Obtain the relation between angular momentum of a particle and torque acting on it.
Obtain the relation between angular momentum of a particle and torque acting on it.
By definition of angular momentum of a particle, $\vec{l}=\vec{r} \times \vec{p}$
Differentiating this equation w.r.t. time $\mathrm{t}, \frac{d \vec{l}}{d t}=\vec{r} \times \frac{d \vec{p}}{d t}+\frac{d \vec{r}}{d t} \times \vec{p}$
But, $\frac{d \vec{p}}{d t}=$ rate of change of linear momentum $=$ force $\overrightarrow{\mathrm{F}}$ and $\frac{d \vec{r}}{d t}=\vec{v}$
$\therefore \quad \frac{d \vec{l}}{d t}=\vec{r} \times \overrightarrow{\mathrm{F}}+\vec{v} \times \vec{p}$
But as $\vec{v}$ and $\vec{p}$ are in the same direction,
$\vec{v} \times \vec{p}=0$
$\therefore \quad \frac{d \vec{l}}{d t}=\vec{r} \times \overrightarrow{\mathrm{F}}=\vec{\tau}$
Thus, "the time rate of change of angular momentum is equal to the torque".
Similar Questions
The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $\mathrm{V}(\mathrm{r})=\mathrm{kr}^2 / 2$, where $\mathrm{k}$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $\mathrm{R}$ about the point $\mathrm{O}$. If $\mathrm{v}$ is the speed of the particle and $\mathrm{L}$ is the magnitude of its angular momentum about $\mathrm{O}$, which of the following statements is (are) true?
$(A)$ $v=\sqrt{\frac{k}{2 m}} R$
$(B)$ $v=\sqrt{\frac{k}{m}} R$
$(C)$ $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^2$
$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$
The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $\mathrm{V}(\mathrm{r})=\mathrm{kr}^2 / 2$, where $\mathrm{k}$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $\mathrm{R}$ about the point $\mathrm{O}$. If $\mathrm{v}$ is the speed of the particle and $\mathrm{L}$ is the magnitude of its angular momentum about $\mathrm{O}$, which of the following statements is (are) true?
$(A)$ $v=\sqrt{\frac{k}{2 m}} R$
$(B)$ $v=\sqrt{\frac{k}{m}} R$
$(C)$ $\mathrm{L}=\sqrt{\mathrm{mk}} \mathrm{R}^2$
$(D)$ $\mathrm{L}=\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{R}^2$
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Where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for ${t}=0$ at time ${t}=$ .....$seconds.$
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