Angular Kinetics 4

Question

A diver is diving from a height of 20 m. First, he makes a rotation (body tucked). At the end of this rotation, he is still at 12 m from the water, in vertical position (head up) with an angular velocity of 2𝜋𝜋/3 rad/s and a vertical velocity equal to 1.8 m/s downwards.

a) How long will it take for him to reach the water?

b) To complete another 2.5 rotations, what constant angular velocity must he have from this point on?

c) What change must he make to his moment of inertia to achieve this? (express If related to Ii)?

Answer 1

Start with the kinematic equation Δy = v_o t + 1/2 at² with v_o = 0 for no initial vertical component of velocity. From 20 down to 12 m, 12 - 20 = (-9.8)/2 t₁², t₁² = -8 /(-4.9) and t₁ = 1.2₇₈ s. Do the same from 20 down to zero (not from 12 to zero because there is no need to calculate the new v_o): 0 - 20 = -4.9 t₂² and t₂ = 2.0₂₀ s. [subscripted digits are not significant, but avoid round-off error in results derived from them].

a) Δt = t₂ - t₁ = 0.74₃ seconds remaining from 12 m until reaching the water.

b) 2.5 rev * (2π rad/1 rev) = 15.₇ radians, ω₂ = 15.₇ /0.74₃ = 21 rad/s.

c) With conservation of angular momentum, L = I₁ ω₁ = I₂ ω₂, I₂ / I₁ = (2π/3) / 21.1 = 0.10. The moment of inertia must be decreased by a factor of 10.