Question

A particle of mass \( 2 \) kg is on a smooth horizontal table and moves in a circular path of radius \( 0.6 \) m. The height of the table from the ground is \( 0.8 \) m. If the angular speed of the particle is \( 12 \) rad/s, the magnitude of its angular momentum about a point on the ground right under the center of the circle is:

• A. \( 14.4 \) kg m\( ^2 \)s\(^{-1} \)
• B. \( 8.64 \) kg m\( ^2 \)s\(^{-1} \)
• C. \( 20.16 \) kg m\( ^2 \)s\(^{-1} \)
• D. \( 11.52 \) kg m\( ^2 \)s\(^{-1} \)

Hint:

For a particle moving in a circular path, angular momentum is given by \( L = m v r \).

Answer 1

The Correct Option is A

{Angular momentum formula} 
\[ L_0 = mvr \sin 90^\circ \] \[ = 2 \times 0.6 \times 12 \times 1 \] \[ = 14.4 \text{ kg m}^2 \text{s}^{-1} \] Thus, the correct answer is 14.4 kg m\( ^2 \)s\(^{-1} \). 
 

Answer 2

Step 1: Given data:
- Mass \( m = 2 \) kg
- Radius of circular path \( r = 0.6 \) m
- Height of table from ground \( h = 0.8 \) m
- Angular speed \( \omega = 12 \) rad/s

Step 2: Angular momentum \( \vec{L} \) about a point is given by:
\( \vec{L} = \vec{r} \times \vec{p} \), where:
- \( \vec{r} \) is the position vector from the reference point to the particle
- \( \vec{p} = m \vec{v} \) is the linear momentum

Step 3: Find the linear speed of the particle:
\( v = r \omega = 0.6 \times 12 = 7.2 \) m/s

Step 4: Position vector of the particle from the point on the ground under the center of the circle:
Since the center is at height 0.8 m and the radius is 0.6 m (horizontal), the position vector from the ground point to the particle has both vertical and horizontal components:
Magnitude of position vector: \( R = \sqrt{r^2 + h^2} = \sqrt{0.6^2 + 0.8^2} = \sqrt{0.36 + 0.64} = \sqrt{1.0} = 1.0 \) m

Step 5: Angular momentum magnitude about the point is:
\( L = m v R \sin\theta \), where \( \theta \) is the angle between \( \vec{r} \) and \( \vec{v} \)
Since \( \vec{v} \) is tangential to the circle and perpendicular to \( \vec{r} \), \( \sin\theta = 1 \)
So,
\( L = 2 \cdot 7.2 \cdot 1 = 14.4 \) kg·m²/s

Final Answer: \( 14.4 \) kg·m²/s