Ch 11: Rotational Inertia & EnergySee all chapters

Sections | |||
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Intro to Moment of Inertia | 30 mins | 0 completed | Learn |

Moment of Inertia via Integration | 19 mins | 0 completed | Learn |

More Conservation of Energy Problems | 55 mins | 0 completed | Learn |

Moment of Inertia of Systems | 23 mins | 0 completed | Learn |

Conservation of Energy in Rolling Motion | 45 mins | 0 completed | Learn |

Moment of Inertia & Mass Distribution | 10 mins | 0 completed | Learn |

Intro to Rotational Kinetic Energy | 17 mins | 0 completed | Learn |

Energy of Rolling Motion | 18 mins | 0 completed | Learn |

Types of Motion & Energy | 24 mins | 0 completed | Learn |

Parallel Axis Theorem | 14 mins | 0 completed | Learn |

Conservation of Energy with Rotation | 36 mins | 0 completed | Learn |

Torque with Kinematic Equations | 59 mins | 0 completed | Learn |

Rotational Dynamics with Two Motions | 51 mins | 0 completed | Learn |

Rotational Dynamics of Rolling Motion | 27 mins | 0 completed | Learn |

Concept #1: Rotational Dynamics of Rolling Motion

**Transcript**

Hey guy! In this video, we're going to talk about the rotational dynamics of rolling motion. Rotational dynamics just means you have a problem of torque and acceleration. Rolling motion just means you have an object that rolls around itself while moving sideways, while the axis of rotation moves, while the object rotates; kind of like a toilet paper that instead of being fixed on the wall, you just roll it on the floor so it does this. That's what rolling motion is. Let's check it out. In some problems, it says, a disc-like object accelerates around a free axis. Again, free axis is the toilet paper that instead of being fixed in the wall it's free to move so it rolls while moving. It says here, in these, the objects will have obviously both rotational and linear motion. For example, in this case this guy is falling down here. It's going to have if you release it from rest, it's going to have an acceleration this way and it's going to have an _ this way. We're going to use both sum of all forces equals ma because it has an a, and some of all torques equals I _ because it has an _. We're going to do this for the same object. That same object has two motions so we can write two equations. Remember, the direction of positive in both will follow the direction of a and _. This is going to be positive and this is going to be positive. Remember also that in rolling motion, rolling motion is the situation we're talking about, there's an extra equation which is that the velocity at the center of mass which is this, Vcm, is tied to your _ by this equation: Vcm = R _. A lot of times you see an equation similar to this. It's V = little r _. This one is V = big R _ because it always has to do with the radius of this object. This is an extra equation that's possible because you have rolling motion. Another equation that's also possible is just like your V = big R _, you have a = big R _. In these cases, in this problem in rolling motion, it's always going to be big R. I want to remind you of some quick things. When you have a disc-like object that is rolling freely, the torque will come from static friction. Let's talk about that real quick. Which torque are we talking about? This guy has an _ and remember, to have an _ you have to have a torque. Acceleration comes from a net force and an _ comes from a net torque. A torque is what's causing an _. In this case, the force that's causing this torque, a torque comes from a force, and in this case the force is static friction. In static friction, if _ is this way, then torque has to be this way which means that friction has to act in a direction that causes this, which would be like this. I hope you can see that if you have a disc and you push in the disc like this, I'm doing basically like this. Imagine I'm doing this to the disc. It's going to cause the disc to do this. That's exactly what friction is doing. Friction goes this way and causes a torque of friction. The force of friction produces a torque of friction, which causes an acceleration. The disc not only falls this way but it also does this as itÕs falling. Two things to remember: If there is an acceleration, in other words if your _ is not zero, there has to be static friction. I like to remember this by using the following short-phrase. You need static friction to _. If you don't have static friction you don't have an _. What that means is that if you're not spinning, you're not going to start up. If you're initially not spinning, you have no way of beginning to spin and if you are already spinning, you have no way of stopping your spinning. You will keep spinning because there's not going to be any change. Another keyword is if without slipping, this means that there is no kinetic friction. Guess what? This is always going to be the case. Always. This is standard language, not slipping. You're not going to have kinetic friction in these problems. If you have acceleration, if your acceleration is not zero, then there has to be static friction because you need static friction to have acceleration. Let's do a problem. It says here when a solid cylinder of mass M and radius R is released from rest, it rolls down without slipping along an inclined plane. Let's draw that. I got an inclined plane like this, just like the picture up above and I released this from rest so the initial velocity is 0. This guy has mass M and radius R and it rolls down. Rolls down means it's going to have an acceleration this way. Without slipping is standard language but it does tell you that there is no kinetic friction for sure, along an inclined plane that makes an angle of _. We want to derive an expression for the angular acceleration of the disc. We want to know what is _. How do we do this? This is an acceleration problem so we would use F = ma, except that there's also rotation. We're going to use F = ma and Torque = I _, in both of them or for the same object because the object has two motions. One object with two motions means two equations. Sum of all forces equals ma and sum of all torques = I _. Let's look at the forces on this thing here. I'm going to draw it over here. I have an mg pulling you down. I have to decompose the mg into mgx + mgy this way. I have a normal like this and I have a friction that acts over here, friction static. Those are all the forces. In terms of torques, the only torque comes from this. Normal on a on a disc that spins on the surface never produces any torque and mg and the center of mass never produces any torque if you remember. The only one here is torque of static friction. Remember also that once you split mg into mgx + mgy, mg is essentially dead. You got rid of it and replaced with two other things. It's really no longer there. Remember also that mgy cancels out with normal. Really the only things you have is these are the two forces in the x axis which is down the plane and this is the only torque. The forces are mgx and friction. Which one is positive and which one is negative? This is the direction of positive so mgx is the positive one and friction is negative that equals ma. Torque, there's only one torque which is the torque of static friction. The moment of inertia I is going to be the moment of inertia of a solid cylinder because it says right there, _ MR^2. _ remember, we want to replace _ with a. If you have a and _, you want to get rid of _ and change it into a. That's so that instead of having a and _, you have a and a. To do this, we use the fact that a = R _, big R because this is rolling motion. I can rewrite _ as a/R. That's what I'm going to do. I'm going to put a/R over here and that's so that this _ becomes an a so I have a and a. Last step is thereÕs not much else to do here. I'll get back to this in a second. I'm going to expand this one more time and this is going to be torque of any force F is Frsin_. The force we're talking about here is static friction. r is the R vector. In this case, it's always going to be the radius as well because notice that friction always acts at a distance of the entire radius. If you draw an r vector here, r vector of course is from the axis of rotation to the point where the force happens. For friction, it's always going to be big R. The torque of friction is always friction big R and the angle is always 90 degrees. The torque due to friction is always F big R. Now that I did it on the side, I'm just going to plug it in here and then look what happens. All the RÕs will cancel which is neat. R cancels with this. This cancels with this and you're left with friction static = _ Ma. I want to warn you not to get too excited with the expanding of these equations and rewrite friction into _ normal. You could do this. You may remember friction is _ normal. You should remember friction is _ normal but you don't want to rewrite, you don't want to expand and that's because guess what, there's a friction here and there's a friction here. All friction is doing similar to what you would see with a tension if you had a cable pulling on a block, all its doing is connecting these two equations. Don't expand friction, just plug it in right there. Let's do that. mg if you remember, mg is mgsin_, that's mgx becomes mgsin_ Ð _ Ma = ma. Notice here that I wrote little m, little m and big M and that's because I usually write F = ma with a little m but I usually write I with a big M because a lot of the equations are like that. It's really the same thing because it's referring to the same object so I can cancel. If you have a single object, the mass will cancel. We're solving for a so I'm going to move this over here. I have gsin_ = a + 1//2 a. a + _ a is 1 + _, 1 _, 1.5 or 3/2 a. a is the 2 goes up here, 2gsin_ and the 3 goes down there. This is the final answer for a. We still got to get _ but I want to again make a point here, notice how this equation looks familiar. It should look familiar. If you have a block that is accelerating down and there's no friction, there's no kinetic friction, it's accelerating down because of mgx. The acceleration of this block is gsin_. You may remember this from back in the day, gsin_. Notice this equation is very similar except that instead of having like a 1 here hiding, you have some fraction in front of it. This fraction is always going to be less than 1. The point here is that in rotation questions that are the rotation equivalent of linear questions, like this is just a rotational equivalent of this. The equation for a here should look similar to this equation. That's how you know you are on the right track, provided of course that you vaguely remember this and you can say, ÒHey that looks familiar.Ó I like to make that point a lot because this way you can know that youÕre generally in the right direction. It gives a little bit more certainty that youÕre correct. Another thing that allows you to do is that if this number here is more than one for whatever reason, which it would be if you screw up your signs, then you know that you're wrong and you have to go back. They should always be less than 1. Let's wrap it up and find _. _ is a/R, so it's just going to be 2gsin_ / 3R. This is our final answer and we're done. That's it for this one. I hope this made sense. Make sure you know how to do this stuff. Let me know if you have any questions and let's keep going.

