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# Circular motion question

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1. 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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2. 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.
3. (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
4. This should be in Relationships

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