Work energy theorem and forces at equilibrium -- Conceptual doubt

[Sourav Suresh]

A pendulum of mass m and length l is suspended from the ceiling of a trolley which has a constant acceleration a. Find the maximum deflection θ of the pendulum from the vertical.

When I used work energy theorem, I got θ = 2 arctan(a/g). But when I took the equilibrium position and equated the opposite forces, I got θ = arctan(a/g). Which is correct & why? Isn’t the bob at equilibrium when it is at its maximum deflection from the vertical?

A book supports the work energy theorem method. There is also a statement in the book saying

"This angle is double to that at the equilibrium."

 

[Orodruin]

I think you have missed a crucial part in your problem statement, which is that the pendulum is originally at rest in the vertical position. Without this information, your problem has no solution.

But when I took the equilibrium position and equated the opposite forces, I got θ = arctan(a/g).

Is a pendulum in a non-accelerating system always at rest at its equilibrium position? Why or why not?

 

[Sourav Suresh]

Sorry, it is originally at rest

 

[Orodruin]

Sorry, it is originally at rest

This still does not address the questions I asked you in my previous post.

 

[Ray Vickson]

A pendulum of mass m and length l is suspended from the ceiling of a trolley which has a constant acceleration a. Find the maximum deflection θ of the pendulum from the vertical.

When I used work energy theorem, I got θ = 2 arctan(a/g). But when I took the equilibrium position and equated the opposite forces, I got θ = arctan(a/g). Which is correct & why? Isn’t the bob at equilibrium when it is at its maximum deflection from the vertical?

A book supports the work energy theorem method. There is also a statement in the book saying

"This angle is double to that at the equilibrium."

If you are standing on the accelerating trolley making physics measurements, it will look and feel just like you are standing still but with the force of gravity pointing both down and a little bit sideways, at angle $\arctan(a/g)$ from the vertical.

So, you have a situation where you are initially at rest and the pendulum is hanging straight down, quietly. Suddenly you switch on a horizontal gravity component $-a$. What do you think will now be the motion of the pendulum?

 

[CWatters]

Sorry, it is originally at rest

Vertically or at angle theta?

 

[Sourav Suresh]

Vertically or at angle theta?

vertically

 

[CWatters]

So it starts off with a deflection relative to the resultant of a and g.