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

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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

Just like we did with Kinematics, we can discus Dynamics for

Â rotating bodies, as well.

Â To start or stop an object from rotating requires a force.

Â But where that force is located, and

Â what direction of points will change how this object will spin.

Â To describe the behavior we use a quantity called torque.

Â Torque is defined, as a perpendicular force multiplied by the lever arm.

Â What do we mean by lever arm?

Â The lever arm is the distance along an imaginary line drawn between the pivot,

Â and where the force is being applied as seen here on this wrench.

Â Let's look at a couple of examples, and

Â calculate the torque that results from each force.

Â Notice that in these three examples,

Â the angle at which the force is applied to the lever arm, changes.

Â Now, let's look at an example in which torque, acts upon the same object.

Â In this question, we have two cylinders, we've got a small one,

Â with radius of 0.05, and a large cylinder with radius of 0.15 meters.

Â It tells us that the smaller, the cylinder with the smaller radius,

Â is the one that's moving counterclockwise.

Â That's important because for us, that means that it's positive.

Â And this one is moving with a tension of 10 Newtons.

Â Then it says, the light string is wrapped around the larger radius, and

Â that one is pulled clockwise, which for us means, that it's moving negative.

Â With the a tension of 7 Newtons, our first task asks us to solve for the net torque.

Â [NOISE] So, for part A, as I solve for the net torque,

Â it's going to be the sum of all of my torques, and

Â remember that that's going to be R times F.

Â And so in this case, I have the smaller one, moving positive.

Â [NOISE] So, I have R, and the force applied to, and

Â I'm going to denote this, as small cylinder.

Â This is positive, and it's being combined with this negative,

Â larger cylinder, or supply to larger one.

Â So, now I can go ahead, and plug in my values.

Â The R over here was 0.05 meters.

Â The force supplied was 10 Newtons or tension.

Â The R over here is 0.15, and

Â then the tension applied to the larger cylinder was 7 Newton.

Â When I solve for this, I end up,

Â with a net torque of 0.55 Newton meters,

Â and it's negative, so I can go ahead and write this out as clockwise.

Â And so that's my first answer.

Â Now part B says, what torque would have to be applied to the system, so

Â it would rotate at a constant angular velocity?

Â Okay.

Â So, when we're looking at constant angular velocity, or just constant velocity in

Â general, well, that tells us that our acceleration equalled zero.

Â Everything is balanced.

Â How is torque related to acceleration?

Â [NOISE] Well, [NOISE] think back to your F equals MA,

Â where some of the forces was directly proportionate to the acceleration.

Â Similarly now, the sum of my torque is

Â directly proportionate to alpha or angular acceleration.

Â So, as I solve for this now, what I need to consider is,

Â if my net torque is 0.55 Newton meters clockwise,

Â then how much torque would completely balance the system out?

Â It has to be equal and opposite,

Â which means that the answer is 0.55 Newton meters counterclockwise.

Â [NOISE].

Â Just like a net force, results in a translational acceleration.

Â A net torque, acting on a body causes an angular acceleration.

Â This is a restatement of Newton's second Law, as can be seen here.

Â The capital I in this equation represents the rotational moment of

Â inertia of an object.

Â Much like mass represents the inertia of an object.

Â For desired angular acceleration,

Â an object with a larger moment of inertia requires a larger net torque,

Â when compared to an object with a smaller moment of inertia.

Â The moment of inertia of an object depends upon how much mass is present, and

Â how far the mass is from the center of rotation.

Â Here, you can see a few equations for the moment of inertia of some common shapes.

Â Notice that a thin hoop, and

Â a point mass have all of their mass are very far away from the center of rotation,

Â and therefore have a much larger moment of inertia than a sphere or even a cylinder.

Â Let's look at an example,

Â where we could determine the moment of inertia of a series of point masses.

Â Remember that when calculating the moment of inertia, both the mass and the distance

Â of that mass from the origin or, or from the center of rotation, is very important.

Â We're talking about point masses here, so I'm going to be using the equation,

Â [NOISE] I equals mr squared, and I can add together the moment of inertia of

Â various shapes to get a total moment of inertia, which is what we'll do here.

Â You draw a quick sketch of the scenario, it's giving me a coordinate system.

Â And telling me that firmly origin,

Â the first mass is located, may be at the point here.

Â 2 meters to the right of the origin, that mass is self however.

Â Its going to be, well, called mass one?

Â 5 kilograms in small point mass.

Â Then only we have another mass located at another point also 2 meters to the right.

Â But also up about 4 meters.

Â [NOISE] And so, we'll call this mass 2.

Â And that will also be [NOISE] 5 kilograms, since it tells me that they're identical.

Â So in this problem, if I want to find the total of inertia of the system.

Â A measure of how hard it is to start the system rotating, or

Â to stop it from rotating, to change its angular velocity.

Â What I'm going to add together all of the point masses together,

Â I'm going to sum them together.

Â So, what I need to do [NOISE] is really find the distance for

Â each of these masses, from the origin.

