14. Torque & Rotational Dynamics

Torque & Acceleration (Rotational Dynamics)

1

concept

## Torque & Acceleration (Rotational Dynamics)

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Hey, guys. So now that you know how to calculate torque in a bunch of different situations, we're gonna look into what happens as a result of a torque, which is that you get an acceleration. So let's check it out. All right, So we're gonna look into torque with acceleration, not just calculate torque, but also maybe calculate the acceleration that results from it and the combination of these two items when you have a knack celebration as a result of the torque, Um, that situation is called. This topic in physics is called rotational dynamics. Okay. Statics typically refers to situations where there is no velocity, right. So you're at rest, and dynamics means that there's going to be velocity or acceleration. Um, so you 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 equivalent of force. You know, this is well, now what I wanna do is since their equivalent of each other, I want to do a compare and contrast between forces and torque. So force, remember, causes linear acceleration, linear acceleration A and the relationship between force and acceleration, a eyes described by by this equation by the sum of all forces equals M A, which we call Newton's second law. This is Newton's second law. Right now, torque is very similar. Instead of causing linear acceleration, a torque causes angular or rotational acceleration. Same thing Alfa, right. And the relationship between, um torque and what it causes, which is Alfa, 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 equivalents. So I just mentioned how the angular or rotation equivalent of forces torque so we're gonna do instead of some of all forces going to some of all torques. Remember the rotation equivalent of mass right in rotation. We don't use mass. What matters is your moments of inertia. So I'm gonna put an eye here, and instead of a in rotation, we're gonna have Alfa. Okay, so these two equations are basically the same thing. Just one is with linear variables, the other ones 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. So you may see your professor or your textbook call this Newton's second Law. And the idea is that both of these guys are Newton's second law. Alright, eso the quantity of inertia, quantity of inertia is how much resistance you have. Tow acceleration to linear acceleration is given by the letter M by mass. So mass is the amount of resistance to change the amount of inertia you have. And the amount of resistance to Alfa is not M. But it is I right. And 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. Um, but if you have Torque and Alfa instead, you have what's called rotational dynamics. That's just the name. It doesn't really matter. But in case you here, these words, you know what's up. All right, so that's the difference. Eso basically you might remember doing a bunch of efficacy may not. You're gonna do a bunch of torque, because Alfa and in fact, in some cases you're gonna two of them combined. So let's do an example here. See how this stuff works. Alright, so here it says, I have a solid disc. Solid discs, Remember, Um, this is a shape of the disk, so I can stop and right at the moment of inertia is half M R square. Um, solid disc of mass 100 radio, too. So I already get those numbers. M equals 100 r equals two is free to rotate, so it can spin around this. A fixed access so 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. Um, that means that if you have the face of the disc, the access points this way, which just means the disks going to spin around itself. Um, and it's friction list. Okay, You push tangentially on the disk. If you push tangentially in the disk. It looks like this, right, Like, sort of at the edge of the disc with the constant force of 50. So let me write this here f equals 50 Newton's, um we want to derive an expression for the angular acceleration that this experiences So part A. We wanna find Alfa. That's angular acceleration. And for part, we want to derive an expression. And for part B, we want to then calculate that so find an expression and then calculate just means plugging all the numbers. Okay, so how do we do this? Well, I'm giving you a force, and I'm asking for an Alfa back in the day. If I give you a force and ask you for a you would use that because I'm a But here, I'm giving you a force, but asking for an Alfa. So instead you're going to use some of all torques equals Scialfa. And that's because you're looking for Alfa. Okay, Now, the only force that causes this a torque that produces the torque on this disk is this force here. So the only torque we have is going to be the torque of F. Now that torque that forces producing a torque that's trying to spin this thing this way, which is a clockwise torque. So it's negative, So I'm gonna put a little negative in front. Okay? The moment of inertia is half m r square. I'm gonna go ahead and write this here and I'm gonna leave Alfa Loan because that's what we're looking for. So now I have to expand torque of F, which is why it is important to know how to calculate a bunch of different torques. So the torque of any force f is f r sine of data where remember R is the distance from the axis and data is the angle between f and R. Here are my our vector looks like this, and the length of the our vector is the entire radius of the wheel. And that's because you're pushing all the way at the edge of the wheel. All right, so this is going to be the force you're applying little r is the radius and then sign of data. The angle between the force theta is the angle between the force and the our vector. The angle between those two is this right here, which is 90. So I get sign of 90. That's what you get on the left side. Let's rewrite the right side again here and you get this. This becomes sign of 90 becomes one. This art cancels with one of the two hours on the other side and we're ready to go. I'm gonna move. I'm solving for Alfa. I'm gonna move everything to the other side. So Alpha's by itself. So we get negative. Negative to f divided by M r. That's going to be our off. Okay, I'm getting noticed. I'm getting negative acceleration to make sense. It's going this way, so the acceleration should be negative. So this is part A for part B. All we're doing is plug in the numbers. So that's easy. Negative two f thief forces 50. The mass is 100 and the radius is two. So this is going to be negative. 05 radiance per second squared. Cool. So that's it? That's it for this one. Hopefully make sense. Let me know if you have any questions. Let's keep going

