Calculating Work and Kinetic Energy in Uniform Circular Motion

[captain.joco]

Homework Statement

A particle of mass m is tied to a string which goes through a hole in a smooth horizontal table. The particle moves in uniform circular motion with speed v0. The radius of the cricle is R0. By puling the string very slowly, the radius of the circular motion is reduced tao R1.
Show that the work done by the string force is equal to the change of Kinetic energy.

Homework Equations

F=Tension(T) = mv^2 /R
K ( kinetic energy ) =1/2mv^2
K1-K0 ( final - initial kinetic energy ) = 1/2m ( v1^2 - v0^2 )
where v1 = R0*v0/R1 ( using angular momentum conservation ).

The Attempt at a Solution

W=∆K ( work energy theorem )
W = ∫Fdr , where F = m*v*v / r ( centripetal force )
I think I need some substitution to get the integral to give the ∆E...

 

[Astronuc]

With the integrand of F = m*v*v/r multiplied by dr, one has units of kg m²/s² (assuming SI) which is units of kinetic energy.

 

[captain.joco]

I know that the units are consistent. I have found the change in Kinetic energy, which is deltaK = 1/2 *v0^2 * [( R0^2 / R1^2 ) - 1 ].
But when I integrate mv1^2/R1 dR ( between R0 and R1 ), I get the very same expression but without the 1/2 factor at the start.. What am I doing wrong?

 

[Doc Al]

What's the force as a function of R?

 

[captain.joco]

I solved the integral and I got the same expression as in for the difference in kinetic energy, so it's got to be right. Thanks for Your help!