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Homework Statement
A 100g (0.1kg) rock is attatched to a 1.0m rope and spun around in a circle with a period of rotation of 1.0s. What is the Radius of the circle that it forms?
Homework Equations
Fc = (mV^2) / r
V= (2∏r/T)
LCosθ = r
The Attempt at a Solution
Im quite stick here, as I am not sure how I could ever find r without knowing the θ of the circle.
First thing I tried was making a triangle like so:
Illustration in 2D
Im going to start with the simple equation TCosθ = Fc. We also know that (mV^2)/r = Fc. If you put those 2 equations together, you get TCosθ = (mV^2)/r. This leaves us with the equation:
TCosθ = (mV^2)/ r
Furthurmore, we know that V = 2πr/P (Where P is period, should normally be T but there's Tension in this equation, so ill just do P)
TCosθ = m (2πr/P)^2 / r
The period of rotation is 1, so the equation simplifies again to:
TCosθ = m (2πr)^2 / r
We can also sub r into the formula we just moved in.
TCosθ = m (2π (Cosθ)^2) / Cosθ
From here, we can make the equation:
T = (m (2π) ^2 * (Cosθ)^2) / (Cosθ)^2 (Correct? I am not sure if that's logical)
Now the (Cosθ)^2 can cancel out, making:
T = m (2π)^2
Sub in the values, we have T = (0.1)(2π)^2
= 0.4π^2
We now know Tension, Which we can use in the triangle.
Triangle illustration
If I use Sin, I can find that the angle is 14.37321*.
We can use this θ in the equation LCosθ = r
(1)(Cos14.37321) = r
r = 0.97 m.
Is that correct?
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