Ch 12: Torque & Rotational DynamicsWorksheetSee all chapters
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Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
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Ch 13: Rotational Equilibrium
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Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Concept #1: Torque & Acceleration (Rotational Dynamics)

Transcript

Hey guys! Now that you know how to calculate torque in a bunch of different situations, we're going to look into what happens as a result of the torque which is that you get an acceleration. Let's check it out. We're going to look into torque with acceleration, not just calculate torque but also maybe calculate the acceleration that results from it and a combination of these two items when you have an acceleration as a result of a torque. That situation is called, this topic in physics is called rotational dynamics. Statics typically refers to situations where there is no velocity, so you're at rest. Dynamics means that there's going to be velocity or acceleration. We already know this that when a force causes rotation, it produces a torque or even when a force tries to cause rotation, it produces a torque. You can think of torque as the rotational equivalence of force. You know this as well. What I want to do is since they're equivalent of each other, I want to do a compare and contrast between forces and torque. Force remember causes linear acceleration, a. The relationship between force and acceleration a is described by this equation, by the sum of all forces equals ma, which we call Newton's second law. This is Newton's second law. Torque is very similar. Instead of causing linear acceleration a, torque causes angular or rotational acceleration, the same thing, _. The relationship between torque and what it causes, which is _, is very similar to the relationship between F and a. In fact, it's the same equation except we're just going to switch the variables to their angular equivalence. I just mentioned how the angular or rotation equivalent of force is torque. We're going to do instead of sum of all forces, weÕre gonna do sum of all torques. Remember, the rotational equivalent of mass right, in rotation we don't use mass. What matters is your moments of inertia, so I'm going to put an I here. Instead of a in rotation, we're going to have _. These two equations are basically the same thing. ItÕs just one is with linear variables the other one is rotational variables. In fact this is also Newton's second law. This is the rotational equivalent or the rotational version of Newton's second law. You may see your professor or your textbook call this Newton's second law. The idea is that both of these guys are Newton's second law. The quantity of inertia is how much resistance you have to linear acceleration is given by the letter m, by mass. Mass is the amount of resistance to change, the amount of inertia you have. The amount of resistance to _ is not m, but it is I. The last point is that when you have force and acceleration, you have this branch of physics called linear dynamics which in the past we may have called it just dynamics because there's only one type. But if you have torque and _ instead, you have what's called rotational dynamics. That's just a name. It doesn't really matter. But in case you hear these words, you know what's up. That's the difference. Basically you might remember doing a bunch of F = ma. Now you're going to do a bunch of Torque = I _. In fact in some cases you're going two of them combined. Let's do an example here. See how this stuff works. Here it says I have a solid disc. Solid discs remember, this is a shape of the disc so I can stop and write that the moment of inertia is _ MR^2. Solid disc of mass 100 and radius 2, so I already get those numbers; M = 100, R = 2, is free to rotate so it can spin around this, a fixed axis. It can rotate around the axis but the axis doesn't move. The disc is fixed in place and it can only spin in place. The axis is perpendicular to the disc. That means that if you have the face of the disc, the axis points this way which just means the disc is going to spin around itself and itÕs frictionless. You push tangentially on the disc. If you push tangentially on the disc, it looks like this, like sort of at the edge of the disc with a constant force of 50. Let me write this here, F = 50 N. We want to derive an expression for the angular acceleration the disc experiences. Part A, we want to find _, that's angular acceleration. We want to derive an expression. For Part B, we want to then calculate that. Find an expression and then calculate just means plug in all the numbers. How do we do this? I'm giving you a force and I'm asking for an _. Back in the day if I give you a force and ask you for a, you would use F = ma. But here I'm giving you a force but asking for an _, so instead you're going to use sum of all Torques = I _ and that's because you're looking for _. The only force that produces a torque on this disc is this force here. The only torque we have is going to be the torque of F. Now, that force is producing a torque that's trying to spin this thing this way, which is a clockwise torque, so it's negative. I'm going to put a little negative in front. The moment of inertia is _ MR^2. IÕm going to go ahead and write this here and IÕm going to leave _ alone because that's what we're looking for. Now I have to expand torque of F which is why it's important to know how to calculate a bunch of different torques. The torque of any force F is Frsin_, where remember r is the distance from the axis and _ is the angle between F and r. Here, my r vector looks like this and the length of the r vector is the entire radius of the wheel and that's because you're pushing all the way at the edge of the wheel. This is going to be the force you're applying, little r is the radius and then sine of _. _ is the angle between the force and the r vector. The angle between those two is this right here, which is 90. I get sine of 90. That's what you get on the left side. Let's rewrite the right side again here and you get this. Sine of 90 becomes 1. This R cancels with one of the two R's on the other side and we're ready to go. I'm solving for _. I'm going to move everything to the other side so _ is by itself. I'm going to get -2F / MR. That's going to be our _. Notice IÕm getting a negative acceleration which makes sense. It's going this way so the acceleration should be negative. This is Part A. For Part B all we're doing is plug it in the numbers. That's easy. -2F, the force is 50, the mass is 100 and the radius is 2. This is going to be -0.5 rad/s^2. That's it. That's it for this one. Hopefully it makes sense. Let me know if you have any questions. Let's keep going.

