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During the launch from a board, a diver's angular speed about her center of mass changes from zero to 6.90 rad/s in 250 ms. Her rotational inertia about her center of mass is 12.0 kg·m2.

a. During the launch, what was the magnitude of her average angular acceleration?
b. During the launch, what was the magnitude of the average external torque on her from the board?

1 Answer

5 votes

Answer:

a) α = 27.6 rad/s²

b) τ = 331.2 N*m

Step-by-step explanation:

Given:

ωinitial = 0 rad/s

ωfinal = 6.90 rad/s

Δt = 250 ms = 0.25 s

I = 12.0 Kg*m²

a) We can use the equation

α = Δω / Δt

α = (ωfinal - ωinitial) / Δt

⇒ α = (6.90 rad/s - 0 rad/s) / 0.25 s

⇒ α = 27.6 rad/s²

b) we apply the formula

τ = I*α ⇒ τ = (12.0 Kg*m²)*(27.6 rad/s²)

⇒ τ = 331.2 N*m

answered
User Raaz
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