2

Problem

ProblemSuppose 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/s^{2} . 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?

A

250 kg•m

^{2}B

285 kg•m

^{2}C

303 kg•m

^{2}D

367 kg•m

^{2}3

concept

## Torque & Acceleration of a Point Mass

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Hey, guys. So in this video, I want to talk about a torque producing an acceleration on a point mass. So 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 have a volume. Eso Let's check out how torques and acceleration works for point masses so real quick. I want to point out that most heart problems are actually going to involve, uh, shapes or rigid bodies. Um, but I want to quickly just do one with point masses because it works just the same. Most of the time you're gonna be here, but this works just the same. Cool. So quick Example you spin a small rock. Small rock is on indication that this is going to be a point mass because they say it's small and I don't give you the shape of the rock. The masses 2 kg at the end. You do this at the end of a light string of length three. If you spin a rock or any object at the end of a string, um, the length of the string will be the distance to the center. Okay, so it will be your little are so little r equals length of the string, which is 3 m and the masses kg. Okay. We want to know what net torque is needed to give the rock and acceleration of four. So if you wanna have an acceleration of four, what is the torque you need? And I want to remind you that net torque is the same thing as some of all torques. Okay, so check this out. I am asking for some of all torques, and I'm giving you Alfa. So I hope that you immediately thought of using, uh, immediate thought of using some of all torques equals I Alfa. And this is what we're looking for. Okay, Now, to calculate this, we just have to figure out these two guys remember point masses have a moment of inertia. Theme, Moment of inertia of point masses is given by M R Square. Where are is a distance to the center. Okay, so we're gonna replace this with m R squared Alfa and we have all of these numbers. The masses to the R is the distance, which is going to be three. And we have to square that. And then Alfa is four. Okay? And if you multiply all of this, you get 72 Newton meters. Remember, Torque is has units of Newton meter. That said four part A Got that done. Um, for part B, it's asking us to find the tangential acceleration while it has that speed. Remember, the tangential acceleration 8 10 is related is related to Alfa. By this equation, 8 10 equals R Alfa. So all we gotta do is multiply. Um, R is the distance here, which is three, and Alfa is a four. So that means that at that point, when you have an Alfa of Four, your tangential acceleration is 12. This is an acceleration A. It's a linear acceleration. So it's going to be 12 m per second square. Cool. Thes two are somewhat unrelated. Um, this is new stuff. This is just plugging it back, um, into a different kind of acceleration. Some old stuff bringing that back, putting it all together. All right, so that's it for this one. Let me know if you have any questions and let's keep going

Additional resources for Torque & Acceleration (Rotational Dynamics)

PRACTICE PROBLEMS AND ACTIVITIES (9)

- A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady hori...
- A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light w...
- CP A 15.0-kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m...
- CP A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a p...
- A machine part has the shape of a solid uniform sphere of mass 225 g and diameter 3.00 cm. It is spinning abou...
- A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg•m^2 about a vertical axle through ...
- CP A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar ...
- The flywheel of an engine has moment of inertia 1.60 kg/m^2 about its rotation axis. What constant torque is r...
- A 12-cm-diameter, 600 g cylinder, initially at rest, rotates on an axle along its axis. A steady 0.50 N force ...