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    Our lecturer made this worksheet for us, and I can't for the life of myself figure out the correct answer. I attached the question paper but I will type it out here as well:
    A seat S is suspended from a fairground ride by an inextensible rope.
    When the ride rotates at an angular velocity of 1.2rad/s , the rope
    makes an angle of θ with the vertical. The distance of S from the axis of
    rotation is 5.0 m.
    (i) the tension in the rope is T. Explain how a component of t provides the
    force needed to keep S in a circular path and equate this component.
    (ii) S stays at a constant height during rotation at constant speed. Write a
    second equation expressing the vertical equilibrium of S
    (iii) Hence, using the two equations from (i) and (ii) calculate the value of θ


    Now I have completed the first part I believe, which I solved to be "Tsinθ" for the component being the centripetal force. However, I cannot figure out the second bit. The answer on the answer sheet reads '0.63 rad' for the last bit.
    Any help would be appreciated, thank you.
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    This question involves resolving forces horizontally and vertically, forming 2 equations.

    You're right that the component of tension providing the force to keep S in a circular path is T sin θ. You need to equate this to the centripetal force, forming an equation T sin θ = mr ω^2.

    Then, for the second part you need to consider the vertical forces - in this case this is the cosine of the tension and gravity, giving T cos θ = mg.

    Using these equations, you can eliminate the mass and the tension to find an expression for the angle θ.
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    (Original post by AlesanaWill)
    This question involves resolving forces horizontally and vertically, forming 2 equations.

    You're right that the component of tension providing the force to keep S in a circular path is T sin θ. You need to equate this to the centripetal force, forming an equation T sin θ = mr ω^2.

    Then, for the second part you need to consider the vertical forces - in this case this is the cosine of the tension and gravity, giving T cos θ = mg.

    Using these equations, you can eliminate the mass and the tension to find an expression for the angle θ.
    Thank you, so much. I was struggling with this
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    This should be in Relationships
 
 
 
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