https://youtu.be/GTsz6mLHalM
PHYS 1101: Lecture Fourteen, Part Seven
This one is just an extra example I’m going to work for you. No quiz questions associated with this. I just wanted to give you another perspective to see how broadly applicable this analysis is.
So, here’s a fun example. This is an amusement ride, a roller coaster. On the Wolverine Wildcat, the section of the ride I’ve got highlighted in pink, a car goes over that section, and it has a speed of 16.6 meters per second right at the very bottom of this dip.
The radius of that dip, if I were to fit a circle into that part of the ride, the radius of that circle is about 8.3 meters. That’s how tight the corner is, tight the turn is.
I have a 75 kilogram person that’s sitting in the car. So, they follow this trajectory. They have this 16.6 meter per second speed when they go through the bottom. What’s their apparent weight? What’s their sensation as they go through that curve?
So, the object to focus on is the person. Again, I get that information by really directly looking at what I’m asked about and what’s the object that that information pertains to. So, I’m going to focus on the person as an object. What’s the person’s acceleration?
Any object following a curve has to have acceleration points to the center of that corner, center of the turn. So, the left side of my equation has to point, as a non-zero value, points up. What are the forces that are on this person? How many would there be?
Well, I know I have gravity pointing down, and the person sitting in the seat… That’s the only thing really in contact with the person. There’s no rope pulling them up. There’s nothing pushing them down. They only are sitting on their seat. In the up/down direction, we know for sure that the seat has to push. The seat can only push. That provides this normal force.
I went out of my way to draw the normal force much bigger, or bigger than the weight, because I know that the visual addition of these has to leave me with an Fnet that’s up. This has to match my acceleration condition. So, how many forces are on the object? Really two.
Right at the bottom here, there’s not much friction that’s needed on that person to keep them in their seat. It’s mostly just an upward force, the seat pushing up to support the person, and it turns out to provide this acceleration, and then gravity pulling down.
Here’s my list of notes for the problem. I was told the person’s speed at this snapshot, the radius of that circular path there, and their mass. I’ve got a picture of a, a picture of f. So, I’ve got my list of knowns. What variable to do I want? Am I focused on?
I was asked for the person’s apparent weight. If you recall, the apparent weight is always the size of the force of whatever agent is supporting the object. What’s holding this person up? What’s in direct contact with the person that’s supporting their weight? Well, it’s the seat. The seat’s providing fn. I need to know the value of fn.
Turns out, this is always where we get our sensation of how heavy we are. It’s not directly from the force due to gravity, but rather, it’s how the surface that we’re standing on or the rope that we’re hanging from is countering this force do to gravity. The surface that’s supporting us, that’s the magnitude we have to go after of that force. We need the variable fn. Okay.
Here’s our force diagram again, with our axis defined. As always, I’m going to start with Newton’s Second Law. I’ve got an ax option. Not helping me here. I don’t even have any horizontal forces or acceleration. The better option is the vertical, or the more helpful option is the vertical vector equation for Newton’s Second Law. Okay. I’m going to work with that.
Ay is the sum of the y components divided by m. Ay is equal to… What do I plug in for my numerator? I have plus fn minus mg. This combination represents my net force. Then, I have to divide that by m. This is the variable that I want. I have a value for m and g. I wasn’t given a value for a directly, but very common as I’ve done before, I know I can make the substitution that the acceleration has to have the value determined by the speed and the radius. Often, these are things that are given in the problem. This is what the speedometer reads. This is the physical radius of that corner. This combination tells me what the meters per second squared is on the left.
So, I’m ready. I just need to do my algebra and solve for fn. I’m going to copy this equation down below. I have more room. Keeping my eye on the ball. You can check my algebra more carefully later, go through it more carefully. But, according to me, the size of my normal force is going to be equal to mg plus m times my acceleration, v squared over r.
Let me factor out an m here, and just read this equation to you, translate it for you. The normal force, if an object is just standing there, just balances the force due to gravity. Then, your apparent weight is equal to your true weight, the force of gravity.
In a ride like this, though, when I’m rounding that corner, that normal force has to be much bigger to provide this acceleration. Said another way: The ground has to significantly push to force this person to make this corner. It’s like that bowling ball being whacked by the mallet. Something has to push or pull this object toward the center of the corner to get it to round the corner. Well, the seat in your ride is doing that, in a carnival ride.
So, the normal force is not just mg, but it also has to be large enough to provide m times your acceleration as you round that corner. Let’s put in some numbers here and see how big this is. The mass of the person was 75 kilograms. Then, I have to multiply 75 times g. 9.8 meters per second squared. Then, I have to add to g, this acceleration, this centripetal acceleration, if you will.
I end up with 75 times 9.8, and then these terms there on the right end up being about 33.2 meters per second squared. So, when you multiply this, or you add together the terms on the right, you end up with 43 meters per second squared.
Let me just point out that this is greater than 4g, meaning the sensation of this normal force is equivalent to something in free fall, feeling the force due to gravity that’s 4 times larger than on the surface of the Earth. That means you feel really heavy as that seat pushes up to make you round that corner.
When you turn this into newtons, you end up with a big number. 3225 newtons. That’s your apparent weight, and that’s why rides like this are so much fun. When you hit the bottom of this corner on this roller coaster, your feeling sensation is something’s pushing you hard, down into your seat, but actually, it’s your seat pushing up really hard to force you around that corner, and to get you going up on the other side. The faster you’re going, the heavier you’re going to feel. The tighter this corner, the heavier you’re going to feel.