Practice: Suppose that piano has a long, thin bar ran through it (totally random), shown below as the vertical red line, so that it is free to rotate about a vertical axis through the bar. You push the piano with a horizontal 100 N (blue arrow), causing it to spin about its vertical axis with 0.3 rad/s2 . Your force acts at a distance of 1.1 m from the bar, and is perpendicular to a line connecting it to the bar (green dotted line). What is the piano’s moment of inertia about its vertical axis?

Concept #2: Torque & Acceleration of a Point Mass

Transcript

Hey guys! In this video, I want to talk about a torque producing and acceleration on a point mass. You may remember that in all these rotation questions, we can have either point masses which are tiny objects with no radius and no volume or we can have rigid bodies or shapes that have radius, that have a volume. Let's check out how torques and acceleration work for point masses. Real quick, I want to point out that most torque problems are actually going to involve shapes or rigid bodies. But I want to quickly just do one with point masses because it works just the same. Most of the time you're going to be here, but this works just the same. Quick example, you spin a small rock. Small rock is an indication that this is going to be a point mass because they say it's small and they donÕt give you the shape of the rock. The mass is 2 kilograms. You do this at the end of a light string of length 3. If you spin a rock or any object at the end of a string, the length of the string will be the distance to the center. It will be your little r. Little r equals length of the string which is 3 meters and the mass is 2 kilograms. We want to know what net torque is needed to give the rock an acceleration of 4. If you want to have an acceleration of 4, what is the torque you need? I want to remind you that net torque is the same thing as sum of all torques. Check this out. I am asking for sum of all torques and I'm giving you _ so I hope that you immediately thought of using sum of all torques equals I _ and this is what we're looking for. To calculate this, we just have to figure out these two guys. Remember, point masses have a moment of inertia. The moment of inertia of point masses is given by mr^2, where r is a distance to the center. We're going to replace this with (mr^2)_ and we have all of these numbers. The mass is 2, the r is the distance which is going to be 3 and we have to square that and then _ is 4. If you multiply all of this, you get 72 Nm. Remember torque has units of Nm. That's it for Part A. We got that done. For Part B, it's asking us to find the tangential acceleration while it has that speed. Remember the tangential acceleration, a tan, is related to _ by this equation: a tan = r _, so all we got to do is multiply. r is the distance here which is 3 and _ is a 4, so that means that at that point when you have an _ of 4, your tangential acceleration is 12. This is an acceleration a. It's a linear acceleration, so it's going to be 12 m/s^2. These two are somewhat unrelated. This is new stuff. This is just plugging it back into a different kind of acceleration, some old stuff bringing that back, putting it all together. That's it for this one. Let me know if you have any questions and let's keep going.