Centripetal Acceleration of laundry dryer

  • Thread starter rambo5330
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  • #1
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Homework Statement


Hi I have a question that involves a laundry dryer, and it must turn at a certain velocity to so that when the object hits 68 degrees to the horizontal it will fall.

the radius is the only given
r = .330m



Homework Equations


the answer is stated as
Fc = N + mg*sin(68) = m * v^2 / r
which equals 1.73 m/s

what I cant make sense of is why it is sin68 when I draw it out I thought weight (mg) has to act directly downward meaning the vector in-line with N and FC is mg/sin(68) but this isnt the case?
the only way I can get the answer to work is if mg is acting radialy why is weight acting radially instead of straight down? my teacher did not explain this well
 

Answers and Replies

  • #2
Doc Al
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Please post the complete problem.
 
  • #3
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Okay complete problem is,

In a home laundry drier, a cylindrical tub containing wet clothes rotates steadily about a horizontal axis, as in Figure P7.63. The clothes are made to tumble so that they will dry uniformly. The rate of rotation of the smooth-walled tub is chosen so that a small piece of cloth loses contact with the tub when the cloth is at an angle of 68.0º above the horizontal. If the radius of the tub is 0.330 m, what rate of revolution is needed?

the figure just shows a clothes dryer with an angle made at 68 degrees above the horizontal
 
  • #4
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I believe that in this case the weight has to components.One radial and one acting tangential to the path.The radial is mgsin68
 
  • #5
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See this picture it may help if i have understood correctly your question.
 

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  • #6
Doc Al
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Okay complete problem is,

In a home laundry drier, a cylindrical tub containing wet clothes rotates steadily about a horizontal axis, as in Figure P7.63. The clothes are made to tumble so that they will dry uniformly. The rate of rotation of the smooth-walled tub is chosen so that a small piece of cloth loses contact with the tub when the cloth is at an angle of 68.0º above the horizontal. If the radius of the tub is 0.330 m, what rate of revolution is needed?
OK, now your question makes sense. And Tzim is correct; mg sin68 is the radial component of the weight, as you suspected. Any vector can be broken into components--weight is no exception. When dealing with centripetal acceleration, you need to look at the radial components of all forces acting. (Since the weight acts downward, it will not have a horizontal component. But it will have a non-zero component in other directions.)
 
  • #7
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Thanks, that does make a lot more sense, I just had trouble visualizing weight being broken into compents on an object that is upside down.

couldnt not make sense of how weight would have a horizontal component when the object is upside down, is there anyway you can help me understand that better?
when an object is on an incline plane I understand the normal force of the surface of the plane helps redirect gravity. I guess in this case like you said its the centripetal force.
 
Last edited:
  • #8
Doc Al
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Thanks, that does make a lot more sense, I just had trouble visualizing weight being broken into compents on an object that is upside down.

couldnt not make sense of how weight would have a horizontal component when the object is upside down, is there anyway you can help me understand that better?
I'm not sure what you mean. Weight will not have a horizontal component. In this problem, you need the radial component of the weight (which produces the centripetal force), not the horizontal component (which would be zero).
 
  • #9
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Sorry I think I worded it wrong,
breaking it into a radial component and a tangential component...
I've drawn it out several ways and have made sense of it. Thanks a lot though, rotational kinematics was covered extremely fast in class.
 

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