fbpx

A rotor consists of four horizontal blades each of length L = 4m and mass m = 90 kg cantilevered off of a vertical shaft. Assume that each blade can be modeled as having its mass concentrated at its midpoint. The rotor is initially at rest when it is subjected to a moment M =\beta t, with \beta = 60N.m/s. Determine the angular speed of the rotor after 10 s.

Step-by-step

From the rotor’s FBD we see that the weight of each blade and the force N do not contribute a moment about the z axis because they are parallel to the z axis. Hence, applying the angular impulse–momentum principle in the z direction we have

\int _{ { t }_{ 1 } }^{ { t }_{ 2 } }{ Mdt={ h }_{ Oz } } ({ t }_{ 2 })-{ h }_{ Oz }({ t }_{ 1 })     (1)

where t1 = 0 and t2 = 10 s and where

{ h }_{ Oz }({ t }_{ 1 })=0 and 4\left[ m\frac { L }{ 2 } (\dot { \theta } ({ t }_{ 2 })\frac { L }{ 2 } ) \right]      (2)

where \dot { \theta } is the angular velocity of the rotor, assumed positive if in the positive z direction, and where we have accounted for the fact that the system starts from rest and that, due to the symmetry of the system, the total angular momentum about the z axis of the system is 4 times the angular momentum about the z axis of a single blade.

Now observe that

\int _{ { t }_{ 1 } }^{ { t }_{ 2 } }{ Mdt } =\int _{ { t }_{ 1 } }^{ { t }_{ 2 } }{ \beta tdt } =\frac { 1 }{ 2 } \beta { ({ t }_{ 2 }^{ 2 }-{ t }_{ 1 }^{ 2 }) }^{ 2 }=\frac { 1 }{ 2 } \beta { t }_{ 2 }^{ 2 }     (3)

Substituting Eqs. (2) and (3) into Eq. (1) and then solving for \dot { \theta } ({ t }_{ 2 }), we have

\dot { \theta } ({ t }_{ 2 })=\frac { \beta { t }_{ 2 }^{ 2 } }{ 2m{ L }^{ 2 } }     (4)

which gives

\dot { \theta } ({ t }_{ 2 })=2.08rad/s

after substituting the following numerical data:\beta = 60N.m/s, t2 = 10 s, m = 90 kg, and L = 4 m.

 

Trending

Online
Courses

Private
Teachers

Hire Our
Professionals

Study
Materials

Writing & Translation

View All
Services