An airplane propeller with a moment of inertia (about its center ) equal to 55 k
ID: 1292001 • Letter: A
Question
An airplane propeller with a moment of inertia (about its center ) equal to 55 kg m^2 starts spinning at 330 rpm. The airplane's engine that takes 25s to accelerate the propeller (at constant acceleration) up to a new rate of 1600 rpm. Express all answers in MKS units. (a). What is the propeller's angular acceleration? (b). If the propeller measures 1.5 m from center to tip, what is the total distance traveled by the tip of the propeller during the acceleration in part (a)? What is the tangential velocity of a point on the tip of the propeller once it reaches 1600 rpms? Express in MPH as well as m/s. (e). After the pilot completes all the flight plans, the engine is shut off. It slows down inaccordance with this expression w(t)=0.010t^3+167.55, where w is in rad/s and t is in seconds. Determine expression for the acceleration and angular position of the propeller. For question (e) how long does the propeller take to stop?
Explanation / Answer
given
MOI I = 55 kg m^2
wi = 330rpm = (330*2*3.14)/(60) =34.54 rad/s
wf = 1600 rpm = (1600*2*3.14)/(60) = 167.47 rad/s
a)
alfa = (wf - wi)/t = (167.47-34.54)/25 = 5.3172 rad/s^2
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b) theta = wi*t + 0.5*alfa*t^2
theta = (34.54*25)+(0.5*5.3172*25*25) = radians
distance = R*theta = 3787.6875 m
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c) Vf = R*Wf = 1.5*167.47 = 251.205 m/s
Vf = 251.205*(1/1609)*(60*60) = 562.05 MPH
d) alfa = dw/dt = 3*0.01*t^2 = 0.03*t^2
angular position = integration of w*dt
theta = 0.01*t^4/4 + 167.55t
theta = 0.0025*t^4 + 167.55*t
e) alfa = wf - wi/t
0.03t^2 = 167.55/t
0.03*t^3 = 167.55
t = 17.74 s