The space capsule has a mass of 1200 kg and a moment of inertia ig = 900 kg·m2 about an axispassing through g and directed perpendicular to the page. if it is traveling forward with a speed vg = 800 m/s and executes a turn by means of two jets which provide a constant thrust of 400 n for 0.3 s, determine the capsule's angular velocity just after the jets are turned off.

can you explain what has done here?

Explanation:

Angular impulse = change in angular momentum

Mg Δt = ΔHg

Mg Δt = Hg₂ − Hg₁

Rearranging, initial angular momentum + angular impulse = final angular momentum

Hg₁ + Mg Δt = Hg₂

Initially, the rocket is not turning, so ω₁ = 0 rad/s.

Hg₁ = I ω₁

Hg₁ = (900 kg m²) (0 rad/s)

Hg₁ = 0 kg m²/s

The total moment applied is the sum of the perpendicular components of each force multiplied by the distance.  Take counterclockwise to be positive.

Mg = (400 N cos 15° × 1.5 m) + (400 N cos 15° × 1.5 m)

Mg = 2 (400 N cos 15° × 1.5 m)

The time of the impulse Δt is 0.3 s.

The final angular momentum is the product of the moment of inertia and the final angular velocity.

Hg₂ = I ω₂

Hg₂ = (900 kg m²) ω₂

Plugging everything in:

0 kg m²/s + 2 (400 N cos 15° × 1.5 m) (0.3 s) = (900 kg m²) ω₂

ω₂ = 0.386 rad/s

what was the answer?

do you have a formula possibly