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$$^{-1}$$, the magnitude of its angular momentum about a point on the ground right under the center of the circle is:

We first note all the data that the question supplies. The mass of the particle is $$m = 2\ \text{kg}$$. The particle moves in a horizontal circle of radius $$r_h = 0.6\ \text{m}$$ on a smooth table. The table is $$h = 0.8\ \text{m}$$ above the ground. Its angular speed is $$\omega = 12\ \text{rad s}^{-1}$$. We have to find the magnitude of the angular momentum of the particle about the point on the ground that lies exactly under the centre of the circle.

Let us choose that ground point and call it $$O$$. Let the centre of the circular path be the point $$C$$, which is vertically above $$O$$. Hence $$OC = h = 0.8\ \text{m}$$. Because the particle is on the circumference of the horizontal circle, at any instant its horizontal displacement from the centre is $$r_h = 0.6\ \text{m}$$. So the particle’s position vector $$\vec r$$, drawn from $$O$$ to the instantaneous position $$P$$ of the particle, has two perpendicular components:

$$OC = 0.8\ \text{m}$$ vertically upward, and $$CP = 0.6\ \text{m}$$ horizontally outward. Since these two components are at right angles, the magnitude of $$\vec r$$ is obtained using Pythagoras’ theorem:

$$|\vec r| = \sqrt{(0.8)^2 + (0.6)^2} = \sqrt{0.64 + 0.36} = \sqrt{1.00} = 1.0\ \text{m}.$$

Now we need the magnitude of the linear momentum $$\vec p$$ of the particle. The speed $$v$$ in uniform circular motion is related to the angular speed by the formula

$$v = \omega r_h.$$

Substituting $$\omega = 12\ \text{rad s}^{-1}$$ and $$r_h = 0.6\ \text{m}$$, we get

$$v = 12 \times 0.6 = 7.2\ \text{m s}^{-1}.$$

Therefore, the magnitude of the linear momentum is

$$|\vec p| = m v = 2 \times 7.2 = 14.4\ \text{kg m s}^{-1}.$$

To find the angular momentum about point $$O$$, we use the vector definition

$$\vec L = \vec r \times \vec p.$$

When the cross-product involves two perpendicular vectors, the magnitude simplifies to

$$|\vec L| = |\vec r|\,|\vec p|\,\sin 90^\circ = |\vec r|\,|\vec p|,$$

because $$\sin 90^\circ = 1$$. In our geometry the position vector $$\vec r$$ is perpendicular to the instantaneous velocity (and hence to $$\vec p$$), so the angle between them is indeed $$90^\circ$$. Hence

$$|\vec L| = (1.0\ \text{m})(14.4\ \text{kg m s}^{-1}) = 14.4\ \text{kg m}^2 \text{ s}^{-1}.$$

Hence, the correct answer is Option A.