h4. Method 1
{toggle-cloak:id=sys1} *System:* {cloak:id=sys1}The object as a [point particle] undergoing [pure rotation].{cloak}
{toggle-cloak:id=int1} *Interaction:* {cloak:id=int1}External forces from the string (tension, non-conservative), the earth (gravity, conservative) and the table (normal force, non-conservative).{cloak}
{toggle-cloak:id=mod1} *Model:* {cloak:id=mod1}[1-D Angular Momentum and Torque] plus [Mechanical Energy and Non-Conservative Work].{cloak}
{toggle-cloak:id=app1} *Approach:*
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{toggle-cloak:id=part1} {color:red} *Consider the Energy* {color}
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The table is level, so the [gravitational potential energy] will be constant. We can set it to zero by taking the height of the tabletop to be zero. The normal force is perpendicular to the motion of the object and so does no work. The only force capable of performing work on the object is the tension, which is equal to the force from the person pulling on the string. Thus, the work done by tension will equal the work done by the person. This work can be computed by finding the change in mechanical energy of the object:
{latex}\begin{large}\[ W^{NC} + E_{i} = E_{f}\]\end{large}{latex}
or, since the height of the object is constant throughout the motion and no spring is present:
{latex}\begin{large}\[ W^{NC} = E_{f} - E_{i} = K_{f} - K_{i}\]\end{large}{latex}
Because the object can be considered to be moving in pure rotation about the center of the circle, we can compute its kinetic energy using the rotational formula:
{latex}\begin{large}\[ K = K_{\rm rot} = \frac{1}{2}I\omega^{2}\]\end{large}{latex}
For a point particle,
{latex}\begin{large}\[ I = mr^{2}\]\end{large}{latex}
and so:
{latex}\begin{large}\[ W = \frac{1}{2}m\left(r_{f}^{2}\omega_{f}^{2} - r_{i}^{2}\omega_{i}^{2}\right)\]\end{large}{latex}
{warning}Note that we have assumed the mass is constant, but we cannot assume the angular speed is constant.{warning}
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{toggle-cloak:id=part2} {color:red} *Consider the Angular Momentum* {color}
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We cannot yet substitute into the equation obtained by considering the object's energy. We do already have the initial and final radius of the motion, but we do not yet have the angular speed. To find it, we consider the torque on the system. We choose the natural location for the axis by locating it at the center of the circle.
!puckonstringfbd.pngjpg!
With this choice, we can see that there is zero net torque on the system, since the [moment arm] for the tension is zero. With the net torque equal to zero, we are in the special case of constant angular momentum, so:
{latex}\begin{large}\[ I_{f}\omega_{f} = I_{i}\omega_{i} \]\end{large}{latex}
By using the formula for the moment of inertia of a point particle, we can show:
{latex}\begin{large}\[ \omega_{f} = \omega_{i}\left(\frac{r_{i}}{r_{f}}\right)^{2}\]\end{large}{latex}
Substituting into the work-energy theorem then gives:
{latex}\begin{large}\[ W = \frac{1}{2} mr_{i}^{2}\omega_{i}^{2}\left(\frac{r_{i}^{2}}{r_{f}^{2}} - 1\right) =\mbox{13.5 J}\]\end{large}{latex}
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