Â One will definitely, be easier than the other.

Â So let's try to set this up.

Â Well, I have mass 1, which again, is 5kg.

Â So, why don't we go ahead, and sub in our values-

Â >> [COUGH].

Â >> Here, at this point.

Â Mass 1 is 5kg.

Â And I also notice that it is exactly 2meters here, from the origin.

Â So, I could just plug that in, as my r from the axis of rotation.

Â The center is the origin in this problem, squared.

Â Plus our m2s also 5 kilograms, and now I reach something of a road block.

Â Notice, [NOISE] that I don't yet, have the distance the that

Â m2 is from the origin, from the center of our rotation.

Â And that's the distance that is our r2 that I need.

Â Noticing, you know, that this a right triangle, though, gives me a hint.

Â I'm going to use our Pythagorean theorem,

Â like distance formula to find that, side of this triangle.

Â If this 4, and this is 2, well, it looks to me like our hypotenus, or hypotenuse.

Â R2 will equal the square root of 2 squared

Â plus 4 squared, that would give me my diagonal.

Â And that would be the distance that we need over here, for r2.

Â So I'm going to sub that in.

Â Square root of 2 squared plus 4 squared.

Â That, all of that needs to be squared.

Â So notice on that right-hand side there, that the square root sign will cancel out,

Â with the square, and it will get rid of it for us.

Â So, reducing this down some 5, 2 squared is 4.

Â And we'll get 5, and on the inside here, what will I, I have left?

Â Well, 4 plus 16.

Â 2 squared and 4 squared.

Â That means our moment of inertia for

Â this system will come out to be 120 kilograms meter squared.

Â And again, if you forget the units for

Â moment of inertia, which is very possible, it's very easy to do.

Â There are lots of units in this section.

Â Just remember to go back, and look at your equation.

Â We're taking a kilogram, a mass, and multiplying it by meters squared.

Â So that gives me the units for our moment of inertia of this system.

Â In a longer problem like this,

Â as usual, try to be very clear about your diagram, and your variables.

Â We have a solid cylindrical pulley that has a light string wrapped around it,

Â and we're attaching a mass.

Â So, why don't we call this mass sub h for hanging mass.

Â And mass sub pulley for our pulley that actually has mass in this problem.

Â And they tell us that the mass of our pulley is 2 kilograms,

Â that the radius of our pulley is 0.1 meters.

Â It also tells me that the hanging mass in this problem is 4 kilograms.

Â I like to draw a lot [NOISE] of my forces,

Â as free body diagrams here in my scenario, so I know that I have an mhg.

Â The weight of this block pulling down, I also know that there's a tension here,

Â we'll call it a T1.

Â Well, because our strings are massless, the tension here will also be the same.

Â T1, that's what pulls down on the cylindrical pulley, and causes it to spin.

Â That will also produce a torque.

Â Because there's a radius here, and

Â it's acting at 90 degree ang, angles, to that lever arm.

Â So, I have that torque that we'll have to make sure we include with the radius,

Â and the tension.

Â Okay.

Â Well, to solve I know that this looks like a forces problem, so

Â why don't we start with that?

Â Let's set up our forces equation for this hanging mass.

Â So, I'm going to do some forces equals an angle like we always start off.

Â Off with our Newton's second Law,

Â our fundamental principle for forces questions like this.

Â Again, we're going to set the positive direction,

Â the direction of acceleration, like we always do, with elevators and pulleys.

Â So down will be positive, clockwise will be positive.

Â That means that the two forces acting on the same mass will be

Â a positive because it's downward.

Â Mhg, and a negative T1 for the tension upward.

Â That should equal, the hanging mass times its acceleration.

Â I can then solve this for T1.

Â [NOISE] I'm moving T1 to the right, and subtracting MA, so

Â that's what I'm going to do.

Â T1 would now equal mhg minus mha.

Â Well, let's take a look at this now.

Â What we're looking for is the linear acceleration, which this equation has?

Â I also know g is 10.

Â I also know m or the hanging mass is 4 kilograms.

Â But I don't know the tension in the string.

Â That means, I need to set up another equation.

Â I have two unknowns, I need two equations.

Â Well, since we're in the rotation section, and

Â I know that this pulley will be rotating, why don't we go ahead and

Â set up an equation for the rotation of this pulley?

Â So, to do that, I'm going to start off with torques.

Â I'm going to change [NOISE] colors here.

Â Sum of torques equals i alpha.

Â All the torques on an object should equal its moments of inertia times

Â this angular acceleration.

Â So let's try to set this up.

Â Remember that torque is the perpendicular component, but

Â our tension is already perpendicular up here to the radius.

Â So T1 times rp.

Â The force times the lever arm, and already perpendicular, so I

Â don't need to worry about sine or a cosine to give me the perpendicular component.

Â [NOISE] That should equal the moment of inertia.

Â Well, this is a solid cylinder, a disc.

Â So that tells me that it needs to be one-half mr squared.

Â M of the pulley, the radius of the pulley squared,

Â and times by the angular acceleration.

