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The body of the satellite shown has a weight that is negligible with respect to the two spheres A and B that are rigidly attached to it, which weigh 150 lb each. The distance between A and B from the spin axis of the satellite is R = 3.5 ft. Inside the satellite there are two spheres C and D weighing 4 lb mounted on a motor that allows them to spin about the axis of the cylinder at a distance r = 0.75 ft from the spin axis. Suppose that the satellite is released from rest and that the internal motor is made to spin up the internal masses at a constant time rate of 5.0 rad/{ s }^{ 2 } for a total of 10 s. Treating the system as isolated, determine the angular speed of the satellite at the end of spin-up.

Step-by-step

The final angular velocity of the internal masses is

{ \omega }_{ i }=(5.00rad/{ s }^{ 2 })(10.0s)=50.00rad/s     (1)

where the subscript i stands for internal. Since the system is isolated, the angular momentum of the system must be conserved throughout the motion of the system. In particular, assuming that the spin axis of the system does not change orientation, then the angular momentum about the spin axis must be conserved, so that we have

({ m }_{ A }+{ m }_{ B }){ R }^{ 2 }{ \omega }_{ satellite }=({ m }_{ c }+{ m }_{ D }){ r }^{ 2 }{ \omega }_{ i }\Longrightarrow { \omega }_{ satellite }=\frac { ({ m }_{ c }+{ m }_{ D }){ r }^{ 2 } }{ ({ m }_{ A }+{ m }_{ B }){ R }^{ 2 } } { \omega }_{ i }     (2)

{ \omega }_{ satellite }=0.0612rad/s

where, in addition to the result in Eq. (1), we have used the following numerical data:

{ m }_{ C }=\frac { 4\quad lb }{ 32.2ft/{ s }^{ 2 } } =0.1242slug

 

{ m }_{ D }=\frac { 4\quad lb }{ 32.2ft/{ s }^{ 2 } } =0.1242slug

 

{ m }_{ A }=\frac { 150\quad lb }{ 32.2ft/{ s }^{ 2 } } =4.658slug

 

{ m }_{ B }=\frac { 150\quad lb }{ 32.2ft/{ s }^{ 2 } } =4.658slug

 

r=0.75ft

R=3.5ft

 

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