Rotational Motion Question 136
Question: A particle is moving along a line $y=x+a$ with a constant velocity $v$ . Find the angular momentum of the particle about the origin.
Options:
A) $mva$
B) $mva\sqrt{2}$
C) $\frac{mva}{\sqrt{2}}$
D) $\frac{mvy}{x\sqrt{2}}$
Show Answer
Answer:
Correct Answer: C
Solution:
[c]
$\angle POQ=45{}^\circ So,OQ=a\cos 45{}^\circ =a/\sqrt{2}$
$L=mv(OQ)=mva/\sqrt{2}$
