How Far is the Pendulum from the Axis on a Rotating Platform?

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The discussion revolves around calculating the distance of a pendulum from the axis on a rotating platform, given an angular velocity of ω = 10 1/s and an angle θ = 20 degrees. The effective gravitational acceleration is described as √(g² + ω²R), leading to the equations of motion involving tension (T) and centripetal acceleration (a_c). A user suggests using the relationships Tcos(θ) = mg and Tsin(θ) = mω²R to derive the formula R = (gtan(θ))/ω². After substituting the values, the calculated distance R is found to be 0.036 meters, confirming the solution's correctness. The discussion highlights the importance of understanding the forces acting on the pendulum in a non-inertial frame.
Luke1121
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Homework Statement


A pendulum is placed on a rotating platform which rotates with angular velocity ω around an axis, at equilibrium the angle between vertical and pendulum is θ
θ= 20 degrees
ω= 10 1/s
how far is the pendulum placed from the axis


Homework Equations


a_c=\frac{v^2}{R}=\omega^2R



The Attempt at a Solution


the frame is non inertial so effective g is \sqrt{g^2 + \omega^2R}

I'm not really sure where to go from here, i thought about doing
\omega=\sqrt{\frac{g}{l}}=\sqrt{\frac{g^2 + \omega^2R}{l}} and solving for R however l isn't given so I'm not really sure where to go next, any ideas?

Thanks, Luke
 
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You are wrong considering a mathematical pendulum instead of conic one.

So you may write down (T is a funicular force):
$$
m a_c = m ω^2 R = T sinθ
$$
$$
0 = T cosθ - m g
$$
 
GregoryS said:
You are wrong considering a mathematical pendulum instead of conic one.

So you may write down (T is a funicular force):
$$
m a_c = m ω^2 R = T sinθ
$$
$$
0 = T cosθ - m g
$$
Ah ok, so:

Tcos\theta=mg (1)<br /> Tsin\theta=m\omega^2R (2)


doing (2)/(1) gives

R=\frac{gtan\theta}{\omega^2}

plugging in the numbers gives 0.036m, is this correct?
 

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