Vectors Question 131
Question: If a particle of mass m is moving with constant velocity v parallel to x-axis in x-y plane as shown in fig. Its angular momentum with respect to origin at any time t will be
Options:
A) $ mvb,\hat{k} $
B) $ -mvb,\hat{k} $
C) $ mvb,\hat{i} $
D) $ mv,\hat{i} $
Correct Answer: B We know that, Angular momentum $ \overrightarrow{L}=\overrightarrow{r,}\times \overrightarrow{p} $ in terms of component becomes $ \overrightarrow{L}=| ,\begin{matrix} \hat{i} & \hat{j} & {\hat{k}} \\ x & y & z \\ p _{x} & p _{y} & p _{z} \\ \end{matrix}, | $ As motion is in x-y plane (z = 0 and $ P _{z}=0 $ ), so $ \overrightarrow{L,}=\overrightarrow{k,}(xp _{y}-yp _{x}) $ Here x = vt, y = b, $ p _{x}=m,v $ and $ p _{y}=0 $ \ $ \overrightarrow{L,}=\overrightarrow{k,}[ vt\times 0-b,mv ]=-mvb,\hat{k} $Show Answer
Answer:
Solution:
BETA
