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Rotational Mechanics — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsRotational Mechanics2 Marks
Question
The torque of a force $\overrightarrow{\text{F}}$ about a point is defined as $\overrightarrow{\text{r}}=\overrightarrow{\text{r}}\times\overrightarrow{\text{F}}.$ Suppose $\overrightarrow{\text{r}}, \overrightarrow{\text{F}}$and $\overrightarrow{\text{r}}$ are all nonzero. Is $\text{r}\times\overrightarrow{\text{r}}\Bigg|\Bigg|\overrightarrow{\text{F}}$ always true? Is it ever true?
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Answer

No, $\overrightarrow{\text{r}}\times\overrightarrow{\text{r}}\Bigg|\Bigg|\overrightarrow{\text{r}}$ is not true. In fact, it is never true. This is because:$\overrightarrow{\text{r}}\times\overrightarrow{\text{r}}$
$=\overrightarrow{\text{r}}\times\Big(\overrightarrow{\text{r}}\times\overrightarrow{\text{F}}\Big)$
Applying vector triple product, we get:$\overrightarrow{\text{r}}\times\Big(\overrightarrow{\text{r}}\times\overrightarrow{\text{F}}\Big)$
$=\Big(\overrightarrow{\text{r}}.\overrightarrow{\text{F}}\Big)\overrightarrow{\text{r}}-\Big(\overrightarrow{\text{r}}.\overrightarrow{\text{r}}\Big)\overrightarrow{\text{F}}$
$\because\overrightarrow{\text{r}}.\overrightarrow{\text{r}}=\text{r}^2$
$=\Big(\overrightarrow{\text{r}}.\overrightarrow{\text{F}}\Big)\overrightarrow{\text{r}}-\text{r}^2\overrightarrow{\text{F}}$
If $\overrightarrow{\text{r}}.\overrightarrow{\text{F}}=0;$ that is, $\overrightarrow{\text{r}}\perp\overrightarrow{\text{F}},$ then:$\overrightarrow{\text{r}}\times\overrightarrow{\text{r}}=-\text{r}^2.\overrightarrow{\text{F}}$
We know that $r^2$ is never negative and $\overrightarrow{\text{r}}\times\overrightarrow{\text{r}}=-\text{r}^2.\overrightarrow{\text{F}}$ This implies that both vectors may be antiparallel to each other but not parallel.

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