A particle of mass m moves in the XY plane with a velocity ' v ' along the straight line A B. If the angular momentum of the particle with respect to origin O is L A when it is at A and LB when it is at B, then

Question

A particle of mass m moves in the XY plane with a velocity ' v ' along the straight line A B. If the angular momentum of the particle with respect to origin O is L A when it is at A and LB when it is at B, then (A) LA>LB (B) LA=LB (C) the relationship between L A and L B depends upon the slope of the line AB (D) LA < LB

Tardigrade Admin · Accepted Answer

From the definition of angular momentum,

L = r \times p = r m v sin ϕ (- \hat{k})

Therefore, the magnitude of

L

is

L = m v r sin ϕ = m v d

where

d = r sin ϕ

is the distance of closest approach of the particle to the origin. As

d

is same for both the particles, hence

L_{A} = L_{B}

.

Q. A particle of mass $m$ moves in the $X Y$ plane with a velocity ' $v$ ' along the straight line $A B$ . If the angular momentum of the particle with respect to origin $O$ is $L_{A}$ when it is at $A$ and $L_{B}$ when it is at $B$ , then

$L_{A} > L_{B}$

$L_{A} = L_{B}$

the relationship between $L_{A}$ and $L_{B}$ depends upon the slope of the line $A B$

$L_{A} < L_{B}$

From the definition of angular momentum,
$L = r \times p = r m v sin ϕ (- \hat{k})$

Therefore, the magnitude of $L$ is
$L = m v r sin ϕ = m v d$
where $d = r sin ϕ$ is the distance of closest approach of the particle to the origin. As $d$ is same for both the particles, hence $L_{A} = L_{B}$ .

Q. A particle of mass m moves in the XY plane with a velocity ' v ' along the straight line AB. If the angular momentum of the particle with respect to origin O is LA​ when it is at A and LB​ when it is at B, then

LA​>LB​

LA​=LB​

the relationship between LA​ and LB​ depends upon the slope of the line AB

LA​<LB​