https://youtu.be/wUzon0IseGo
PHYS 1101: Lecture Fourteen, Part Four
I am going to walk you through a few examples now. And we are going to start that with an example based on watching this animation. So here is our object, in red. It’s attached this red mass, ball by a string and think this as a hockey puck on frictionless table. This is an ice table, no friction. This object, the hockey puck, we are looking at it from above. And it’s just going to follow a circular path.
So there it goes; it’s undergoing uniform circular motion. How do I know it’s uniform? Well watching it the speed appears to be nice and constant as it follows the circle. And, in fact, if I check this box below, it’ll leave the timestamps so I have a view of the motion diagram. So sure enough I see nice equal spacing between all of those points.
So for that motion, so for that motion, we’re going to calculate what the tension is for that rope. Perhaps you have the sensation if you were holding this rope and swinging this hockey puck around. The faster you made it go, the stronger the tug you would feel or the tension in the line on your fingers at this end.
Let’s go calculate the value of this tension. It’s not arbitrary; it’s going to be related to the radius and the speed of this object and the mass. Here we have the values; just a minute let me go paste our picture in. Okay, there’s the top view; just a copy, a paste of the picture from our past animation. And a little later I’ll go paste over the values. We were given the speed, the mass, and the radius of that circular path.
First point, we’re going to solve this problem, determine what the tension is in the line. We viewed the motion from above but in order to do this analysis I want to encourage you about picturing the side view.
In this case, for this problem, it turns out it doesn’t matter. You can view all of the information you need from this top view. But I encourage you to get in the habit of using the side view because often from the side you’re going to see the gravitational force and a normal force and sometimes you need information about those forces. For example, if there’s friction involved to help you figure out all the force effects that you have in this radius diction, or radial direction.
So let me highlight for you. Get in the habit of considering the side view. Okay, here’s a quiz question for you, about that problem, when you view it from the side, this hockey puck was going around clockwise when you are standing in the front viewing it from the side. By showing green here, roughly at an angle with that trajectory looks like. What the direction of acceleration for this, at this instant, at that snapshot when the object is at the far left side?
So notice what we are doing. We are following the problem solving steps to work out the tension. Next thing to consider are what are the forces. Again viewed from the side, what would be the right, free-bodied diagram for this hockey puck going around in this circle?
Here’s the top view picture, again, if you want a reminder. And what it looks like, again, it’s it at the snapshot when the object is right here as it’s going around. So picture snapshot right there. It’s a side view; remember this is a hockey puck that’s sitting on a sheet of ice.
Ok what’s the right free body diagram, question 9. And now our problem-solving step would lead us to writing a list of the knowns. I am going to go back to our animation here and I’m just going to copy a picture of the knowns that we were given for that problem. There they are.
And what do we want? We were asked what the tension was in the line. That’s the variable, T. We have to be a little careful now, since we have a fancy T that I write with two little pieces here at the top hat of the t to try and make it a fancy T when we are talking about the period, the time to go around once.
T is also the variable we naturally want to use to represent tension. The magnitude of the force caused by a rope and that’s what we want to solve for. That’s the variable T we want in this problem.
Okay, well how do we start? After making our list of knowns, and the variable we want, we always start Newton’s Second Law. And to do that I like to have a fresh picture of the free body diagram at that snapshot and I know I have to have a coordinate system come into play.
So Newton’s Second Law, the equations I start with is in the x direction the acceleration component is the sum of the horizontal forces divided by m and I potentially could use the sum of the y forces, if I needed to solve the problem.
Given that I want the magnitude of the tension, I consider the forces on my object from that side view. The only horizontal force I have, the only thing directly in contact with the object is the rope that putting a horizontal force on that object. Ropes only pull, it’s to the right.
Is that consistent to our acceleration? Well, the acceleration points to the center of the circle for any object undergoing circular motion. Maybe I’ll draw it up there. So it looks good. My picture tells me that my y component forces balance so Ay is 0. I’m in equilibrium vertically that’s why the hockey puck is restricted to motion horizontally. In that horizontal plane, though, as that object rounds the corner.
At this instant, acceleration points dead to the right. My rope is aligned horizontally. The tension in it means its pulling to the right. This is the force that’s responsible for m acceleration. So this is the equation that is going to get me there. I start with the horizontal Ax is equal the sum of all my horizontal forces. Well, I only have T, one force, it’s in the positive direction and I have to divide by m.
