Gymnast's Floor Routine: Angular Velocity & Time

[ussjt]

A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from 3.00 to 4.65 rev/s while rotating through one-half of a revolution. How much time does this maneuver take?

ok I understand that I am going to be using one of the equations of kinematics for rotational and linear motion. I just don't understand what I am suspose to use the "one-half of a revolution" for.

 

[d_leet]

What angle is one half of a revolution?

 

[kyle8921]

The "one-half of a revolution" refers to the angle through which the gymnast rotates during the tumbling run. This information is important because it allows us to calculate the time it takes for the gymnast to complete the maneuver. Using the equation for angular velocity, we can set up the following equation:

ω = Δθ/Δt

Where:
ω = angular velocity (in rev/s)
Δθ = change in angle (in rev)
Δt = change in time (in s)

We know that the initial angular velocity (ωi) is 3.00 rev/s and the final angular velocity (ωf) is 4.65 rev/s. We also know that the change in angle (Δθ) is one-half of a revolution, which is equivalent to 0.5 rev. Substituting these values into the equation, we get:

4.65 rev/s = 0.5 rev / Δt

Solving for Δt, we get:

Δt = 0.5 rev / 4.65 rev/s
Δt = 0.1075 s

Therefore, the maneuver takes approximately 0.1075 seconds to complete. Keep in mind that this is an ideal calculation and does not account for factors such as air resistance and the gymnast's body movements, which may affect the actual time taken.