Question

A particle of mass \[ m=0.25\ \text{kg} \] is moving along a straight line parallel to the \(x\)-axis with a constant velocity \[ v=5\ \text{m s}^{-1} \] as shown in the figure. What is the angular momentum of the particle with respect to the origin?

• A. \(0\)
• B. \(1.25\ \text{kg m}^2\text{s}^{-1}\) along \(+z\)-axis
• C. \(1.25\ \text{kg m}^2\text{s}^{-1}\) along \(-z\)-axis
• D. \(1.25\ \text{kg m}^2\text{s}^{-1}\) along \(x\)-axis

Hint:

For a particle moving in a straight line, \[ L = p \times (\text{perpendicular distance from origin to line of motion}). \] Use the right-hand rule to determine the direction.

Answer 1

The Correct Option is C

Concept: Angular momentum of a particle about the origin is \[ \vec L=\vec r\times\vec p \] where \[ \vec p=m\vec v. \]

Step 1: Calculate the linear momentum. \[ m=0.25\ \text{kg} \] \[ v=5\ \text{m s}^{-1} \] \[ p=mv \] \[ p=0.25\times5 \] \[ p=1.25\ \text{kg m s}^{-1} \]

Step 2: Use the perpendicular distance from the origin. From the figure, \[ b=1.0\ \text{m} \] Therefore, \[ |\vec L| = pb \] \[ = 1.25\times1 \] \[ = 1.25\ \text{kg m}^2\text{s}^{-1} \]

Step 3: Determine the direction. \[ \vec r=(x\,\hat i+b\,\hat j) \] \[ \vec p=p\,\hat i \] Thus \[ \vec L = \vec r\times\vec p \] \[ = (x\hat i+b\hat j)\times(p\hat i) \] \[ = bp(\hat j\times\hat i) \] \[ = -bp\,\hat k \] Hence the direction is along the negative \(z\)-axis. \[\begin{aligned} \boxed{ 1.25\ \text{kg m}^2\text{s}^{-1} \text{ along }-z\text{-axis} } \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.