A diver performing a double somersault spins at an angular speed of 4.5 π rad/s precisely 0.70s after leaving the platform.
Assuming the diver begins with zero initial angular speed and accelerates at a constant rate, what is the diver’s angular acceleration during the double somersault?
Answer in units of rad/s^2.
ω = ε•t
ε = ω/t= 4.5•π/0.7=20.2 rad/s^2
To find the angular acceleration of the diver during the double somersault, we can use the following equation:
ωf = ωi + αt
Where:
ωf is the final angular speed (4.5π rad/s),
ωi is the initial angular speed (0 rad/s),
α is the angular acceleration (unknown),
t is the time interval (0.70 s).
Rearranging the equation, we have:
α = (ωf - ωi) / t
Substituting the given values, we get:
α = (4.5π rad/s - 0 rad/s) / 0.70 s
Simplifying, we find:
α = (4.5π rad/s) / 0.70 s
α ≈ 20.45 rad/s^2
Therefore, the diver's angular acceleration during the double somersault is approximately 20.45 rad/s^2.
To find the diver's angular acceleration during the double somersault, we need to use the equations of rotational motion. The equations we'll need are:
1. θ = ω*t + (1/2)*α*t^2
2. ω = ω0 + α*t
Let's break down the information given:
- Initial angular speed (ω0) = 0 rad/s
- Final angular speed (ω) = 4.5π rad/s
- Time elapsed (t) = 0.70 s
From equation 2, we can rearrange it to solve for the angular acceleration (α):
α = (ω - ω0) / t
Plugging in the given values:
α = (4.5π rad/s - 0 rad/s) / 0.70 s
Simplifying the expression:
α = (4.5π / 0.70) rad/s^2
Therefore, the diver's angular acceleration during the double somersault is approximately 19.091 rad/s^2.