Explanation
The centre of mass would lie at the point of contact. Therefore, required distance = 0.
The moment of inertia is maximum about axis 3, because rms distance of mass is maximum for this axis.
Let x be increase in length of the spring. The particle would move in a circular path of radius (l +x).
Centripetal force = force due to the spring
As mass decreases, moment of inertia I decreases. Since L = I ω is constant, therefore ω increases.
As the plate is a uniform square plate, its moment of inertia about any axis in its plane passing through the centre of plate must be the same.
The cord is most likely to break at the orientation, when mass is at B as tension in the string at this point is maximum.
As P lies on the line along which the particle moves, angular momentum of the particle about P is L = mυ × r = mυ × 0 = 0.
In uniform circular motion, the only force acting on the particle is centripetal force (towards the centre). Torque exerted by this force about the centre is zero. Therefore, angular momentum about centre remains conserved.
The centre of mass of bodies B and C taken together does not shift as no external force is involved.
When liquid of mass m is dropped at the centre, moment of inertia I increases. As Iω angular momentum is constant, therefore, ω decreases initially. However, when the liquid falls finally, I reduces and hence ω increases to the initial value.
Please disable the adBlock and continue. Thank you.