Kinematics Question 262
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. it’s 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} $
Show Answer
Answer:
Correct Answer: B
Solution:
We know that, Angular momentum $ \overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p} $ in terms of component becomes $ \overrightarrow{L}= \begin{vmatrix} \hat{i} & \hat{j} & {\hat{k}} \\ x & y & z \\ p _{x} & p _{y} & p _{z} \\ \end{vmatrix} $
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}=mv $ and $ p _{y}=0 $ \ $ \overrightarrow{L}=\overrightarrow{k}[ vt\times 0-bmv ]=-mvb\hat{k} $
BETA
