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UrsaMajor
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The problem statement:
A flyball governor is a device used to regulate the speed of steam turbines in steam power plants. The rotation of the shaft causes the two balls move outward. As the balls move out, they pull on the bearing A. The position of the slider A is linked to a valve admitting steam into the turbine. This way, the flow of speed is automatically restricted when the shaft starts rotating too fast.
What is the force (in N) pulling on the slider A if the shaft rotates at the speed given below? Ignore friction and the weights of the components other than the two balls.
http://s589.photobucket.com/user/johnnigan/media/flyballgovernor.jpg.html[/URL]
Given data:
L = 300 mm
Angle gamma = 45 degrees
mass of each ball = 1 kg (the mass of each ball)
RPM = 900 revolutions per minute
The attempt at a solution:
I understand the problem. As the shaft spins, the balls are subjected to centripetal acceleration which causes them to move away from the shaft and draw the attached rods away from the shaft thus subjecting the bearing A to a vertical pulling force.
F=mω2r=m(2πN)2r
Calculating the angular velocity works out to be ω=94.248 rads/s.
The radius works out to be r = cos(45)*0.3m, r = 0.212 m
So plugging in values, the "outwards force" due to centripetal acceleration is 1883.129 N.
Since there are two balls, the total force acting outwards is 3766.259 N.
We work out the pulling force on the rods attached to the bearing A by:
cos(45)*3766.259 N = 2663.147 N
Unfortunately, this is apparently the wrong answer. What am I doing wrong? Do I also subtract the weight force of the balls from the total force?
A flyball governor is a device used to regulate the speed of steam turbines in steam power plants. The rotation of the shaft causes the two balls move outward. As the balls move out, they pull on the bearing A. The position of the slider A is linked to a valve admitting steam into the turbine. This way, the flow of speed is automatically restricted when the shaft starts rotating too fast.
What is the force (in N) pulling on the slider A if the shaft rotates at the speed given below? Ignore friction and the weights of the components other than the two balls.
http://s589.photobucket.com/user/johnnigan/media/flyballgovernor.jpg.html[/URL]
Given data:
L = 300 mm
Angle gamma = 45 degrees
mass of each ball = 1 kg (the mass of each ball)
RPM = 900 revolutions per minute
The attempt at a solution:
I understand the problem. As the shaft spins, the balls are subjected to centripetal acceleration which causes them to move away from the shaft and draw the attached rods away from the shaft thus subjecting the bearing A to a vertical pulling force.
F=mω2r=m(2πN)2r
Calculating the angular velocity works out to be ω=94.248 rads/s.
The radius works out to be r = cos(45)*0.3m, r = 0.212 m
So plugging in values, the "outwards force" due to centripetal acceleration is 1883.129 N.
Since there are two balls, the total force acting outwards is 3766.259 N.
We work out the pulling force on the rods attached to the bearing A by:
cos(45)*3766.259 N = 2663.147 N
Unfortunately, this is apparently the wrong answer. What am I doing wrong? Do I also subtract the weight force of the balls from the total force?
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