Amerry-go-round rotates at the rate of
0.28 rev/s with an 84 kg man standing at
a point 2.3 m from the axis of rotation.
what is the new angular speed when the
man walks to a point 0 m from the center?
consider the merry-go-round is a solid 35 kg
cylinder of radius of 2.3 m.
answer in units of rad/s.

Explanation:

Angular momentum is conserved.

I₁ ω₁ = I₂ ω₂

(½ Mr² + md²) ω₁ = (½ Mr²) ω₂

(½ (35 kg) (2.3 m)² + (84 kg) (2.3 m)²) (0.28 rev/s) = (½ (35 kg) (2.3 m)²) ω

ω = 1.624 rev/s

ω = 10.2 rad/s

Round as needed.

think its the third option

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day