Solving Object's Velocity at Bottom of Frictionless Incline

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SUMMARY

The discussion focuses on calculating the velocity of an object at the bottom of a frictionless incline, where the angle θ decreases at a constant rate w. The correct formula for the final velocity is derived as Vf = (√Lg(1-cos(θ)))/(√(sin(θo - θo cos(θo))). The initial attempt using the equation Vf² = [Vo]² + 2a∆x was incorrect due to the assumption of constant acceleration, which does not apply in this scenario. Integration of the equations of motion is necessary to solve the problem accurately.

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Homework Statement


An object rests at the top of a frictionless incline with length L and angle θo. At the moment the object is released the angle begins to decrease at a constant rate w. Thus the angle as a function of time is θ(t)= θo-wt. Value of w is defined to be the rate such that at the instant the object reaches the bottom of the incline, θ(t)=0. Find the objects velocity when it reachs the bottom of the incline in terms of L, θo, and the gravitational constant.


Homework Equations



〖Vf〗^2=[Vo]^2+2a∆x

The Attempt at a Solution


I tried solving this by using the equation 〖Vf〗^2=[Vo]^2+2a∆x, with Vo=0 and ∆x=L.
I solved for a by adding up the forces moving in the x direction giving me: a= gsinθ. I plugged this in and got the answer Vf=(2(gsin(θ))(L))^(1/2) but I was wrong. The answer is (√Lg(1-cos(θ)))/(√(sin⁡〖θ_o-θ_o cos⁡〖θ_0 〗 〗 )
I don't understand where to even begin with this problem, please help me!
 
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You have to set up and integrate the equations of motion. The equation you are using only applies to constant acceleration.
 

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