Cart with Pendulum | Tension expression

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SUMMARY

The discussion focuses on deriving the expression for tension (T) in a massless bar attached to a cart moving in the x-direction, with the bar pivoted at the cart. The initial expression derived is T - mg*cos(θ) = m * (θ (dot))² * L, which accounts for centripetal acceleration. However, it is established that the cart's acceleration must also be considered, specifically the term M*x(double dot) * sin(θ), as it contributes to the radial forces acting on the system.

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AAO
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I am trying to derive an expression for the Tension T in the massless bar in the given photo. Where there is a cart that is moving only in the x-direction and the bar is rotating around a point pivoted at the cart with angle theta. The expression that I have is deduced from:

∑ F (towards center) = Centripetal Acceleration

Which gives:

T - mg*cos(θ) = m * (θ (dot))2 *L

Is this true? or I should add another force term that comes from the cart acceleration it self:
M*x(double dot) * sin(θ)?

Thanks!
 

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AAO said:
∑ F (towards center) = Centripetal Acceleration
You have not stated the whole problem. From the diagram, it looks as though the cart is accelerating, or at least, free to accelerate in response to the tension.
If you think of the bob's acceleration as that plus acceleration relative to the cart, the latter is the centripetal acceleration, yes? But the acceleration of the cart has a component in the radial direction, so it would not be true that the centripetal acceleration can be found by summing the radial components of the forces.
 

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