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A particle of mass 2 kg is moving along a straight line $3x+4y=5$ with speed $8 m/s$. What is the angular momentum of the particle about the orgin.

\[\begin {array} {1 1} (a)\;8\;kg\;m^2/s & \quad (b)\;16\;kg\;m^2/s \\ (c)\;20\;kg\;m^2/s & \quad  (d)\;24\;kg\;m^2/s \end {array}\]
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$4 y= -3x+5$
$y= -\large\frac{3}{4}x +\frac{5}{4}$
Slope $= -\large\frac{3}{4}$
$\qquad =\tan \theta$
Angular momentum about the origin
$L= mvd$
$[p \times perpendicular \;distance]$
$\qquad= 2 \times 8 \times d$
$\qquad= 16 \;d$
$Y=0,x=\large\frac{5}{3}$
$Y=0, x= \large\frac{5}{3}$
In $OCB$,
$\sin \theta=\large\frac{OC}{OB}$
$\qquad= \large\frac{d}{OB}$
From $\Delta OAB, \tan \theta=\large\frac{5/4}{5/3}$
$\qquad= \large\frac{3}{4}$
=> $\sin \theta= \large\frac{3}{5}$
$\therefore \large\frac{3}{5}=\large\frac{d}{5/3}$
$d= 1 m$
$\therefore L=16 \;kg\;m^2/s$
answered Nov 28, 2013 by meena.p
edited Jun 21, 2014 by lmohan717
 

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