A force \(\vec{F}=\boldsymbol{\alpha}^{\hat{i}}+3 \hat{j}+6 \hat{k}\) is acting at a point \(\vec{r}=2 \hat{i}-6 \hat{j}-12 \hat{k}\). The value of α for which angular momentum about origin is conserved is

Concept:

The angular momentum (L) about the origin is given by the cross product of the position vector (r) and the linear momentum (p). For a force F acting on a particle, the time rate of change of angular momentum is given by:

dL/dt = r × F

For angular momentum to be conserved, the time derivative must be zero:

r × F = 0

If the above condition is satisfied, the angular momentum about the origin is conserved.

Calculation:

Given the force F = αi + 3j and the position vector r = 2i - 6j - k.

The cross product of r and F is:

r × F = (2i - 6j - k) × (αi + 3j)

Using the properties of the cross product and calculating each component:

r × F = (2 × 3 - (-6) × α)i + ((-6) × α - 2 × 3)j + ((2 × 3) - (-6) × α)k

r × F = (6 + 6α)i + (-6α - 6)j + (6 + 6α)k

For angular momentum to be conserved, r × F = 0, which gives the following system of equations:

6 + 6α = 0,

-6α - 6 = 0,

6 + 6α = 0.

Solving these equations, we find α = -1.

∴ The correct answer is option 4: α = -1.