A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v₀ in the x-y plane. At a time t

Question

A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v₀ in the x-y plane. At a time t<(v₀ sin θ)/g, the angular momentum of the particle about the origin is (where ̂i, ̂j and ̂k are unit vectors along the x, y and z axes respectively)

Accepted Answer

(4) -1/2mgv₀t² cos θ ̂k. Write the position and velocity at time t, with uₓ=v₀ cos θ and u_y=v₀ sin θ. x=uₓt, y=u_yt-1/2gt²; vₓ=uₓ, v_y=u_y-gt. In the plane, L_z=m(xv_y-yvₓ). L_z=m[uₓt(u_y-gt)-(u_yt-1/2gt²)uₓ]. =muₓ[u_yt-gt²-u_yt+1/2gt²]=muₓ(-1/2gt²). ⃗L=-1/2mgv₀t² cos θ ̂k. The sign says the particle turns clockwise about the origin, which is right for a projectile launched into the first quadrant. Cross-check with torque: τ⃗=⃗r×(-mĝj) has z-component -mgx=-mgv₀t cos θ, and integrating that from 0 to t gives exactly the L_z above.

A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v₀ in the x-y plane. At a time t<(v₀ sin θ)/g, the angular momentum of the particle about the origin is (where ̂i, ̂j and ̂k are unit vectors along the x, y and z axes respectively)

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