First slide

Angular momentum

Question

A Particle of mass M is moving in a straight line with uniform speed v parallel to X-axis in X- Y plane at a height ‘h’ from the X-axis. Its angular momentum about the origin is

Moderate

A

zero
B

mvh and is directed along positive z-axis
C

mvh and is directed along negative z-axis
D

mvh and is directed along positive x-axis

Solution

$\large \vec v=v\hat i\\\\\vec r= (x\hat i+y\hat j)=(x\hat i+h\hat j)$
Here P (x,y) is the instantaneous position of the particle.
$\large \therefore \vec L_0=\vec r \;X\;m\vec v=m[(x\hat i+h \hat j)\;X\;v\hat i]\\\\=mvh(\hat j\;X\;\hat i)=-(mvh)\hat K$

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