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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA particle of mass \( m \) is moving in a circular path of radius \( R \) with an angular velocity \( \omega \). A constant tangential retarding force \( F \) acts on it. The time required for the angular momentum of the particle about the center of the path to reduce to half of its initial value is:
Correct answer · Option 3
\( \frac{m \omega R}{2F} \)
Explanation
Initial angular momentum \( L_i = m v R = m (\omega R) R = m \omega R^2 \). Retarding torque \( \tau = F R \). From \( \frac{dL}{dt} = -\tau \), \( \Delta L = L_i - \frac{L_i}{2} = \frac{m \omega R^2}{2} = \tau t = (FR)t \implies t = \frac{m \omega R}{2F} \).
