A shaft is turning at 71.6rad/s at time zero. Thereafter, its angular acceleration is given by
α = –10.8rad/s2 – 4.42 t rad/s3
where t is the elapsed time. Calculate its angular speed at t = 2.73s.
and
How far does it turn in these 2.73s
dw/dt = -10 - 4.42 t
so
w = -10 t - 2.21 t^2 + constant
find constant
at t = 0
71.6 = -10(0) - 2.21(0)^2 + c
so c = 71.6
w = 71.6 - 10 t - 2.21 t^2
put in t = 2.73
Now the angle
dTheta/dt = w = 71.6-10 t -2.21 t^2
so
Theta = 71.6 t - 5 t^2 - (2.21/3)t^3
To calculate the angular speed of the shaft at t = 2.73s, we need to integrate the angular acceleration function over the given time period.
The angular acceleration function is given by α = –10.8 rad/s^2 – 4.42t rad/s^3, where t is the elapsed time.
To find the angular speed, we need to integrate α with respect to time to obtain the angular velocity ω:
ω = ∫ α dt
To integrate the function, we need to break it down into two separate parts: –10.8 rad/s^2 and –4.42t rad/s^3. Integrating each part separately:
∫ –10.8 dt = –10.8t (integration of a constant term)
∫ –4.42t dt = –2.21t^2 (integration of t)
Now, let's calculate the angular velocity ω at t = 2.73s:
ω(t) = –10.8t – 2.21t^2
ω(2.73) = –10.8(2.73) – 2.21(2.73)^2
ω(2.73) = –29.484 – 15.228
ω(2.73) ≈ –44.712 rad/s
The angular speed of the shaft at t = 2.73s is approximately –44.712 rad/s (rotating in the negative direction).
To calculate how far the shaft turns in these 2.73s, we need to integrate the angular velocity function over the time period:
θ = ∫ ω dt
Integrating the angular velocity ω = –10.8t – 2.21t^2:
∫ –10.8t dt = –5.4t^2 (integration of t)
∫ –2.21t^2 dt = –0.737t^3 (integration of t^2)
Now, let's calculate the angular displacement θ at t = 2.73s:
θ(t) = –5.4t^2 – 0.737t^3
θ(2.73) = –5.4(2.73)^2 – 0.737(2.73)^3
θ(2.73) = –38.0718 – 36.1239
θ(2.73) ≈ –74.1957 radians
The shaft turns approximately –74.1957 radians in these 2.73 seconds.