Engineering
Physics
Relative Motion
Moment of Inertia
Angular Momentum and Conservation of Angular Momentum
Question

Match the column

Column-I Column-II
(A) Increases (P) A small particle projected upwards at an angle (<90°) with horizontal in a projectile motion. Its angular momentum relative to the point of projection ...... with time.
(B) Decreases (Q) A train is moving along equator with a constant speed in the direction from east to west. If it changes its direction and moves from west to east without changing the speed then the force it exerts on the rails ......... after changing the direction.
(C) Remains same (R) A uniform triangular plate is made to rotate about an axis along the smallest side and then about an axis along the longest side if the torque applied is same in both cases then by changing the axis its angular acceleration ..........
  (S) A rough uniform disc of mass m and radius R rotating with angular velocity ω is connected coaxially with another uniform ring of mass 2m and radius R rotating with angular velocity ω/2 then in the absence of any external torque during connection, the angular momentum of system about the axis ..........
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Solution
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This question tests understanding of angular momentum (L), torque (τ), and moment of inertia (I). The key principle is conservation of angular momentum (L = Iω) when net external torque is zero. For rotational dynamics, τ = Iα.

For (P): Angular momentum L = m(r × v). The perpendicular distance from the point of projection changes with time, so L increases.

For (Q): The force on rails relates to the apparent weight. On equator, g' = g - ω²R. Changing direction changes the relative ω, thus the centripetal force and the normal reaction (force on rails) changes.

For (R): Angular acceleration α = τ/I. Changing the axis changes the moment of inertia I. Since τ is same, α changes.

For (S): No external torque, so angular momentum is conserved. Linitial = Ldisc + Lring = (½mR²)ω + (2mR²)(ω/2) = (½ + 1)mR²ω = (3/2)mR²ω. Lfinal = Itotalωf = (½mR² + 2mR²)ωf = (5/2)mR²ωf. Setting equal, ωf = (3/5)ω, so L remains same.

Final Answer: (A) matches (P), (B) matches (Q), (B) matches (R), (C) matches (S).