Rotation Ques 38
- A mass $m$ is moving with a constant velocity along a line parallel to the $X$-axis, away from the origin. Its angular momentum with respect to the origin
(1997C, 1M)
(a) is zero
(b) remains constant
(c) goes on increasing
(d) goes on decreasing
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Answer:
Correct Answer: 38.(b)
Solution:
- $|\mathbf{v}|=v=$ constant and $|\mathbf{r}|=r$ (say)
Angular momentum of the particle about origin $O$ will be given by
$$ \begin{array}{ll} & \mathbf{L}=\mathbf{r} \times \mathbf{p}=m(\mathbf{r} \times \mathbf{v}) \\ \text { or } \quad & |\mathbf{L}|=L=m r v \sin \theta=m v(r \sin \theta)=m v h \end{array} $$
Now, $m, v$ and $h$ all are constants.
Therefore, angular momentum of particle about origin will remain constant. The direction of $\mathbf{r} \times \mathbf{v}$ also remains the same (negative $z$ ).
NOTE Angular momentum of a particle moving with constant velocity about any point is always constant. e.g. Angular momentum of the particle shown in figure about origin $O$ will be
$L=m v h=$ constant
BETA
