A shaft is turning at angular speed ω at time t = 0. Thereafter, its angular acceleration is given by
α = A + Bt
(a) Find the angular speed of the shaft at time t. (Use any variable or symbol stated above as necessary.)
(b) Through what angle does it turn between t = 0 and t? (Use any variable or symbol stated above as necessary.)
I got a. right.
ω(t) = At + B(t^2)/2 + ω
I can't get b, though, and I will never get b right since I've used up my allotted tries. That wouldn't bother me so much if it would just show what the correct answer is. I'm starting to think about it more (even though it's absolutely pointless since I can NEVER get that one point), and I think you take the integral of the angular velocity at time (t) and do that from 0 to t to find the change in angle (Δθ). Is that correct?
Preemptive huge thanks for an answer. I am incredibly angry right now and this anger will not subside until I get a freaking answer.
α = A + Bt
(a) Find the angular speed of the shaft at time t. (Use any variable or symbol stated above as necessary.)
(b) Through what angle does it turn between t = 0 and t? (Use any variable or symbol stated above as necessary.)
I got a. right.
ω(t) = At + B(t^2)/2 + ω
I can't get b, though, and I will never get b right since I've used up my allotted tries. That wouldn't bother me so much if it would just show what the correct answer is. I'm starting to think about it more (even though it's absolutely pointless since I can NEVER get that one point), and I think you take the integral of the angular velocity at time (t) and do that from 0 to t to find the change in angle (Δθ). Is that correct?
Preemptive huge thanks for an answer. I am incredibly angry right now and this anger will not subside until I get a freaking answer.
-
Hello,
ω(t) = At + B(t^2)/2 + ω
Well, lets see....ω is your initial angular velocity, right?
Yes, your logic seems sound, the integral would be θ(t) = (A/2)t^2 + (B/6)t^3 + ω t + θ
Taking the actual integral from t=0 to t only does one thing, it gets rid of your initial offset angle θ,
θ = {(A/2)t^2 + (B/6)t^3 + ω t + θ} - [(A/2)0^2 + (B/6)0^3 + ω 0 + θ]
So θ = (A/2)t^2 + (B/6)t^3 + ω t
Good Luck
ω(t) = At + B(t^2)/2 + ω
Well, lets see....ω is your initial angular velocity, right?
Yes, your logic seems sound, the integral would be θ(t) = (A/2)t^2 + (B/6)t^3 + ω t + θ
Taking the actual integral from t=0 to t only does one thing, it gets rid of your initial offset angle θ,
θ = {(A/2)t^2 + (B/6)t^3 + ω t + θ} - [(A/2)0^2 + (B/6)0^3 + ω 0 + θ]
So θ = (A/2)t^2 + (B/6)t^3 + ω t
Good Luck