A way to think about this problem that clarified it for me: The top of the slinky is experiencing tension equal to its whole weight. The bottom is experiencing upward tension equal to its weight, so at the instant when it is dropped, it experiences no net force—until the upper parts of the slinky start traveling downward, removing the upward tension. (The speed-of-information-in-materials wasn’t really convincing to my intuition, because it feels like, since the bottom of the slinky is in a gravitational field , it should already “know” it is falling… )
For a mantra: How about something like, “What do you know? What don’t you know? How do you connect them?”
I’m not really sure what you mean by “upward tension”, sorry. Tension in one dimension is just a scalar. The very bottom of the spring is under no tension at all, and the tension increases as the square root of the height for a stationary hanging slinky.
By “upward,” I just meant to emphasize that it was opposing gravity, e.g., positive. But of course, now that I think about it for a minute, I see that I was wrong, it is under no tension at all. Oops.
I think I see what you mean. To clarify, though, tension doesn’t have a direction. In a rope, you can assign a value to the tension at each point. This means that if you cut the rope at that point, you’d have to apply that much force to both ends of the cut to hold the rope together. It’s not upward or downward, though. Instead, the net force on a section of rope depends on the change in the tension from the bottom of that piece to the top. The derivative of the tension is what tells you if the net force is upward or downward. This derivative is a force per unit length.
In general, tension is a rank-two tensor, and is just a name for when the pressure is negative.
A way to think about this problem that clarified it for me: The top of the slinky is experiencing tension equal to its whole weight. The bottom is experiencing upward tension equal to its weight, so at the instant when it is dropped, it experiences no net force—until the upper parts of the slinky start traveling downward, removing the upward tension. (The speed-of-information-in-materials wasn’t really convincing to my intuition, because it feels like, since the bottom of the slinky is in a gravitational field , it should already “know” it is falling… )
For a mantra: How about something like, “What do you know? What don’t you know? How do you connect them?”
I’m not really sure what you mean by “upward tension”, sorry. Tension in one dimension is just a scalar. The very bottom of the spring is under no tension at all, and the tension increases as the square root of the height for a stationary hanging slinky.
The tension gradient is upward, indicating an upward force per length.
By “upward,” I just meant to emphasize that it was opposing gravity, e.g., positive. But of course, now that I think about it for a minute, I see that I was wrong, it is under no tension at all. Oops.
I think I see what you mean. To clarify, though, tension doesn’t have a direction. In a rope, you can assign a value to the tension at each point. This means that if you cut the rope at that point, you’d have to apply that much force to both ends of the cut to hold the rope together. It’s not upward or downward, though. Instead, the net force on a section of rope depends on the change in the tension from the bottom of that piece to the top. The derivative of the tension is what tells you if the net force is upward or downward. This derivative is a force per unit length.
In general, tension is a rank-two tensor, and is just a name for when the pressure is negative.