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• Hi. It’s Mr. Andersen and this AP Physics essentials video 52. It is on torque. Torque

• is simply the product of the force, which is perpendicular to the lever arm. And this

• only works in rotational motion. So what is rotational motion? Imagine that we have a

• wrench that is sitting here. And I have marked the center of gravity. And so let’s say

• it is just sitting in space to make this easier. And let’s say I apply a force right at that

• center of gravity. What we are getting is linear, or we call that translational motion.

• We apply a force in one direction and we are seeing an acceleration in that direction.

• Let’s say I do the same thing but now I am going to do it off center. So I am applying

• a force not at the center of gravity. And watch what we get. We get two forms of motion.

• We are getting that translational motion but we are also getting rotational motion. And

• it is hard to separate those two. And so an easy way to look at rotational motion is to

• pin it down. So now I have this same thing. We have the wrench. But now it is centered

• around this nut on one side. And so this is not moving. So it is going to rotate around

• that point. Now let me apply that same force on one direction. And now we are just getting

• that rotational motion. Let me return the wrench for a second. And we can start to talk

• about what torque is. And so this is going to be the lever arm. The lever arm is going

• to go from this point in the middle, that axis of rotation. It is going to go out from

• that. It is perpendicular to that axis of rotation. And it is just like a lever that

• you would use as a simple machine. And then we are applying a force perpendicular to that

• lever arm. And so what is torque? We are multiplying these two values. We are multiplying the force

• that we apply, perpendicular to the lever times the lever arm itself. And so it is two

• things and the units are not just going to be newtons, it is going to be newton meters.

• And so in a rotational system like this, where we have an axis of rotation, what is torque?

• It is simply the product of that lever arm which is going to be a vector, it has to be

• perpendicular to that axis of rotation. Again the axis of rotation is coming straight at

• us through the middle and we are going to multiply that times the force. The force has

• to be perpendicular to that lever arm. And so if we multiply those two values then we

• are going to have ourselves a torque. And this is our equation right here. Tau, which

• stands for torque, is equal to that lever arm, perpendicular and then that force. And

• so if we have a system that is not moving, like our rotational system is not moving anymore

• we know that all of those forces must be balanced. In other words all the torques must be balanced.

• And so the net torque on a balanced system is equal to 0. Have you ever noticed that

• a door will always have the door knob on the outside or far away from the hinge? Well the

• reason that is just deals with torque. And so what I have here is the door itself. This

• is the door hinge right here. And here is the door knob on the outside. And so if I

• apply a force out here what am I going to get? Torque. We have the lever arm and then

• we are applying a force perpendicular. And so we are going to get an acceleration or

• rotational acceleration like that. What happens if I apply a greater force? What am I going

• to have? I am going to have a greater torque. And so we are going to see a faster acceleration

• like that. What happens if I take that smaller force however and move it to the inside? What

• happens if I put the doorknob on the inside of the door? If I try to pull on it with that

• small force nothing will happen. In other words I have decreased that torque so much,

• since I have decreased that lever arm, that I would not have enough torque to open that

• door. And so that lever arm distance is incredibly important. Just like a lever in a lever system,

• it is giving us a mechanical advantage to open up that door. So in torque we are applying

• that lever arm times the force. That is how we calculate the torque. And the equation

• looks like this. Tau is equal to r perpendicular F. And so let’s add some numbers in here

• and I will show you how to calculate that. Let’s say the doorknob is 75 centimeters

• from the hinge and we apply a 6.7 newton force. What is going to be our torque in this situation?

• Well I am going to plug in those values. Again I had to convert the centimeters into meters.

• We have to use SI units. And now I simply multiply those values. What are my units?

• It is now in newton meters. You can see I have too many significant digits. And so we

• would get a 5.0 newton meter torque if the doorknob is at the outside. Now watch what

• happens if we move it 15 centimeters from the hinge. Watch what happens to our torque.

• Again the force is going to be the same. But now we are going to get a 1.0 newton meter

• torque. It is 1/5 of what it was before, which is not surprising, because the lever arm is

• now 1/5 of what it was before. Let’s take a second to look at a balanced system. And

• a see-saw or a teeter-totter is a great example of that. We have the axis of rotation right

• here in the middle. And we can have a lever arm on this side and a lever arm on the other

• side. So what we can do is we can apply a torque on one side and a torque on the other

• side. And if those torques are equal, then we are going to have a balanced system where

• you should not see any movement. And so watch what happens when I remove the supports. Since

• the torque is the same on either side, nothing happens. Let me apply a 5 kilogram weight

• on either side, if I remove the supports, it is totally balanced. Torque is equal on

• either side. Let’s say I add a 10 kilogram weight to the right side. And now remove the

• supports. What happens? Well we have greater torque on the right side and so we are getting

• rotational motion in that direction. Now a good question I might ask you is where could

• I move that 5 kilogram mass on the left side so the torque on the left equals the torque

• on the right? How could I balance this system out? Well if we look at our equations again,

• it is simply the product of the lever arm times the force. And so if I throw in some

• values here and set them equal to each other, on the left side we have the unknown lever

• distance times that 50 newton force. Where do I get the 50 newton force? I am simply

• taking 5 kilograms times about 10 for the acceleration due to gravity. So it is a 50

• newton force down on that side. On the other side it is going to be 100 newton force. So

• I could solve for this. We would have a torque of 400 newton meters on the right side. And

• so to solve for that we should have a distance on the left side equal to 8 meters. And so

• if I take that 5 kilogram mass, move it out to 8 meters, what are we going to get? A balanced

• system. The torque on either side is going to be exactly the same. And so did you learn

• the relationship between a force and a torque? Again we have to multiply the force times

• that lever arm. Do you see what happens when we apply different forces? We increase the

• torque. Or what happens if we move that force that in? We are decreasing the torque because

• we are decreasing the lever arm. Can you design an experiment that would allow you to kind

• of manipulate these balanced forces? Again we used a teeter-totter to do that. And finally

• can you calculate torque in a two-dimensional system like this?

• I hope so. And I hope thatwas helpful.

Hi. It’s Mr. Andersen and this AP Physics essentials video 52. It is on torque. Torque

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# Torque

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Wayne Lin posted on 2015/03/28
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