System of Particles and Rotational Motion - Result Question 49
52. Angular momentum is
[1994]
(a) vector (axial)
(b) vector (polar)
(c) scalar
(d) none of the above
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Answer:
Correct Answer: 52. (a)
Solution:
- (a) Angular momentum $\vec{L}$ is defined as $\vec{L}=\vec{r} \times m(\vec{v})$
So, $\vec{L}$ is an axial vector.
(a)
$y=x+4$ line has been shown in the figure when $x=0, y=4$. So, $O P=4$.
The slope of the line can be obtained by comparing with the equation of the straight line $y=m x+c$
$m=\tan \theta=1$
$\Rightarrow \theta=45^{\circ}$
$\angle O Q P=\angle O P Q=45^{\circ}$
If we draw a line perpendicular to this line, length of the perpendicular $=O R$
$ \begin{aligned} & \text{ In } \triangle O P R, \frac{O R}{O P}=\sin 45^{\circ} \\ & \begin{aligned} \Rightarrow O R & =O P \sin 45^{\circ} \\ & =4 \times \frac{1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2 \sqrt{2} \end{aligned} \end{aligned} $
Angular momentum of particle going along this line $=r \times m v=2 \sqrt{2} \times 5 \times 3 \sqrt{2}=60$ units
