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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA particle of mass \( m = 0.5 \text{ kg} \) rotates in the \( xy \)-plane in a circle of radius \( r = 2 \text{ m} \) centred at the origin with constant angular speed \( \omega = 4 \text{ rad s}^{-1} \). If the origin is shifted along the \( z \)-axis such that the position vector is \( \mathbf{r} = \mathbf{r}_{\text{in-plane}} + 3\mathbf{\hat{k}} \), the magnitude of the component of angular momentum perpendicular to the axis of rotation (\( |\mathbf{L}_{\perp}| \)) is:
Correct answer · Option 3
\( 12 \text{ J s} \)
Explanation
From \( \mathbf{L} = \mathbf{r} \times \mathbf{p} = (\mathbf{r}_{\text{in-plane}} + z\mathbf{\hat{k}}) \times m\mathbf{v} \), the perpendicular component is \( \mathbf{L}_{\perp} = z\mathbf{\hat{k}} \times m\mathbf{v} \). Since \( v = \omega r = 4 \times 2 = 8 \text{ m s}^{-1} \), \( |\mathbf{L}_{\perp}| = z m v = 3 \times 0.5 \times 8 = 12 \text{ J s} \).
