A particle -m- starts with zero velocity along a line y - 4d- -Where d is a constant- The position of particle -m- varies as x - A-sin-t- At -t - -2- its angular momentum with respect to the origin is- - - - - - - mA- d-m- d-A-mAd-zero-

The correct option is D zero At ω t = π/2 v = dx/dt = A ω cos ω t = A ω cos(π/2) = 0 So the angular momentum about origin = 0. ...

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Standard X

Physics

Angular Analogue of Linear Momentum

A particle 'm...

Question

A particle ' $m$ ' starts with zero velocity along a line $y$ = $4 d$ . (Where $d$ is a constant). The position of particle ' $m$ ' varies as $x$ = $A s i n ω t$ . At $ω t$ = $\frac{π}{2}$ , its angular momentum with respect to the origin is?

$m A ω d$

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$\frac{m ω d}{A}$

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$\frac{m A d}{ω}$

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zero

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Solution

The correct option is D
zero

At $ω t$ = $\frac{π}{2}$

$v$ = $\frac{d x}{d t}$ = $A ω c o s ω t$ = $A ω c o s (\frac{π}{2})$ = 0

So the angular momentum about origin = 0.

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A particle 'm' starts with zero velocity along a line y = 4d. (Where d is a constant). The position of particle 'm' varies as x = A sin ωt. At ωt = π2, its angular momentum with respect to the origin is?

The correct option is D zero At ωt = π2 v = dxdt = Aωcos ωt = Aωcos(π2) = 0 So the angular momentum about origin = 0.

A particle ' $m$ ' starts with zero velocity along a line $y$ = $4 d$ . (Where $d$ is a constant). The position of particle ' $m$ ' varies as $x$ = $A s i n ω t$ . At $ω t$ = $\frac{π}{2}$ , its angular momentum with respect to the origin is?

The correct option is D
zero

At $ω t$ = $\frac{π}{2}$

$v$ = $\frac{d x}{d t}$ = $A ω c o s ω t$ = $A ω c o s (\frac{π}{2})$ = 0

So the angular momentum about origin = 0.