# Homework Help: Dynamics of a Falling Rod

1. Nov 16, 2007

### kataya

1. The problem statement, all variables and given/known data

Suppose there is a rod balancing on a table. The rod tips past vertical and begins to fall. Derive an equation for the force on the bottom of the rod as a function of the angle between the rod and the table. Assume that the rod does not slip, even if this implies a very large coeffecient of static friction.

The rod has length L and mass M. Its mass is evenly distributed along its length.

2. Relevant equations

sum of forces, sum of moments. Parallel axis theorem.

3. The attempt at a solution

Here is a free body diagram of the rod/table.

http://www.prism.gatech.edu/~gtg857w/PforumMisc.bmp [Broken]

Taking the sum of the forces in the vertical direction yields:

N = M(a + g). a is the vertical acceleration of the center of mass.

This doesn't make any sense to me, because it seems to be implying that the pendulum is falling faster than under free fall! How can this be? I know that the normal force is non-zero because the force of friction must counteract the moment caused by it (in order to not slip at the point of rotation) but why on earth is it greater than g?

Last edited by a moderator: May 3, 2017
2. Nov 16, 2007

### Bill Foster

It's not.

What is the force when it is upright? $$N=mg$$

What is the force when it is lying down? $$N=0$$

So at any angle, the force will be between 0 and mg.

3. Nov 16, 2007

### kataya

Yes that makes sense intuitively, but it does not explain why the summation of forces in the vertical direction gives the wrong answer. Please expand your response to include newton's laws or another aptly derived relation. Thank you for your reply.

-Tyler

4. Nov 16, 2007

### ozymandias

$$ma_x = +N -Mg$$

leads to the conclusion $$a_x<0$$, as N<Mg. Hence a+g<g.

Regarding the problem: here is a hint. The table is meaningless. You are describing the rotation of a pendulum released from an inverted position, since you only have one point of contact (a pivot). The normal force only tells you "half the tale", mind you, as there is also a force (in this case friction) that also constraints the rod in the y-direction. So saying "N=0, N=Mg" isn't enough to understand what's going on.

Assaf.
http://www.physicallyincorrect.com/" [Broken]

Last edited by a moderator: May 3, 2017
5. Nov 16, 2007

### Bill Foster

In the vertical, the forces are $$N=mg$$

At any angle, the force is $$N=mgsin{\theta}$$

6. Nov 16, 2007

### Bill Foster

The end of the rod does not move.

Therefore acceleration on the end of the rod is always 0.

Therefore, the net force on the end of the rod is always 0.

Therefore, see post #5.

7. Nov 16, 2007

### kataya

WOW that's embarrassing.

I believe I can derive the rest of the relation.

Bill Foster, the sum of forces on a solid body is equal to the product of the mass and acceleration of the center of mass of the body. You can't just sum it up at a point and say that point isn't moving if it's part of a larger body.

Anyways, thank you ozymandias for pointing out that the acceleration is negative. This gives the relation:

$$N = m(g - a)$$

a here is the component of the acceleration in the vertical direction, which is also equal to:

$$\frac{L\alpha}{2}cos(\theta)$$

If you put this into a above, you get

$$N = m(g - \frac{L\alpha}{2}cos(\theta))$$

Where alpha is the angular acceleration of the rod's center of mass.

8. Nov 16, 2007

### ozymandias

Don't sweat it, I have made much worse mistakes in my career :rofl:
At any rate, you should let the "mathematical machinery" take care of minus signs. Newton's 2nd law for the CM, once we agree "up" is "positive", is:

$$MA = +N - Mg$$

or:

$$N = M(A+g)$$

You should just keep in mind that A can be both negative and positive.

The angular acceleration doesn't simplify your problem because it is also (1.) unknown, (2.) not a constant. You need to derive an expression for it using the other constants of your problem.

Hint: it makes much more sense (to me) to analyze this problem in polar coordinates, since in such coordinates the force exerted by the pivot is always central, which makes the tangential part easy to solve.

Hint #2: write the (CM) acceleration in polar coordinates - look at Kleppner & Kolenkow if you have it, p. 36.

Good luck

Assaf.
http://www.physicallyincorrect.com/" [Broken]

Last edited by a moderator: May 3, 2017
9. Nov 16, 2007

### kataya

A few additions/corrections, and so we can maybe add the *solved* tag:

$$\alpha$$ in the previous equation was the angular acceleration of the rod about its pivot point. Not about the center of mass.

The sum of the moments on the rod about the point of rotation will give alpha.

$$\alpha = \frac{1}{I}[Mgcos(\theta)]$$

So the entire equation becomes:

$$N = Mg(1 - \frac{ML}{2I}cos^{2}(\theta))$$

Where I is the moment of inertia of the rod spinning about its end.

This does not give N = 0 for theta = 0, which seems wrong, but perhaps it is legit.

Check my work anyone?

Last edited: Nov 16, 2007