The position of a particle is given by vec r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The angular momentum $\vec L$ is given by the cross product of position vector $\vec r$ and momentum vector $\vec p$: $\vec L = \vec r 	imes \vec

Vedclass · Accepted Answer

$x-axis$

Vedclass · Answer

(A) The angular momentum $\vec L$ is given by the cross product of position vector $\vec r$ and momentum vector $\vec p$: $\vec L = \vec r 	imes \vec p = \begin{vmatrix} \hat i & \hat j & \hat k \ 1 & 2 & -1 \ 3 & 4 & -2 \end{vmatrix}$ $= \hat i(2(-2) - (-1)(4)) - \hat j(1(-2) - (-1)(3)) + \hat k(1(4) - 2(3))$ $= \hat i(-4 + 4) - \hat j(-2 + 3) + \hat k(4 - 6)$ $= 0\hat i - 1\hat j - 2\hat k$ $= -\hat j - 2\hat k$ Since the angular momentum vector $\vec L = 0\hat i - 1\hat j - 2\hat k$ has no component along the $x$-axis (the coefficient of $\hat i$ is $0$),the vector $\vec L$ lies in the $yz$-plane. Therefore,the angular momentum is perpendicular to the $x$-axis.

The position of a particle is given by $\vec r = (\hat i + 2\hat j - \hat k)$ and its momentum is $\vec p = (3\hat i + 4\hat j - 2\hat k)$. The angular momentum is perpendicular to:

Similar Questions

Can an object moving in a straight line have angular momentum?

$A$ particle moves along a circular path with decreasing speed. Hence,

$A$ mass $m$ is moving with a constant velocity $v$ along a line parallel to the $X$-axis. Its angular momentum with respect to the origin or the $Z$-axis is:

$A$ particle of mass $0.5\, kg$ is rotating in a circular path of radius $2\, m$ and centripetal force on it is $9\, N$. Its angular momentum (in $J\cdot s$) is:

Which graph represents the relationship between angular momentum $L$ and angular velocity $\omega$ for a rigid body rotating about a fixed axis?

Vedclass Test Series

Exam Paper Generator

Online Exam Module