11.2 Angular momentum  (Page 2/8)

Here we have used the definition of $\overrightarrow{p}$ and the fact that a vector crossed into itself is zero. From Newton’s second law, $\frac{d\overrightarrow{p}}{dt}={\displaystyle \sum \overrightarrow{F}},$ the net force acting on the particle, and the definition of the net torque, we can write

Note the similarity with the linear result of Newton’s second law, $\frac{d\overrightarrow{p}}{dt}={\displaystyle \sum \overrightarrow{F}}$ . The following problem-solving strategy can serve as a guideline for calculating the angular momentum of a particle.

Angular momentum and torque on a meteor

Strategy

Solution

We write the velocities using the kinematic equations.

1. The angular momentum is
   
   $\begin{array}{cc}\hfill \overrightarrow{l}& =\overrightarrow{r}\times \overrightarrow{p}=(25.0\text{km}\widehat{i}+25.0\text{km}\widehat{j})\times 15.0\text{kg}(0\widehat{i}+v_y\widehat{j})\hfill \\ & =15.0\text{kg}[25.0\text{km}(v_y)\widehat{k}]\hfill \\ & =15.0\text{kg[}2.50\times {10}^4\text{m}(\mathrm{-2.0}\times {10}^3\text{m}\text{/}\text{s}-(2.0\text{m}\text{/}{\text{s}}^2)t)\widehat{k}].\hfill \end{array}$
   
   At $t=0$ , the angular momentum of the meteor about the origin is
   
   ${\overrightarrow{l}}_0=15.0\text{kg}[2.50\times {10}^4\text{m}(\mathrm{-2.0}\times {10}^3\text{m}\text{/}\text{s})\widehat{k}]=7.50\times {10}^8\text{kg}·{\text{m}}^2\text{/}\text{s}(\text{−}\widehat{k}).$
   
   This is the instant that the observer sees the meteor.
2. To find the torque, we take the time derivative of the angular momentum. Taking the time derivative of $\overrightarrow{l}$ as a function of time, which is the second equation immediately above, we have
   
   $\frac{d\overrightarrow{l}}{dt}=\mathrm{-15.0}\text{kg}(2.50\times {10}^4\text{m})(2.0\text{m}\text{/}{\text{s}}^2)\widehat{k}.$
   
   Then, since $\frac{d\overrightarrow{l}}{dt}={\displaystyle \sum \overrightarrow{\tau }}$ , we have
   
   ${\displaystyle \sum \overrightarrow{\tau }}=\mathrm{-7.}5\times {10}^5\text{N}·\text{m}\widehat{k}.$
   
   The units of torque are given as newton-meters, not to be confused with joules. As a check, we note that the lever arm is the x -component of the vector $\overrightarrow{\text{r}}$ in [link] since it is perpendicular to the force acting on the meteor, which is along its path. By Newton’s second law, this force is
   
   $\overrightarrow{F}=ma(\text{−}\widehat{j})=15.0\text{kg}(2.0\text{m}\text{/}{\text{s}}^2)(\text{−}\widehat{j})=30.0\text{kg}·\text{m}\text{/}{\text{s}}^2(\text{−}\widehat{j}).$
   
   The lever arm is
   
   ${\overrightarrow{r}}_{\perp }=2.5\times {10}^4\text{m}\widehat{i}.$
   
   Thus, the torque is
   
   $\begin{array}{cc}\hfill {\displaystyle \sum \overrightarrow{\tau }=}{\overrightarrow{r}}_{\perp }\times \overrightarrow{F}& =(2.5\times {10}^4\text{m}\widehat{i})\times (\mathrm{-30.0}\text{kg}·\text{m}\text{/}{\text{s}}^2\widehat{j}),\hfill \\ & =7.5\times {10}^5\text{N}·\text{m}(\text{−}\widehat{k}).\hfill \end{array}$