Practice: A hollow sphere 10 kg in mass and 2 m in radius rolls without slipping along a horizontal surface with 20 m/s. It then reaches an inclined plane that makes 37° with the horizontal, as shown. If it rolls up the incline without slipping, how long will it take to reach its maximum height? (Hint: You will need to first calculate its acceleration)

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Concept #1: Rotational Dynamics of Rolling Motion

Practice #1: Time for sphere to reach top of hill

A thin-walled hollow sphere with mass M = 2.00 kg and radius R = 0.05 00 m has a moment of inertia of I = 2/3 MR2 for rotation about an axis through its center. Initially the sphere is rolling without slipping on a level horizontal surface and its center of mass has a translational speed of vcm = 8.00 m/s. The sphere then rolls without slipping up a ramp that is inclined at 37° above the horizontal. What is the magnitude of the friction force that the ramp exerts on the sphere while the sphere is rolling up the ramp?

A solid uniform cylinder (I = 1/2 MR2) with mass 2.00 kg and radius 0.120 m is released from rest at the top of a ramp that is inclined at 36.9° above the horizontal. The length of the ramp is 1.80 m. What is the magnitude of the static friction force required for the cylinder to roll without slipping?

A spool is being unwound by a constant force F pulling on a string. The spool rolls without slipping. The spool has a radius r, mass m and moment of inertia I =2/3 mr2. If the spool rolls without slipping, what is the ratio of the magnitude of the applied force F to the magnitude of the friction force f, F/f?
1. 2/3
2. 1/5
3. 3/2
4. 1/7
5. 1
6. 1/3
7. 7
8. 5
9. 5/3
10. 3/5

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