A particle of mass 'm' is moving in time 't' on a trajectory given by
r⃗ = 10α t²î + 5β (t - 5)ĵ
Where α and β are dimensional constants.
The angular momentum of the particle becomes the same as it was for t = 0 at time t = ____________ seconds.

Solution

Physics sir · 4 steps

Find L as a function of time, notice it is zero at t = 0, and ask when it becomes zero again.

1. Velocity first. Differentiate the position: v⃗ = dr⃗/dt = 20α t î + 5β ĵ.
2. Cross product. Both vectors lie in the xy-plane, so L⃗ = m r⃗ × v⃗ points along k̂ with size m(x v_y - y vₓ).

   L_z = m[10α t² (5β) - 5β(t-5)(20α t)] = 50 mαβ t (10 - t)

3. Starting value. At t = 0, L_z = 0. Here r⃗ = -25β ĵ and v⃗ = 5β ĵ are parallel, so the cross product vanishes.
4. So. Setting 50 mαβ t(10 - t) = 0 gives t = 0 or t = 10. The angular momentum is back to its t = 0 value at t = 10 s.

The idea to keep. For motion in a plane, L = m(x v_y - y vₓ). Write it as a function of t, then set it equal to its starting value and solve for the other root.

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