Â But we're not solving for angular acceleration.

Â Right now, we have another variable in there that we don't want to keep.

Â But I do remember my bridge equation.

Â And since, the string isn't slipping,

Â there's a relationship between the linear acceleration of the falling mass.

Â And the linear acceleration of the outside of that pulley.

Â They should be the same.

Â So I can use this bridge equation.

Â The pulley's not being, the rope's not slipping, so they will move together.

Â They will have the same acceleration.

Â And so, let's go ahead,

Â and kind of do some algebra here with our multiple equations.

Â I'm going to bring the this down, and sub m for T1.

Â That gets rid of that variable I didn't know.

Â I'm also going to come, and sub in for alpha, here.

Â to get rid of it there, as well.

Â So let's see what our equation looks like once that occurs.

Â [NOISE] I'm going to go ahead, and change colors one more time.

Â So, mhg minus mha multiplied by the radius of the pulley.

Â That should equal one-half mass of the pulley,

Â radius of the pulley squared, and inside alpha here will be a over r of the pulley.

Â Notice that a lot of times in these problems, the radius will cancel out.

Â One radius will cancel out one of these, and there's a division here,

Â which will cancel out the second.

Â So rp, the radius of the pulley, is completely gone from our set up.

Â I can even give myself a little bit more room, and solve this for a.

Â Remember, we're solving it for the linear acceleration.

Â That means, when you get 2s here, to one side of the equation.

Â So I'm going to move all the as, the terms with a's, to the right-hand side.

Â And begin to solve by pulling a out.

Â So, just not showing all my algebra here, I get to this step.

Â [NOISE] And feel free to pause, if you need to see where this comes from.

Â [NOISE] But I've moved all the terms with a's over to the right.

Â And I've pulled a out.

Â And then, I can divide if I want to see,

Â maybe what the equation looks like, solving it for a and variables first.

Â Although, this isn't necessary.

Â You can also sub in your variables at this point, or your numbers at this point.

Â One-half mass of the pulley plus mass of the hanging mass.

Â Now let's go ahead, and separate our numbers.

Â They told me the hanging mass was 4 kilograms, g, for

Â us, is 10, that's perfectly acceptable on the AP exam.

Â One-half the mass of the pulley, they told us, was 2 kilograms, and,

Â again, the hanging mass was 4 kilograms.

Â [NOISE] So

Â when I plug all that into my calculator, I get 8 meters per second squared.

Â That's a good point to notice though.

Â This object isn't in free fall.

Â So of that force is and energy is going into making the pulley spin and rotate.

Â I should end up with an acceleration here, less than the acceleration due to gravity.

Â Cause gravity is the applied force here doing the work and

Â making this se, scenario move.

Â Okay. Let's see what it asks for

Â next because we're not quite done although we're getting there.

Â And then asks, let me go back up here.

Â Calculate the linear acceleration of mass in the falls.

Â Then calculate the angular acceleration of the pulley.

Â Now, the pulley is speeding up, and it's spins.

Â It started at rest, so that's something we can calculate.

Â And it wants to know through what angle does the pulley rotate in just

Â the first 3 seconds.

Â So, [NOISE] let's see where that can take us.

Â again, I'm going to try to jump to my bridge equation whenever possible.

Â If I'm looking for an angular acceleration, but I've already solved for

Â linear acceleration, I'm going to try to use this whenever possible.

Â We just solved for the linear acceleration of the outside of that pulley.

Â We also know it's radius is 0.1 of a meter.

Â And so very quickly here I've already solved for the angular acceleration.

Â The 80 radians per second squared.

Â [NOISE] That's the nice thing about Kinematic equations typically, the further

Â along you get, the shorter the problem becomes because you know, and more.

Â You get more, and

Â more options that you're opening up, as you know more about the scenario.

Â It also wants to know through what distance it turned.

Â And you have to be careful, because this is an acceleration scenario.

Â So, I need to use a Kinematic equation.

Â I can't just do something like,

Â angular displacement equals angular velocity times time.

Â That angular velocity was changing.

Â So, I went ahead, and

Â chose this one, although there are many options that will work.

Â [NOISE] Some more quickly, than others.

Â This is our equivalent to x equals v naught t plus one-half at squared,

Â except now we're talking about angular quantities.

Â It started from rest, so that pulley had no initial angular speed.

Â We just solved for the angular acceleration, so

Â let's solve for how much radians this pulley spins through.

Â One-half 80, it asks me in the first 3 seconds, so

Â I'm going to make sure that term is squared.

Â When I put that number into my calculator, I get 360 radians.

Â Be careful with that answer.

Â What some people will see, when they see 360 is they think, oh,

Â this objects spun, or rotated only one time.

Â Be careful when radians, not degrees.

Â Radians, degrees would be yes, the 360 degrees would tell us, one whole rotation.

Â 360 radians, remember there's a,

Â two pi radians in one rotation, something about 6.28.

Â That means there are lots of rotations here.

Â This is something like 60 circles this objects going to make in it's rotation, so

Â just keep that in mind.

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