The FULL story behind the slinky drop

This paper presents a "fully involved" (as they would say at Firehouse Subs) mathematical model of the behavior of a slinky under the influence of gravity. YES: There's a lot of math above the level of this course. BUT: What can you take away from the analysis offered by this author?

Sample Problem: Toy Train

Question: Suppose a child pulls on the toy train below with a force of Fpull = 6 newtons and the cars have masses of m1 = 3 kg, m2 = 2 kg, and m3 = 1 kg. If the friction between the floor and the cars is very small, what is the acceleration of the train, and what is the tension (T12 and T23) in each cable between the cars?

Answer: We can think of this problem two ways: We can think of the child's force Fpull acting on the collection of cars as a whole (and thus ignore the internal forces of tension), and we can think of each car individually (and include the forces of tension through which they interact). By utilizing the first technique, we can easily calculate the acceleration.

Thinking of the collection of cars as a whole, Newton's Second Law (net force on the object = mass of the object * acceleration, where, in this case, the "object" is the entire train) is

Fpull = (m1 + m2 + m3) a,

so a = Fpull / (m1 + m2 + m3) = 6 N / (3 kg + 2 kg + 1 kg) = 6 N / 6 kg = 1 m/s^2 is the train's acceleration.

We can now think of each car individually. For example, we know that the smallest car has an acceleration of 1 m/s^2 (just like the other two). The only force acting on this car is the tension T23 pulling to the right. So, Newton's Second Law for this car is T23 = m3 * a = (1 kg)*(1 m/s^2) = 1 N.

Now that we know T23, we can look at Newton's Second Law on the medium-sized cart. The net force on this cart is T12 - T23, since T12 is pulling right and T23 is pulling left. So, Newton's Second Law gives us T12 - T23 = m2*a. We can rearrange to get T12 = T23 + m2*a = 1 N + (2 kg)*(1 m/s^2) = 3 N.

At this point, we're technically done with the problem, but if we take just a few extra seconds to apply Newton's Second Law to the largest car, we can check our work! The net force on this car Fpull - T12 (since Fpull is pulling right and T12 is pulling left), so Newton's Second Law is Fpull - T12 = m1*a. We can rearrange for a to make sure that we get 1 m/s^2: a = (Fpull - T12) / m1 = (6 N - 3 N) / (3 kg) = 1 m/s^2. Check! Now we know we were correct!

How would the tensions and acceleration change if there were friction between the blocks and the floor? Would the acceleration be larger or smaller? Would the tensions be larger or smaller?

Sample Problem: Mountain Climbers

Problem: Suppose five mountain climbers are hanging  from the edge of a cliff, each holding onto the arm of the climber above, forming a chain like that in the figure below (with red circles representing the climbers and the black lines representing their arms).

If each mountain climber (plus his equipment, clothing, etc.) has a weight of 200 pounds, how much force does each mountain climber need to exert?

Answer: The bottom is likely a good place to start, especially since the bottom mountain climber isn't holding up anyone but himself. He thus needs to exert 200 pounds to hold onto the arm of the climber above.

The second-from-the-bottom climber must provide enough upward force to counteract the downward force caused by the bottom climber as well as his own weight. The only downward force caused by the bottom climber is his weight, so the second-from-the-bottom climber must exert a force of 200 + 200 = 400 pounds to hold up the bottom mountain climber.

The middle climber must must provide enough upward force to hold up himself and to counteract the downward force caused by second-from-the-bottom climber, who is providing 200 pounds due to his weight and 200 pounds to support the climber beneath him. Thus, the middle climber must provide 200 + 200 + 200 = 600 pounds of upward force to support the second-from-the-bottom climber.

By extending this reasoning, we see that the next climber up must exert 4*200 pounds = 800 pounds of upward force to support the climbers beneath him, and the top climber must exert 5*200 pounds = 1000 pounds of upward force to support the four climbers beneath him.

The top climber's fingers that are grabbing the ledge must thus provide 1000 pounds of frictional force. We'll talk more about friction next class!

Slinky drop simulator

This demo simulates dropping a slinky (and you control the length of the slinky)!

Here's a picture of what you'll see:

"Reset" and "Pause/play" do pretty much what you'd expect them to do.

At what point does each atom in the slinky begin to "fall?"

How does the "hang time" for the bottom atom change as you make the slinky longer?

You can also have some fun by clicking and dragging the atoms to stretch and compress the slinky both vertically and horizontally! Also try the "Random atoms" setting! How does the behavior you observe demonstrate the concepts of forces we've discussed in class?

Slinky drop in slow motion!

Since the demo in class was likely to quick to view, here's a slow motion video:



Here's a similar video with a more dramatic angle (and some narration):



And here's a third video with a SUPER-LONG slinky, just to see how the length impacts the result!



How does this behavior demonstrate the concepts of forces we discussed in class?

Student slinky drop videos!

Here are some slinky drop videos recorded by physics students!





Were each of these slinkies at rest upon release? If not, how did that impact the behavior of the bottom rim?

If you record your own video, post a link to it in the comments!