https://youtu.be/65GsLD2vbM8
PHYS 1101: Lecture Twenty-Two, Part Six
All right. Let’s just do an example here. Try to put some of these things together. We have here a cylinder that’s rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of .085 meters, an angular speed of 88 radians per second. This tells me I have an Omega information here, and it has a moment of inertia of .85 kilogram meters2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force. The frictional force reduces the angular speed of the cylinder by a factor of 2, during a time of 5.2 seconds. So, it starts out at 88 radians per second, and 5.2 seconds later, it’s reduced by a factor of 2.
Let’s first start out and just draw a picture of this object and get a clear view of the rotation axis. Here’s my cylinder. I’m not a very good drawer. If it’s not a game of Pictionary, where the stakes are high, I just can’t draw very well. Here’s my rotation axis. Okay. Let’s make this thing rotating in the counterclockwise direction. So, at least my Omegas, I can start out with them being nice positive values, et cetera. So, there’s the cylinder. It’s rotating about the axis that passes through the center of each round circular end.
What are some of my knowns? A moment of inertia, that’s the variable I, that’s .85 kilogram meters2. It has an initial angular speed of 88 radians per second. I know that 5.2 seconds later, so at T equals 5.2, that my final angular velocity is half. It’s gone down by a factor of 2. So, it’s one half of this. It’s gone down to 44 radians per second. I need to find some different things here. For part A, what’s the magnitude of the angular deceleration of the cylinder?
Let’s see. The other thing, I guess, I want to add to my sketch here is… Let me sketch here. This is supposed to be my brake pad, brake shoe. The whole purpose to the brake shoe, as it says here, is that by pressing against the surface of the cylinder, it applies a tangential frictional force. If the wheel is rotating this way, the frictional force has to oppose it. I’m going to do this and label it ft. It’s my tangential frictional force. For part A, what variable do I want to solve for? Find the magnitude of the angular deceleration of the cylinder. The variable I want there is Alpha.
The angular deceleration… Deceleration just emphasizes, again, that the force is working to slow it down. So, if my Omega0 starts out in that direction, Alpha is opposite to that. Or, if I draw Alpha over here on this side, I see that Alpha, therefore at, is in the direction of ft. It’s opposite to the direction it’s going. That will decelerate it. So, my Alpha is going to have a negative value, opposite to the sign of my Omega0. So, for part A, I want to find Alpha. My first reaction is that that’s kinematic information. I know time. I know by how much my angular velocity is changed. So, I’m inclined to try my kinematic equation number one, which is the final angular velocity is the initial plus Alpha times t. I know the time. I know both of these values. I’m after Alpha.
If I rearrange that algebraically, I end up with Alpha is equal to Omega minus Omega0 divided by t, and I can plug those numbers in. 88 minus 44 is going to be 44 radians per second, divided by 5.2 seconds gives me 8.46 radians per second2. Let me be a little bit more careful here. Omega final is 44 minus the initial, which was 88, gives me actually -0.846 radians per second2. The negative is important. It’s opposite sign to my angular velocities, which is why the thing slows down. I go from 88 to 44. It wants me to report the magnitude. Therefore, I would just report 8.46.
Now, for question B, it asks: What’s the magnitude of the friction force applied to the brake? I definitely am after, ultimately, the variable fT, the magnitude of this force, but what equation do I use? What captures the implications of this force to this angular deceleration? Well, that’s our new equation, the new material we’re learning in this chapter. I have to use my Newton’s Second Law analog. I have to use that this angular deceleration is a result of the net torque divided by the moment of inertia. Well, the only force I have applying a torque to this wheel is fT. So, the torque I include is only the torque do to this. The moment of inertia, that’s a number that I was given. I have the angular deceleration. So, I’m going to be set. I just have to…
Let’s start customizing this equation. Alpha equals sum of the torques. Well, the only torque I have is the torque caused by this force fT, f tangential from that brake shoe. Then, I have to divide by the moment of inertia. What’s the expression for torque? Every torque is related to or given by the distance from where the force is applied to the center r times the part of the force that’s perpendicular. Well, if the brake shoe is pressing straight into the wheel, all of that frictional force is going to be nice and perpendicular to r. So, this is just going to be fT. So, let me substitute that into my equation that sets this constraint. I’ve always got to go back to this equation.
Alpha equals rfT over I. Okay. Now, I do have all of these values. This is the variable I need to solve for. Do a little rearranging there, and I have that fT is equal to 8. Let me write it out here. Alpha I divided by r. I should be a little more careful with the sign here and with our sign convention and the direction here of these torques. I should represent that the torque that this friction does is in the clockwise, and so the negative, direction. So, I really should have put a negative sign in there to be careful.
When I do my rearranging, then, that negative sign is still there, so that when I go to plug in numbers, I end up with the value of this force as just being a positive number. In other words, I’m going to have negative minus 8.46 radians per second squared times my moment of inertia, which was given to me as .850 kilogram meters2. Then, I have to divide by the radius of this wheel, which was 0.085. So, the negatives here cancel, and my force value ends up being a positive number. That’s as it should be, because I already took the proper vector information into account by explicitly writing this as a negative torque. The value of that frictional force, that tangential force from the brake shoe, is 84.6 Newtons.