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Find the angle theta

  • Thread starter Radarithm
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  • #1
Radarithm
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Homework Statement


A mass m = 5.500 kg is suspended from a string of length L = 1.570 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 2.874 m/s. What is the angle theta between the string and the vertical?
PIC OF DIAGRAM: http://gyazo.com/fc4a226a7d1a3af6c298b9782de0d7fe


Homework Equations



a = v2/r

The Attempt at a Solution



I am totally stumped. I don't want anyone to give me an answer; all I want is a hint so I can get started on this problem (I assume I'll have to use sin / cos somewhere in the formula). Thanks in advance.
 

Answers and Replies

  • #2
CAF123
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Start by drawing a free body diagram for the mass and resolve the forces in the radial and +z direction, +z pointing upwards.
 
  • #3
Radarithm
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I believe the Tension and weight forces cancel with respect to the y direction, that means there is a centripetal force towards the center of the circle in the x direction and since the mass is tilted along an angle theta, then the tension also has an x component. I guess I'm stuck again, maybe that's because its getting late; I'm not sleeping until I solve this problem :P
 
  • #4
CAF123
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I believe the Tension and weight forces cancel with respect to the y direction, that means there is a centripetal force towards the center of the circle in the x direction and since the mass is tilted along an angle theta, then the tension also has an x component. I guess I'm stuck again, maybe that's because its getting late; I'm not sleeping until I solve this problem :P
What equations did you get for the two directions?
 
  • #5
Radarithm
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What equations did you get for the two directions?
Still didn't get any. I have no idea how to get θ.
Any tips? :confused:
I did get this though (and I'm pretty sure its wrong): Fnet = Tx + Fcentripetal cos θ × m(v2/r)
 
  • #6
CAF123
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Still didn't get any. I have no idea how to get θ.
Any tips? :confused:
I did get this though (and I'm pretty sure its wrong): Fnet = Tx + Fcentripetal cos θ × m(v2/r)
It is dimensionally inconsistent, so just by looking at it, it must be incorrect.

There is a component of tension in the vertical direction. What is this in terms of θ? What other force acts in the vertical direction? Write Newtons second law for this direction.

There is also a component of tension in the horizontal direction. What is this in terms of θ? This force component is providing the centripetal force required for the mass to revolve. So you may equate this to the general expression for a centripetal force.

This will give you two equations. What you said was correct in #3, now you have to translate those ideas into equations.
 

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