A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x- y plane as shown in the figure: Which of the following statements is false for the angular momentum L about the origin? (a) L = mv[R/√(2) + a]k when the particle is moving from B to C. (b) L = (mv)/(√(2))Rk when the particle is moving from D to A. (c) L = -(mv)/(√(2))Rk when the particle is moving from A to B (d) L = mv[R/√(2) - a]k when the particle is moving from C to D.

Answer: (b). A is at ((R)/(√2),(R)/(√2)). Use Lz=m(xvy-yvx) on each side. A→B ( vx=v, y=(R)/(√2)): L=-(mvR)/(√2) k — (c) true. B→C ( vy=v, x=(R)/(√2)+a): L=mv((R)/(√2)+a) k — (a) true. C→D ( vx=-v, y=(R)/(√2)+a): L=mv((R)/(√2)+a) k — (d) is false. D→A ( vy=-v, x=(R)/(√2)): L=-(mvR)/(√2) k — (b) is false. So (b) is false (as per the key); (d) is also false.

A particle of mass m is moving along the side of a square of side… | JEE Main 2016 ((one Or More Correct))

Q69JEE Main 2016 · (One or more correct)Medium

Angular Momentum & its Conservation

A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x- y plane as shown in the figure:

Which of the following statements is false for the angular momentum $\vec{L}$ about the origin?

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Answer: (b)

A is at $(\frac{R}{\sqrt2},\frac{R}{\sqrt2})$ . Use $L_z=m(xv_y-yv_x)$ on each side.
A→B ( $v_x=v$ , $y=\frac{R}{\sqrt2}$ ): $L=-\frac{mvR}{\sqrt2}\hat k$ — (c) true.
B→C ( $v_y=v$ , $x=\frac{R}{\sqrt2}+a$ ): $L=mv(\frac{R}{\sqrt2}+a)\hat k$ — (a) true.
C→D ( $v_x=-v$ , $y=\frac{R}{\sqrt2}+a$ ): $L=mv(\frac{R}{\sqrt2}+a)\hat k$ — (d) is false.
D→A ( $v_y=-v$ , $x=\frac{R}{\sqrt2}$ ): $L=-\frac{mvR}{\sqrt2}\hat k$ — (b) is false.
So (b) is false (as per the key); (d) is also false.

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Angular Momentum & its Conservation question from JEE Main 2016 ((one Or More Correct))

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