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## Homework Statement

The puck in the figure shown below has a mass of 0.120 kg. Its original distance from the center of rotation is 40.0 cm, and it moves with a speed of 80.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck.

## Homework Equations

## L = I \omega##

##KE = 1/2 \cdot I \cdot \omega^2 ##

## The Attempt at a Solution

Consider angular momentum. Note it's conserved.

Therefore, ## \frac{1}{2}mr_i^2 \omega_i = \frac{1}{2} mr_f^2 \omega_f \Rightarrow r_i^2 \omega_i =r_f^2 \omega_f ##

Note that ##\omega_i = 80 cm/s = 2 rad/s## (because 2pi radians is one revolution, which is 80pi cm, so 1 radian is 40 cm)

Threrefore, ##0.40^2 \cdot 2 = 0.25^2 \cdot \omega_f \Rightarrow \omega_f = 5.12##

Note that Work Done is the change in kinetic energy. KE initial is equal to ## 1/2 (1/2 \cdot 0.12 \cdot 0.4*0.4) \cdot 2^2 ## (1/2 Iw^2 = 1/2 (1/2mr^2)w^2). KE final is simialrly equal to ## 1/2 (1/2 \cdot 0.12 \cdot 0.25 \cdot 0.25 \cdot 5.12^2##.

We can do the calculation, to get the change as 0.029952. But the book says answer is 5.99*10^-2.

What's wrong?