Question

Angular momentum of a single particle moving with constant speed along circular path :

• A. is zero
• B. remains same in magnitude but changes in the direction
• C. changes in magnitude but remains same in the direction
• D. remains same in magnitude and direction

Hint:

For uniform circular motion, torque about the center is zero (\(\vec{r}\) and \(\vec{F}_c\) are anti-parallel), which implies angular momentum is conserved.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Angular momentum of a particle about the center of a circular path is given by \(\vec{L} = \vec{r} \times \vec{p}\), where \(\vec{r}\) is the position vector and \(\vec{p} = m\vec{v}\) is the linear momentum.
Step 2: Detailed Explanation:
1. Magnitude: In a circular path with constant speed \( v \) and radius \( R \), the velocity vector is always perpendicular to the radius vector. Thus, \( L = m v R \sin(90^\circ) = m v R \). Since \( m, v, R \) are constants, the magnitude remains same.
2. Direction: By the right-hand rule, the direction of \(\vec{L}\) is perpendicular to the plane of the circle. As the particle moves in the plane, the plane itself does not tilt, so the normal to the plane (the direction of \(\vec{L}\)) remains fixed.
Step 3: Final Answer:
Both the magnitude and direction of angular momentum remain constant.