https://youtu.be/GFExINFdfho
PHYS 1101: Lecture Twenty-Two, Part Two
Let’s start our new material again with my big picture, our tree perspective for you. Chapter eight was all about kinematics. But now it is about rotational motion and rotational kinematics. As I described, you were working on really a subset of this kinematic area for rotating objects. Now we’re going to go do a modification and realize for Newton’s laws there’s a part of that that’s applicable to circular motion. What causes something to wind up and wind down?
This is the analog to these equations for rotational motion. Here’s our kinematic analog. Now I’m going to write down for you our force analog. We’re going to work with the equation that the angular acceleration is equal to the sum of the torques divided by the moment of inertia, “I,” where each torque we calculate as “r” times “F” tangent. I’m going to define this carefully for you in a minute, but this, in essence, becomes our basic starting equation for the set of problems where it really falls in this category of Newton’s laws. Right in here for circular motion.
Now we are doing the analog, as I said, of chapter four, where we’re asking what causes motion to change? But now it’s the context of an object that might be rotating. The statement is still true that we’ve learned up until now, which is that for linear motion for an object to follow a general trajectory, we just need to now the direction of the net force to predict the total acceleration vector. That then and knowing what direction the initial velocity is in, we can predict the trajectory. That’s called the translational motion of the object.
To get the complete picture, though, an object can be translating and it can be rotating. If you take your pen and you hold it at one end and you throw it across the room you’re going to see that it will spin and rotate as it’s flying through the air. The flying through the air is a projectile motion. That’s the trajectory of a projectile, the translational motion. The rotation is now rotational motion that we have to describe in addition to that translation. And you need tools to understand and describe both of those to give a complete description of the motion.
Our focus is now on the rotational part. And we need to define this idea of torque in order to do that. At the heart of it is still the role that forces play in causing something to start spinning faster or slow down. We’re going to need to know the size of the forces and the direction of the forces, but now there’s one extra piece of information that’s very important. We now have to know where that force is applied to the object.
Let me note here for you. We need the total force information, which means both magnitude and direction, but we also need the location that the force is applied to the object. This would be the actual contact point where that force is acting. You can think of this as the location of that contact point or directly where the force is acting on this object.
I think you can appreciate this in the context of just opening a simple door. Here’s a top view of a door. The hinge is on the left. Here’s the door. Here’s my doorknob. This is what someone sees looking down on the door. It’s definitely rotation that occurs in this problem, right? The door swings and rotates about this hinge. Your intuition, I think, would tell you that where you push on the door has a strong impact of how quickly the door is going to open. Which is the best approach? When you go to open a door, do you push and apply with your hand a force that is close to the hinge to open the door? Do you push with a force that’s near the end of the door, far away from the hinge? Or do you push kind of into the door in hopes of opening the door?
I hope for sure recognize that this makes no sense. You’re not going to open very well at all if you try to apply a force like this. Try this. Compare these two the next time you go to open the door. Turn the knob so that it’s not latched, so all you have to do is push it to get it to swing and compare pushing close to the hinge compared to pushing right at the edge of the hinge. You’ll notice a huge difference.
Question three: if you need to open a door, what’s the best approach? We need to appreciate that a torque, which relates to how much action, how much rotation action we’re going to get because of this force, means I have to think about where the force is applied the distance from the rotation axis and then the size of the force and the direction. And here’s what’s important. If I have a rotating object, and it’s clear what the rotation action is, so it’s given to you in the problem or it’ll be like a door, it will be an obvious hinge somewhere. If a force is applied to this object at a distance, r, so the line from the rotation axis out to where the force is applied will be my distance, r. I need, then, to find the component of the force that is perpendicular or tangent to this line, R,” the radius.
And then once this object starts to rotate, it’s going to follow a circular arc and fT here is going is the part of the force that is going to be tangent to that circular path. It’s the part of the force that’s perpendicular to this radius line. That’s the part that I need because that’s the only component of f that helps to rotate this object. The component of the force that’s in the same direction as r is analogous to pushing straight into the door, and that’s not at all going to help rotate the door.
I need the part of the force that’s tangent to this, the rotation. This is the only part that’s going to contribute the torque. Tau, it’s the Greek variable Tau we’re going to use for that. The torque is equal to the product of r and this force but, again, it’s only part of this force that is helping with the rotation. Highlight this, very important equation.
Now what we are going to have is problems where maybe it’s a door, maybe it’s a bridge, it’s a record, it’s a merry-go-round, rotating objects where I can have multiple torques potentially applied to it. Some of these torques are going to want to rotate the object in one direction and other torques, perhaps the other. I’m going to use the sign of this torque to again make that distinction and to take into account these possible competing directions.
I need a convention. We’re going to go with the same sign convention that we use for our angular quantities. If a torque would tend to cause an object to rotate in a counterclockwise direction, we’re going to assign a positive value to that torque. Conversely then, if a force wants it to rotate in the clockwise direction, it’s going to be a negative torque.