Question

the puck in the figure below has a mass of 0.120 kg. its original distance from the center of rotation is 44.0 cm, and the puck is moving with a speed of 78.2 cm/s. the string is pulled downward 17.7 cm through the hole in the frictionless table. determine the work done on the puck.

Answer 1

Apply the law of conservation of angular momentum:

$$ I_{1} \omega_{1}=I_{2} \omega_{2} $$

Here, \(I\) is moment of inertia and ais angular velocity

$$ \begin{array}{c} \frac{1}{2} m r_{1}^{2} a_{1}=\frac{1}{2} m r_{2}^{2} a_{2} \\ r_{1}^{2} \omega_{1}=r_{2}^{2} \omega_{2} \end{array} $$

Use the relation between angular velocity and linear velocity

\(\omega=\frac{v}{r}\)

$$ \begin{array}{c} r_{1}^{2} \frac{v_{1}}{r_{1}}=r_{2}^{2} \frac{v_{2}}{r_{2}} \\ r_{1} v_{1}=r_{2} v_{2} \\ (44 \mathrm{~cm})(78.2 \mathrm{~cm} / \mathrm{s})=(44 \mathrm{~cm}-17.7 \mathrm{~cm}) v_{2} \end{array} $$

Solve for \(v_{2}\)

$$ \begin{aligned} v_{2} &=130.82 \mathrm{~cm} / \mathrm{s} \\ &=1.3082 \mathrm{~m} / \mathrm{s} \end{aligned} $$

The work done is equal to change in kinetic energy

$$ \begin{aligned} W &=\Delta K E \\ &=\frac{m}{2}\left(v_{2}^{2}-v_{1}^{2}\right) \\ &=\frac{0.120 \mathrm{~kg}}{2}\left[(1.3082 \mathrm{~m} / \mathrm{s})^{2}-(0.782 \mathrm{~m} / \mathrm{s})^{2}\right] \\ &=0.0659 \mathrm{~J} \end{aligned} $$

Round off the result to 3 significant digits

$$ W=0.0660 \mathrm{~J} $$