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

• A. $mvb\widehat{k}$
• B. $- mvb\widehat{k}$
• C. $mvb\widehat{i}$
• D. $mv\widehat{i}$

Answer 1

Answer: (B)

$- mvb\widehat{k}$

Answer 2

We know that, Angular momentum

$\overset{\rightarrow}{L} = \overset{\rightarrow}{r} \times \overset{\rightarrow}{p}$ in terms of component becomes

\widehat{i} & \widehat{j} & \widehat{k} \\ x & y & z \\ p_{x} & p_{y} & p_{z} \end{matrix} \right|$$ As motion is in x-y plane (z = 0 and $P_{z} = 0$), so $\overset{\rightarrow}{L} = \overset{\rightarrow}{k}(xp_{y} - yp_{x})$ $\overset{\rightarrow}{L} = \overset{\rightarrow}{k}(xp_{y} - yp_{x})$) Here x = vt, y = b, $p_{x} = mv$ and $p_{y} = 0$ ∴ $\overset{\rightarrow}{L} = \overset{\rightarrow}{k}\lbrack vt \times 0 - bmv\rbrack = - mvb\widehat{k}$