This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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From the course by University of Houston System

Preparing for the AP Physics 1 Exam

35 ratings

University of Houston System

35 ratings

This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Rotational Mechanics

Topics include torque, rotational kinematics and energy, rotational dynamics, and conservation of angular momentum. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

In this unit we will begin our discussion in physics with a topic we called kinematics. In that module we discussed translational motion only, that is the motion in which the object does not rotate. To describe rotating objects we must define several new terms. Here's one: Angular displacement, reported in units of radians, measures the amount of rotation of an object about some axis. Another term is angular velocity. Much like linear velocity, it's defined as an object's angular displacement over time.

The faster an object rotates about an axis, the larger it's angular velocity in radians per second. Angular acceleration is defined as the change in angular velocity divided by time. And it is measured in radians per second squared. Since these quantities are vectors the direction for each is important. We will use customary convention in which counter-clockwise direction is considered positive.

The new terms can be related to our old translational quantities. Consider for a moment a solid disk that rotates at a constant rate of one rotation every second. Notice, that points A and B on this disk make one rotation every second.

The two points, however, do not have the same tangential velocity. Point B on the edge of the disk travels a much larger circle. And it does so because it is further from the axes of rotation. It must still complete the rotation in the same one second interval, but point A will travel a much smaller distance during this time. We need a way to relate these linear and rotational quantities that fits this behavior.

Sometimes called the bridge between translational and rotational quantities. Notice that while points a and b in this example had the same angular velocity their larger radius from the axis of rotation resulted in a larger linear velocity. Make sure that you are using radius and not degrees before performing calculations with angular quantities.

>> Looking at a question like this, it's important to realize that the object rolls without slipping. What that tells me is that, that we can use our bridge equations here to relate the linear speed of a rolling object to its rotational speed. Remember that if there's a wheel on a car for instance, slipping in the mud, it can spin at any rate and the wheel may not move forward at all. There's not a relationship there that we can use. So looking at what they've told me, they give me the radius of this ball as 0.2 meters, already in our base units, so I don't have to do any conversions there. It wants me to calculate the angular velocity of that ball. And so we know the radius, we know that the center of mass has a linear speed forward of 5 meters per second. But it's also rotating, and so there's a relationship there. I'm going to use my bridge equation rather than looking at this in terms of geometry. Which you could also do.

And use this expression. The linear velocity of the center of mass here will equal the radius of the sphere in this case, this ball, multiplied by the angular velocity. So I could just plug in my values here. The linear speed was five. The radius 0.2 meters. And I solve for my angular velocity omega. Which in this case, when I divide, gives me 25 radians per second. Remember the units for angular velocity are radians per second.

It then asks if the ball rolls for a full ten meters, through what total angle has the ball rotated? Again, you might tempted to use the geometry of the situation and say something like, well, I know that every time it rolls once, it rolls two pi radians for one complete roll. And you might look at how many rolls it took to go ten full meters, maybe looking at the circumference. Well, that method worked, it's very valid, but again, I'm going to try to stick with my physics relationships here since those are especially going to be helpful later on. And use this bridge equation that the linear distance covered should equal the radius times the total angle of rotation. So I know that it's gone a linear distance of ten meters. I know the radius of my object to be point two and I'm solving for theta, which is the total rotation of this object. Solving for theta then, I find this object rotated to a total of 50 radians.

Many, rotations, remember, radians, two pi, about six radians per one rotation. This took several rotations to complete this task.

As usual on a problem like this you will need to start with a diagram. What I have is a ball attached to a string that is being rotated in a circle. Imagine for a moment that that actually looks like a circle.

And it tells me several pieces of information about this rotational motion. It starts with an angular velocity of 0.5 radians per second but there's an acceleration and it tells me that it travels a total angle of 20 radians. And at the end of that motion it has this speed, this angular speed. And wants to know what acceleration do we have. Well, you might be tempted to solve for something like this using your angular acceleration equals your change in angular velocity over time. Except that they never told me how long this took. And so I'm going to use one of our kinematic equations here, but that one's probably not my best bet. Instead, I have a kinematic equation that does not include time which will allow me to solve in one step. It's the one where our final angular velocity squared will equal our initial angular velocity squared plus two alpha beta. So I have our two speeds, our acceleration, and our displacement, our angular displacement. And I can plug in the numbers they gave me if they're in base units appropriately. So, for instance, a final angular speed of three radians per second squared started off initially with only 0.5, and that needs to be squared, plus two. I'm solving for alpha, which is our radians per second squared, our angular acceleration, and I know it traveled for a total angular displacement of 20 radians. So, to solve this, I would subtract the 0.5 squared to the left-hand side of the equation. I would divide by 20. I would divide by 2. And I'd get an angular acceleration of 0.219 radians per second squared.

So it sped up. I have a positive acceleration in the direction of its motion. In this case, it doesn't really tell me which way it's rotating. It doesn't tell me if it's clockwise or counterclockwise. So for this problem, I just assumed counterclockwise, so direction didn't matter so much in this.

Let's see what else it asks. It says, now, calculate the tangential acceleration. Now, remember there's a difference. Angular acceleration is how many radians per second, change per second. So, how your angular speed changes. Tangential acceleration is measured in meters per second squared, and it's tangent to the circle. So, I want to know, how much more linear distance per second is being covered per time.

And so I'm going to solve either by using the linear quantities of linear kinematic equations which would work. Or I could use a bridge equation which is what makes them so very helpful here.

Be careful in circular motion like this. There is a centripetal acceleration to the center due to some net force inward. This problem has nothing to do with that. So there are really three types of acceleration: there's a tangential,

along and tangential to the circle that's in meters per second squared, a centripetal, also in meters per second squared due to a net force inward, and then there's an angular acceleration in radians per second squared. So, this equation, this bridge equation, gives you a tangential acceleration. It is not the same as your b squared over r for centripetal acceleration.

So, let's see. Using the numbers that we found earlier, it told me the radius towards the end of that problem, was 1.2 meters. We just determined the angular acceleration 0.219. And multiply those two numbers together to get our tangential acceleration. Something along the lines of 0.263 meters per second squared.

>> So, this first part a, where I'm asked to solve for angular acceleration, I consider what I've been given. Well, it tells me that this disk starts at rest. Remember back from when we emphasized how important that term was, at rest. That tells me that my initial velocity, in this case I'm looking at initial angular velocity, is zero.

Then they tell me that this reaches an angular velocity of two radians per second, so that's my final angular velocity over here, 2 rad/s, and all of this occurs any time of four seconds. So now that that's all for acceleration, I'm going to use my equation, and these equations will look very familiar to you from your kinematics equations.

I'm solving for alpha over here for acceleration or angular acceleration. So, again, the initial angular velocity was zero. And what I end up with here is 2 equals

alpha multiplied by the time, which is 4. And so when I solve for Alpha, I end up with 0.5 radians per seconds squared.

So then we use another equation you may have seen from one of your kinematics ones, but we're looking at it in terms of angular. So, I have angular displacement equals initial angle of velocity multiplied by time plus one-half alpha multiplied by time squared. So, from here, again, it started from rest, so this part cancels out.

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