Question #115563
A machine part rotates at an angular speed of ωi. Its speed is then increased to ωf at an angular acceleration of α. If both the initial and final angular speeds are now doubled and the angular acceleration remains the same, by what factor is the angular displacement changed?
1
Expert's answer
2020-05-13T10:12:48-0400

The problem is similar to determining the distance for the movement with constant acceleration.

Let us write the dependence of angular velocity on time:

ω(t)=ωi+αt.\omega(t) = \omega_i + \alpha t. Therefore, displacement is Δφ=ωit+αt22\Delta \varphi = \omega_i t + \dfrac{\alpha t^2}{2} .

Let t0t_0 be the time of movement, therefore ωf=ωi+αt0.\omega_f = \omega_i + \alpha t_0. So t0=ωfωiα.t_0 = \dfrac{\omega_f-\omega_i}{\alpha}. Therefore,

Δφ=ωit0+αt022=ωf2ωi22α.\Delta \varphi = \omega_i t_0 + \dfrac{\alpha t_0^2}{2} = \dfrac{\omega_f^2-\omega_i^2}{2\alpha}.

If we double the velocities and the acceleration remains the same, then

Δ2φ=4ωf24ωi22α=4ωf2ωi22α.\Delta_2 \varphi = \dfrac{4\omega_f^2-4\omega_i^2}{2\alpha} = 4\dfrac{\omega_f^2-\omega_i^2}{2\alpha}. So the factor is 4.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS