Question

Instructions: Provide full solutions to the following problems:
A propeller consists of two blades, each 3.00 m in length and with a mass of 120 kg. The propeller can be modeled as a single rod rotating about its center of mass. The propeller starts from rest and rotates up to 1200 rpm in 30. s at a constant rate.
A)Find the angular momentum of the propeller at t = 10 s and t = 20 s.
B)Find the torque on the propeller.

Answer 1

According to the problem,The propeller is a rod of length 6 m and mass 2*120=240kg

Moment of inertia of rod is given by: ml²/3 where m is mass of the rod and l is its length

So, moment of inertia I for given rod is:I=240*6*6/3=2880 kg-m²

A)Under uniform angular acceleration:w_f=w_i+at where w_f is the final angular velocity, w_i is the initial angular velocity,a is angular acceleration and t is time interval.

The propeller starts from rest and rotates up to 1200 rpm in 30 seconds.

So,w_i=0 rad/s (as propeller starts from rest)

w_f=1200 rpm= 1200*2/60 rad/s=125.66 rad/s

t=30 s

So, 125.66=0+a*30=>a=125.66/30=4.19 rad/s²

Now, at t=10 s,again using the above eqn of motion,we get

w(10)=0+4.19*10      (here w(10)=angular velocity at t=10s,w_i=0 rad/s,a=4.19 rad/s²,t=10 s)

So,w(10)=41.9 rad/s

Also,at t=20s,w(20)=0+4.19*20      (here w(20)=angular velocity at t=20s,w_i=0 rad/s,a=4.19 rad/s²,t=20 s)

So,w(20)=83.8 rad/s

Angular momentum is given by:L=I*w,where L is angular momentum,I is moment of inertia,w is angular velocity

So,angular momentum at t=10s=I*w(10)=2880*41.9=120672 kg-m²/s

Angular momentum at t=20s=I*w(20)=2880*83.8=241344 kg-m²/s

B)=Ia, where =torque, I=moment of inertia,a= angular acceleration

Here,I=2880 kg-m²,a=4.19 rad/s² (from part A)

So,=2880*4.19=12067.2 kg-m