Question

A particle of mass \(m\) is moving around the origin with a constant speed \(v\) along a circular path of radius \(R\). When the particle is at \( (0, R) \), its velocity is \( \mathbf{v} = -v \hat{\mathbf{i}} \). The angular momentum of the particle with respect to the origin is :

• A. \( mvR \hat{\mathbf{k}} \)
• B. \( -mvR \hat{\mathbf{k}} \)
• C. \( mvR \hat{\mathbf{j}} \)
• D. \( -mvR \hat{\mathbf{j}} \)

Hint:

Angular momentum \( \mathbf{L} = \mathbf{r} \times \mathbf{p} \). Identify the position vector \( \mathbf{r} \) and the linear momentum vector \( \mathbf{p} = m \mathbf{v} \) from the given information. Then compute the cross product using the properties of unit vectors.

Answer 1

The Correct Option is A

Step 1: Angular Momentum Formula
The angular momentum L of a particle with respect to the origin is given by the cross product of its position vector r and its linear momentum vector p = m v: L = r × p = r × (m v).

Step 2: Position and Velocity Vectors
The position of the particle is given as (0, R), so the position vector is: r = 0 î + R ĵ = R ĵ.
The velocity of the particle is given as: v = -v î.
The linear momentum vector is: p = m (-v î) = -m v î.

Step 3: Compute the Cross Product
Now, we compute the cross product: L = (R ĵ) × (-m v î) = -m v R (ĵ × î).

Step 4: Cross Product of Unit Vectors
We know that the cross product of unit vectors follows the cyclic order: î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ. Also, ĵ × î = - (î × ĵ) = - k̂.

Step 5: Final Expression for Angular Momentum
Substituting this into the expression for L: L = -m v R (-k̂) = m v R k̂.

Therefore, the angular momentum of the particle with respect to the origin is: L = m v R k̂.