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(B) The angular momentum $\vec{L}$ of a particle about the origin is given by the cross product of its position vector $\vec{r}$ and its linear moment

Vedclass · Accepted Answer

$-m v b \hat{k}$

Vedclass · Answer

(B) The angular momentum $\vec{L}$ of a particle about the origin is given by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p} = m\vec{v}$.$\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Let the position of the particle be $\vec{r} = x \hat{i} + b \hat{j}$,where $x$ is the horizontal distance from the $y$-axis and $b$ is the constant vertical distance from the $x$-axis.The velocity of the particle is $\vec{v} = v \hat{i}$ (since it moves parallel to the $x$-axis).Therefore,$\vec{L} = (x \hat{i} + b \hat{j}) 	imes (m v \hat{i})$.Using the distributive property of the cross product:$\vec{L} = (x \hat{i} 	imes m v \hat{i}) + (b \hat{j} 	imes m v \hat{i})$.Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$,we get:$\vec{L} = 0 + m v b (-\hat{k}) = -m v b \hat{k}$.Thus,the angular momentum is $-m v b \hat{k}$.

$A$ particle of mass $m$ is moving along a straight line parallel to the $x$-axis with a constant velocity $v$ at a distance $b$ from the origin. Find the angular momentum about the origin in vector form.

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