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Inverted pendulum 
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#1
Dec2208, 07:06 PM

P: 101

1. The problem statement, all variables and given/known data
I was reading a book and it had this problem: "The input u is an acceleration provided by the control system and applied in the horizontal direction to the lower end of the rod. The horizontal displacement of the lower end is y. The linearized form of Newton's law for small angles gives [tex]mL\ddot{\theta} = mg\theta  mu[/tex] . . . (1) ... (then it proceeds to part (a) of the problem.)" 3. The attempt at a solution My question is about equation (1). Suppose I sum up the forces at the mass m: First, I see two axes. One is pointing from the ball to the hinge. The other is tangential to the circular movement. The forces pointing to the hinge will be cancelled by the normal force from the rod. Therefore I ignore that. The forces on the tangential direction (clockwise) are: F = ma = mg sin(θ) + mu cos(θ) . .. . (2) Here, I moved the acceleration u to the ball because this acceleration is equivalent to one applied at the center of mass of the ball. Now, since a is the tangential acceleration, I could translate it into angular acceleration a = Lα, this gives me: mLα = mg sin(θ) + mu cos(θ) . . . (3) This is almost the same as (1). Now when I use small angle approximation: mLα = mgθ + mu . . . (4) Why is my sign for mu flipped compared to (1)?  Thanks 


#2
Dec2308, 06:00 AM

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(have a mu: µ ) (how come you had α and θ but not µ? ) Sorry, but your analysis is all wrong … there are two forces on the ball, the gravitational force and the force along the rod … you want to ignore the latter, so you take components perpendicular to the rod, giving a = gθ … but acceleration of ball = acceleration of ball relative to hinge + acceleration of hinge ie a = Lα + µ … so Lα + µ = gθ (or, without maths, if there's negligible gravity, and you pull the hinge to the right, obviously the rod will swing to the left ) 


#3
Dec2308, 12:25 PM

P: 101

mu was not a greek letter µ, it was the mass m times the acceleration u.
How do I determine whether I should include u to the acceleration or as part of a force? The original setting was Fig 1. In Fig 4, the ball is attached rigidly to a box that is accelerating at u. From the reference frame of the joint, the ball is not accelerating, so a_{Joint} = 0. From the reference frame of absolute coordinate, the ball is accelerating at u, the same as the box, so a_{Ball In Absolute} = a_{Ball in Joint Frame} + u = u. In Fig 3, the ball is sitting inside a frictionless box. The box is accelerating at u. Since there is no force applied to the ball in the Absolute frame, the ball cannot accelerate: a_{Absolute} = 0. Meanwhile, from the reference frame of the box, the acceleration of the ball is a_{Ball in Box Frame} = a_{Ball in Absolute Frame}  a_{Box in Absolute Frame} = 0  u = u. In Fig 2, a force F_{u} is applied to point A in a no gravity environment. In the frame of the center of mass, the pinshaped object got a torque and would start to spin about the center of mass with angular acceleration α = a/L = F_{u}/mL. In the absolute frame, the object got a push and its center of mass would translate at a_{CM in Absolute} = F_{u}/m. In Fig 1, in the reference frame of A, the ball is falling to the right: a_{Ball in Frame A} = gsinθ. In the absolute frame, a_{Ball in Absolute Frame} = a_{Ball in Frame A} + a_{A in Absolute Frame} = gsinθ + u. This is the wrong result. But why? In Fig 1, with small angles, the ball is approximately resting on a frictionless surface. In the absolute frame, the ball is accelerating at a_{Ball in Absolute} = gsinθ due to gravity, in the frame of A, a_{Ball in Frame A} = a_{Ball in Absolute}  a_{A in Absolute} = gsinθ  u. This is the same as (1), but is the explanation valid? The implication seems to be that in small angles, the cart does not move the ball sideway, it only moves the ball up or dowm by moving hinge A underneath the ball. Here is another attempt: The orientation of the ball is affected by g and u. The angular acceleration due to g is α_{g} = gsinθ/L. . . . (5) The angular acceleration due to u is α_{u} = ucosθ/L. . . . (6) The total angular acceleration is α = gsinθ/L  ucosθ/L. In small angle approximation, α = gθ/L – u/L. In this explanation, how do I justify (5) and (6)? 


#4
Dec2408, 04:03 AM

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Inverted pendulum
Hi saltine!
you can only find the acceleration from good ol' Newton's second law, and for that you need an inertial frame, which the hinge isn't … that's all there is to it! You must use an inertial frame, and in this case the only candidate is the stationary frame, so you use … 


#5
Dec2608, 07:46 PM

P: 101

I am confused because u is not a force but just acceleration.
In Fig 1, there are two forces acting on the ball. The component of mg that is parallel to the rod is cancelled by the normall force from the rod, so the net force at the ball is only mg sinθ. You said: acceleration of ball = acceleration of ball relative to hinge + acceleration of hinge And that the acceleration of the ball is a = F/m = g sinθ. Therefore g sinθ must also equal Lα + u. In Fig 5, the situation is similar except that u is now from an external force, F_{u} = Mu. Is this situation the same? Why is/isn't F_{u} part of the net force acting on the ball?  Thanks 


#6
Dec2908, 09:21 PM

P: 101

I found another way to explain the original situation:
When a rocket accelerates sideway, the perceived gravity is in the opposite direction of the acceleration. Therefore with the applied the acceleration u to the cart, the "gravity" that the ball perceives is the sum g + u. Now I could take the tangential component due to this perceived gravity: Net Tangential Force = ma_{T} = mg sinθ  mu cosθ Since a_{T} = α L, m α L = mg sinθ  mu cosθ. With small angle approximation: [tex]mL\ddot{\theta} = mg\theta  mu[/tex] Which is equation (1) ! This is the only way I understand it at the moment. 


#7
Dec3008, 05:44 AM

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Hi saltine!
you add it to the other (rotational) acceleration. The net force on the ball is mg sinθ plus the force from the rod. Nothing is cancelled (what is it to be cancelled with?) … the reason why you leave out the mg cosθ is not because its cancelled, but simply because you're only looking at tangential components. you must decide which body you're applying Netwon's second law to, and then only use the forces on that body … your book is using the ball as the body, so the only forces on it are mg and the force from the rod … any force on anything else is irrelevant. (The tangential force is mg sinθ, and the tangential acceleration (times mass) is mαL + mu cosθ) This isn't necessary … and it's risky, because it's so easy to get the ± sign wrong. Your book just uses the stationary frame, and uses only tangential components, so that there's only one component of force on the ball, and the acceleration is split into two parts. 


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