Two kids walk up to a small histrorical airplane with a simple propeller of mass

Question

Two kids walk up to a small histrorical airplane with a simple propeller of mass 5.5 kg. The kids apply two forces, to try to get the propeller to turn. The propeller is quite rusty and there is some frictional torque that must be overcome. The first kid (F1) applies 42 N at the angle of 45 degrees and the second kid (F2) applies 21 N perpendicularly. The length of the propeller is 1.2 meters, and you can approximate it as a uniform rod rotating about its center.

a) Calculate the resulting torques for both kids. Which kid produces a greater torque on the propeller?

b) If they get the propeller to accelerate at a rate of 32 rad/s^2, what must be the frictional torque in the bearings?

c) What is the tangential acceleration at the edge of the propeller once the kids get it going as described in part b.

Explanation / Answer

r = l/2 = 24/2 = 12 cm = 0.12 m

initial angular momentum = 0

final angular momentum of projectile = m*v*r

final angular momentum of rod = I*w = (1/12)*M*l^2*w

from momentum conservation Lf = Li

(1/12)*M*l^2*w + m*v*r

(1/12)*0.04*0.24*0.24*w = (0.005*3*0.12)

w = 9.375 rad/s

r = L/2

a) torque Q1 = F1*r*sin45 = 42*0.6*sin45 = 17.82 Nm

torque Q2 = F2*r*sin90 = 12.6 Nm

Q1 > Q2

the first kid

b) net torque = I*alfa

I = (1/12)*M*L^2

(Q1 +Q2 +Qf)*r = I*alfa

(17.82+12.6+Qf) = (1/12)*5.5*1.2*1.2*32

Qf = -9.3 Nm

#c) a_tan = r*alfa = 0.6*32 = 19.2 m/s^2

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