angular momentum of the block.
(b) Find the kinetic energy of the block.
(c) Find the tension in the string.
(d) A student under the table now slowly pulls the string downward. How much work is required to reduce the radius of the circle from r0 to r0/2?

Homework Equations



The equations to use are a, b, and c.

The Attempt at a Solution



I need help with (d).

(a) rmv0
(b) 0.5mv02
(c) mv02/r0

(d) My first answer was 9/16mv02. Kf-K0 = W. So, Kf = Lf2/2If and K0 = L02/I0. Using equation a Lf = 0.5r0mv0 so Lf2 = 0.25r02m2v02. I rearranged the equation like this: Lf-L0(1/If-I0). Lf2-L02 = -0.75r02m2v02 and the parenthesis = 1/(-4/3)mr02). When I simplify I get 9/16mv02.

This answer turned out to be wrong, so I looked up an online solution and it turned out to be wrong as well; the answer was -2/3mv02. They had similar work to mine. Lf2/2If - L02/2I0 was simplified to L02/2If - L02/2I0. They simplified so that angular momentum was conserved (I don't know why though because the torque is not constant [force changes as r changes])? I think I am wrong, though. However, after doing the algebra the answer is -2/3mv02. This answer was not accepted either.

Sorry if it appears confusing, but I am assuming I can skip most of the algebra since that is not where I am going wrong. Although, if the second solution is actually correct, I want to know why angular momentum remained conserved for the problem.
 
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Interesting, I had this same problem in my classical mechanics course in September.

I would recommend that you integrate the force of tension from r to r0, its simpler than using the change in energy.
 
Thanks! I integrated and got 3/2mv02. Although, the problem from before was an algebraic one (subtracting fractions [forgot common denominator]). I realized this when I was doing the integral. =)
 

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