Okay, I have a value for m. I want T. What can I say about a? Remember it’s not 0; it’s value is not arbitrary. I have my kinematic relationship that I know that tells me, hmm, if I need to I can substitute in that the acceleration has to have the value given by v squared over r.
So let me substitute that into this equation. By doing that I know I have values that everything that I, everything else, all the other variables except the thing that I want so I am ready to do my algebra where my goal is to solve for T.
So I am going to multiply both sides by m and switch the order of my equations. So then I have that the tension is equal to m v squared over r. Ready to plug in numbers.
My mass was given to me as 150, note the units though, that’s in grams. I have to convert that to SI units of kilograms, which means I have to divide by 1000. So I have 0.50 kilograms. My speed was 19.5 meters per second. That has to be squared and then I have to divide by the radius and the radius was 1 meter.
The units look good to me. I am going to get kilograms. One of these meters is going to cancel with one of these. So I am going to end up with kilograms, meter, and then I’ll have seconds squared in the denominator and that’s a Newton.
Now when I plug that in, I end up with 57 newtons. Okay, see the strategy, we aren’t doing anything new. We are starting with Newton’s Second Law and it’s always starting with this equation and then plugging things into it as we customize it to our particular problem.
Here are a few quiz questions for you to get you to think about that mathematical relationship between this tension and the various physical quantities. So go back to that equation that we just derived to solve our previous problem, and look at it and mathematically check and see, imagine if you were to change the velocity. If you were to increase it by a factor of two; so take the value v and multiply it by 2. If everything else stays the same, by what factor will that tension change? This literally would be how much change you would sense if you were holding that rope, the other end of the rope, if you were swinging that hockey puck around.
Question 11. Ask yourself the same thing but in regards to the mass. How does the tension depend on the mass as described, determined by that equation?
Question 12. Now change. Question 12. Now change the radius of that circular path. How does the tension depend on the radius?
The next few quiz questions I want you to think of the different aspects of circular motion. In question 13, you want to be sure, be sure, you understand the direction of the velocity and the acceleration vectors during this motion.
Okay so I want you to watch this animation real quick to help visualize this. Because we know when an object is undergoing circular motion, it does have acceleration. There is a force directing toward the center that is always pulling the object around toward the center. If I look at the, as this motion is going on, the acceleration is always changing direction ’cause it has to continue to change because it has to continue to point to the center.
The velocity vector as always has to point, literally, in the direction that that object is headed at that instant. So in this animation you see that it’s leaving a record of these velocity vectors at each of these instances. So the velocity is always along the motion. It’ll end up being a line tangent to this circle that’s being mapped out. The acceleration always points to the center. So notice that these two vectors are always at right angles to each other.
Okay. Question 13, then, perhaps won’t be very difficult for you. I want you to think about if you are viewing circular motion from the top as viewed from the top. What’s the direction of the velocity vector at this instant? We know that the speed is constant, the magnitude of the velocity stays the same the whole way around but the velocity vector direction changes.
Question 14. I want you to try and predict or picture the direction of the trajectory of this object if all of a sudden you were to cut the line, cut the rope, and then watch the motion of the object. So let’s say that this is an object going around horizontally, so you are looking down on the object, it’s like a hockey puck sliding on a table and right at this instant, picture cutting the rope, and then I want you tell me what is the trajectory that this object is going to follow.
To give you a hint, remember, that you can predict a trajectory with knowing two vectors. You need to know at the beginning that you are supposed to start following this motion; predicting this trajectory. What’s the initial velocity and then what’s the acceleration because remember this acceleration predicts what, or tells you what the little delta-vs are that cause this to change every second. So as viewed from the top, as soon as I cut this rope, what does my acceleration vector become as viewed from the top? So what’s v0. What path is the object going to follow?
Question 15 changes it a bit. Now someone is swinging a ball vertically, up and down. So now gravity would be at every instant, gravity is pointing down. It’s a downward force on this ball. Now there’s no longer a normal force at all from this view as long as I have this rope and the object has this speed. It’s going to go around in a circular path.
The instant I cut this rope though, ask yourself what’s my initial velocity. So what would be my starting velocity at this instant and what is my acceleration because again that will tell you what your delta-v is and, again, that will help you predict what your trajectory is. Cut the string at this instant, what path do I take?