If L⃗ and P⃗ represent the angular momentum and linear momentum respectively of a particle of mass ' m ' having position vector as r⃗=a(î cos ω t+ĵ sin ω t). The direction of force is

Solution

Physics sir · 3 steps

Differentiate the position twice and you get -ω²r⃗, so the force points back at the centre.

1. Velocity first. Differentiate r⃗ = a(î cos ω t + ĵ sin ω t) once to get v⃗ = aω(-î sin ω t + ĵ cos ω t). Its magnitude is a constant aω, so this is uniform circular motion of radius a.

   F⃗ = md²r⃗/dt² = -mω² a(î cos ω t + ĵ sin ω t) = -mω²r⃗

2. Check every vector. P⃗ = mv⃗ is tangential. L⃗ = r⃗×P⃗ = ma²ω k̂ is perpendicular to the plane. Using k̂×î = ĵ and k̂×ĵ = -î gives L⃗×P⃗ = -m²a²ω² r⃗.

   Vector	Direction	Angle with F⃗
   r⃗	radially outward	180°, exactly opposite
   L⃗	along +k̂, out of the plane	90°
   P⃗	tangential	90°
   L⃗×P⃗	radially inward	0°, same direction

3. So. The force is centripetal and points opposite to r⃗, which is option (c).

The idea to keep. Whenever r⃗ has the form a(cos ω t, sin ω t), the acceleration is -ω²r⃗. The net force in uniform circular motion always points opposite to the position vector from the centre.

Answer from the NTA final answer key. Solution by PrepWiser.