# Acceleration without rotation (intertial, nonintertial)

1. Jan 8, 2009

Hey all,

I'm looking at an example in John Taylor's Classical Mechanics that I have some questions about.
The example states:

Consider a simple pendulum (mass m, length L) mounted inside a railroad car that is accelerating to the right with constant acceleration A. Find the angle at which the pendulum will remain at rest relative to the accelerating car and find the frequency of small oscillations about this equilibrium angle.

First question: what small oscillations is he talking about? Does he mean the bob will oscillate about the angle? Or that it will oscillate in and out, effectively altering the angle slightly?

He goes on to say that in the inertial reference frame, there are two forces, the tension and 'mg'.
Net F = T + mg.
In the noninertial frame we introduce the inertial force, -ma
Net F = T + mg - mA.

Setting g(eff) = g - A, the noninertial net force is
F = T +mg(eff)

Second question:
He has an illustration of the car, with the pendulum inside swinging to the left, and a second illustration beside it of the triangle made by the pendulum. The hypotenuse is labeled as g(eff), the vertical side as 'g' and the bottom as '-A'. Why isn't the tension included on the hypotenuse?

The figure has this caption:
A pendulum is suspended from the roof of a railroad car that is accelerating with constant acceleration A. In the noninertial frame of the car, the acceleration manifests itself through the inertial force -mA, which in turn, is equivalent to the replacement of g by the effective g(eff) = g-A.

2. Jan 8, 2009

### Staff: Mentor

He's talking about the usual small oscillations that a pendulum can exhibit if you displace it from equilibrium and get it swinging. (What's the frequency of a swinging pendulum?)

OK.

Because he's just talking about the effective weight force. Of course, it will be equal and opposite to the tension (when the pendulum is in equilibrium).

OK.

The point here (at least part of it) is that in the inertial frame there is a net force on the pendulum bob. But in the accelerated frame, the net force (if you include the inertial force) is zero, since in that frame the acceleration is zero.

3. Jan 8, 2009

Meaning, the angle changes slightly, right?

4. Jan 8, 2009

### chrisk

Yes, the angle is slightly changing so that sin(theta) = theta and the equation of motion can be solved similar to a simple pendulum in a fixed frame of reference but you are using g(eff) rather than g.

5. Jan 8, 2009

### Staff: Mentor

If a pendulum swings, then its angle with respect to its equilibrium position must change.

If the car was not accelerating, the pendulum will hang straight down (like usual). Displace it a bit, and it will oscillate like a pendulum does.

If the car is accelerating, the pendulum now hangs at some angle with respect to the vertical. Displace it from that angle, and it will oscillate about that angle.

Make sense?

6. Jan 8, 2009

### chrisk

Yes, makes sense. The pendulum will make small oscillations about the equilibrium angle created by the inertial force.

7. Jan 13, 2009

### jojo12345

The beautiful thing about changing into non-inertial reference frames is that you see an equivalence between various problems in physics. A lot of times, one of the equivalent problems is easier to solve (as in this case). Another interesting thing you can do with this problem is once you have expressed everything in the accelerated box car frame, move into the frame that rotates with the pendulum. If you do so correctly, you will obtain the Euler-Lagrange equations where the angle the pendulum makes with g is the generalised coordinate.