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

[Mmm_Pasta]

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. K_f-K₀ = W. So, K_f = L_f²/2I_f and K₀ = L₀²/I₀. Using equation a L_f = 0.5r₀mv₀ so L_f² = 0.25r₀²m²v₀². I rearranged the equation like this: L_f-L₀(1/I_f-I₀). L_f²-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. L_f²/2I_f - L₀²/2I₀ was simplified to L₀²/2I_f - 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.

 

[Schr0d1ng3r]

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.

 

[Mmm_Pasta]

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. =)