https://youtu.be/CGdkismRv6Q
PHYS 1101: Lecture Twenty-One, Part Three
Let’s say we have a 220-kilogram speedboat that’s negotiating a circular turn. During the turn, the engine causes a net tangential force, that’s FT, of 550 Newtons. That’s applied to the boat. So, the boat is our object. The initial tangential speed going into the turn is 5 meters per second. Find the tangential acceleration for part A and part B. After the boat is 2 seconds into the turn, I have to find the centripetal acceleration. Okay. Quiz question 8: What tool, what mathematical equation, is the best place to start to find the answer to A?
I’m after the tangential acceleration. If I were to use this equation, would that work for me? How about equation B? There is aT, but do I know r? Do I know Alpha? Here’s an equation. Is the thing I’m after related to a force divided by the mass? Do I have that information? Question nine: For part B, what’s the final variable that you need to focus on or solve for to answer that question?
So, let’s walk through this problem. See how this would work. As always, I want to start with a sketch. In red, I’m drawing my boat. I know that that boat is navigating a turn. It has a certain radius. It starts into the turn at some instant. That’s my T0, the initial scope of my problem. Then, I know after it’s two seconds into the turn, I’m going to have to answer some information here about the centripetal acceleration. So, here’s my scope.
I know the mass of the boat and the radius of this turn. So, let’s write here. What things do I know? Mass is 220 kilograms of this object. I know that the radius of this circular motion is 32 meters. I know that the variable that I’m going to call FT, for the tangential force, is 550 Newton’s. I know that the other number here, I know that it’s in meters per second. That’s the initial tangential speed. At this instant, that’s my vT0. Five meters per second.
Okay. For part A, what’s the variable that I want? Find the tangential acceleration. I need to go after the variable aT. For part B, what do I want? After the boat’s 2 seconds into the turn, find the centripetal acceleration. That’s aC. I need, for part A, the component of the acceleration that’s along the circular path, tangent to that circular path. Part B, I need to find the component that’s directed toward the center. Let’s go after part A. Given my list of knowns and what I’m after, I recognize that for any acceleration, if I know the force that’s causing it, that this value is given by Newton’s Second Law. I’m going to go with that, because I directly have the values to plug that in.
So, aT is going to be 550 Newtons divided by 220 kilograms. That’s 2.5 meters per second squared. That’s part A. Part B, I have to find the centripetal acceleration after the boat is 2 seconds into the turn, aC. What equations do I know about aC? If you scroll back previous in the lecture to the areas that are highlighted, I know from this chapter and from chapter five that the magnitude of ac is the speed at that instant squared divided by r. The magnitude of my tangential velocity squared divided by r. I have r. I don’t know what that tangential speed is, or that tangential velocity is, 2 seconds into the turn. How do I get that?
Let’s think about our [kinematic] equations for this motion along the circular arc. ]Well, here I’ve copied them for us. So, along this circular path that this boat is taking, these equations connect the distance along the circular path, my position coordinates s, my initial and final vT, et cetera. This is the variable that I want. What’s the final velocity for this one-dimensional motion, given that I know now a tangential acceleration? I now know aT. I know the initial velocity. I like equation 1, given the things that I know.
I’m going to try equation one. That final velocity is going to be equal to my initial plus acceleration times time. I have numbers to directly plug in for all of that. I have 5 meters per second plus 2.5 meters per second squared times two seconds. That gives me 10 meters per second. Okay. Let me emphasize that the vector nature of these equations still holds. I have to carefully think about the sign of the velocities compared to the acceleration. I left at as a positive number. My initial velocity was positive, so this would only be consistent if the object were speeding up, and it does in this case. The speedboat increases its velocity as it goes around that corner, increases speed. If it were slowing down, I would need to make one of these negative. I’d need to pick a sign convention in these vector quantities along this line of motion would have to have opposite sign. Okay.
Here’s another example for us. A Ferris wheel has a radius of 38 meters, and it completes a full revolution every two minutes. That’s when it’s operating at its maximum speed. So, that’s its steady rate that it goes around at, after you’ve turn it on, it’s gotten up to speed. If the wheel, the Ferris wheel, were uniformly slowed from this maximum speed to a stop in 35 seconds, what would be the magnitude of the tangential acceleration at the outer rim of the wheel during this deceleration?
Question ten: What’s the final variable to solve for? Question 11: What is the value, from what’s given, for the variable vT0? Question 12: What’s the value, given that you need, assigned to the variable vT? Let me draw this scenario. Let’s work out this problem.
Here’s my Ferris wheel, my big Ferris wheel. I’m going to focus on one of the cars on this Ferris wheel, so I can be sure I stay focused on a single object as it completes this circular motion. Here’s the center. I’ve got my spokes of the Ferris wheel. I know that the radius is 38 meters. I know that to do one revolution, it takes two minutes. That’s the period of the motion. Two minutes is 120 seconds. I know that, well, this information alone is enough to tell me, and you’ll see in a minute, what the transverse, I’m sorry, the tangential velocity has to be when it’s going around at a steady pace, vT.
For the scope of my problem, I’m looking at the scenario where the Ferris wheel slows down and comes to a stop. So, this is going to be my initial velocity. Okay. What is that velocity? Constant speed. Do you remember when, chapter 5, we talked about when it object goes around one time, it travels a distance 2 pi r. Distance, over the time it takes to do it, is equal to the speed. Velocity is distance over time. One revolution takes 2 minutes. So, that should give me my value for vT0.
I have to multiply 2 by pi by 38 meters. Then, I have to divide that by 120 seconds. I end up with 1.99 meters per second. That’s my operating speed out here for this car going around the circular path. That’s what I start with for the scope of this problem, because I want to wind down, slow down, and come to a stop. So, I’m going to start here. Then, I have some initial time, initial position, initial velocity. Then, at the end scope of my problem, I have time, position, and velocity. I have aT, my deceleration, between those two snapshots for the duration of my problem.
What’s the magnitude of aT at the outer rim during its deceleration? The variable that I want is the variable aT. Okay. From my list of knowns, I’m after aT. Let’s be sure I’ve got all my knowns. I can start my clock at zero. I can say that this is my origin. So, these two quantities, I’m going to set to zero. My initial tangential velocity I have, 1.99 meters per second. As this thing slows down, eventually, it comes to a stop. That happens at time t later, which I know is 35 seconds. I don’t know how far that car has moved along this path. My final speed is zero.
So, here I have tidied it up a bit for us and made a clear list of our knowns and the variable that we want. All right. Can we use our kinematic equations to determine at? Here they are again. I copied them for us. We are after at. These are the things that we know. aT is in all three of these equations. I like looking at the shortest one first to see if it would work. Do I know the beginning and the ending velocity? Yes. Do I know the time? I do. Equation one will work.
vT, at the end, is equal to the initial velocity plus at times time. I end at rest. So, my final velocity is zero. I’m trying to solve for aT. Let me just do some algebra and rearrange for that. I conclude that at has to be minus vT0 divided by time. That’s minus 1.99 meters per second divided by 35 seconds. That’s negative .057 meters per second squared. Negative sign there tells me direction. My velocity is in the direction of motion. It’s always, then, in this counterclockwise direction by my sketch. So, my tangential acceleration at every instant points opposite to that. It’s pretty small, .057, but this is the deceleration that I need to bring the Ferris wheel to a stop in 35 seconds.