Question

Redo the analysis, calculating torque and angular momentum relative to a fixed location in the ice anywhere underneath the string (similar to the analysis of the meter stick around one end). Show that the two analyses of the puck are consistent with each other.

Short answer

The two analyses of the puck are consistent with each other.

Step-by-step solution

Definition of angular momentum

Angular momentum is a property that describes an object or a system of items' rotational inertia in motion around an axis that may or may not pass through the object or system.

To characterize the motion of the hockey puck, calculate the torque and angular momentum about the end of the puck in contact with the ice and the end of the puck where the force is applied. To calculate the rate of change of angular speed, we first find the rate of charge of angular momentum and set it to zero. The torque around the fixed point is 0 because there is no distance between the place where the force is applied and the fixed point.

Figure shows a puck with string wound around it about point A and pulled with a constant tension $F_T$.

Proof

Let us consider a hockey puck of mass M and radius R .

As we are pulling with constant force$F_T$. It is moving with angular speed$\omega$

Applying momentum principle about location A , we get

$\frac{d{\overrightarrow{L}}_A}{dt}={\overrightarrow{\tau }}_{net.A}$

When it is moving towards right and component of angular momentum into the page, then the net torque is

$\frac{d}{dt}\left[RM{v}_{CM}-\frac{1}{2}M{R}^2\omega \right]={\overrightarrow{\tau }}_{net.A}$

Since, the torque ${\overrightarrow{\tau }}_{net.A}$is zero about a point under the string, we get

$\frac{d}{dt}\left[RM{v}_{CM}-\frac{1}{2}M{R}^2\omega \right]=0$

$\frac{d}{dt}\left(RM{v}_{CM}\right)-\frac{d}{dt}\left(\frac{1}{2}M{R}^2\omega \right)=0$

$\frac{d{v}_{CM}}{dt}\left(RM\right)=\frac{d\omega }{dt}\left(\frac{1}{2}M{R}^2\right)$

$\frac{d{v}_{CM}}{dt}=\frac{d\omega }{dt}\left(\frac{R}{2}\right)$

Thus, the rate of change of angular speed is

$\frac{d\omega }{dt}=\left(\frac{2}{R}\right)\frac{d{v}_{CM}}{dt}$

From Newton’s second law of motion, $F=ma$

When the puck is moving with a speed $v_{CM}$about center of mass, the above equation can rewrite as

$$\begin{gathered} \frac{d\omega }{dt}=\left(\frac{2}{R}\right)\frac{d{v}_{CM}}{dt}\left(\frac{M}{M}\right) \\ =\left(\frac{2}{R}\right)\left(\frac{M{a}_{CM}}{M}\right) \\ =2\left(\frac{F_T}{MR}\right) \end{gathered}$$

The rate of angular speed expressed above accords with the analysis in which we used torques around the puck's center of mass. As a result, the puck's two analyses are mutually exclusive and consistent with each other.