Practice: A flywheel is a rotating disc used to store energy. What is the maximum energy you can store on a flywheel built as a solid disc with mass 8 x 10^{4} kg and diameter 5.0 m, if it can spin at a max of 120 RPM?

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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: Intro to Rotational Kinetic Energy

**Transcript**

Hey guys! In this video, we're going to talk about rotational kinetic energy, which is the energy associated with the motion of spinning. Let's check it out. If you remember, if you had linear speed which is V, you had kinetic energy. There's going to be two types of kinetic energy so we're going to specify that this is linear kinetic energy. You're used to this equation KL = _ mv^2. I put a little L there to indicate this is the linear type of kinetic energy that's because now, we have a new one which is if you have rotational speed, instead of V, it's w or Omega. You have rotational kinetic energy and instead of KL, we call it KR. Now, the equation is very similar, it's half. Instead of using m, we're going to use the rotational equivalent of m which is I, moment of inertia and instead of V, we're going to use the rotational equivalent of V, which is w (omega) so I get this. So if you remember the first equation, it should be easy to remember the second one. On a special case, there's a special situation when you're moving and rotating so you have a V and a w. This is called rolling motion. One example of this is if you have a toilet paper roll that is sort of moving this way while rolling around itself. It's a toilet paper that's rolling on the floor. It has both kinds of motion, therefore it has both kinds of kinetic energy. IÕm going to say that the K total is KL plus KR. The last thing I wanna remind you, weÕll do a quick example is that for point masses, point masses are tiny objects that don't have a shape that have negligible size and radius. They have no volume. The moment of inertia, I, is mr^2 (squared) where r is a distance between the object and the axis. Remember also that if you have shape or a rigid body, an object with non-negligible radius and volume, we're going to get the moment of inertia from a table lookup. For example if you have a solid cylinder or a solid disk, same thing, the equation for that is _ mr2 (squared). Point mass is always this and some sort of shape will have a different equation each time. Let's do a very quick example here. I have a basketball player that spins a basketball around itself on top of his finger. I'm going to try to draw this, it's going to come out terrible. Here's a basketball player. Here's his finger, exaggerating some stuff and here's a basketball and he's rotating the basketball around itself so it looks kind of like this. The basketball is spinning around itself on top of your finger. It says here the ball has a mass of 0.62, a diameter of 24 centimeters so point 0.24 meters and it spins at 15 radians per second. Radians per second is angular velocity omega, 15. We want to know the ballÕs linear rotational and total kinetic energy. In other words, we want to know what is KL, what is KR and what is KTotal. First things first, you may already have caught this in physics. We never use diameter we always use radius, so when you see the diameter you immediately convert it to radius. Radius is half so it's 0.12. We're going to plug into the equation here. Kinetic energy is _ mv^2 (squared) and this ball has no kinetic energy, no linear kinetic energy I should say and that's because it spins in place. It's rotating but it's not actually moving. It has rotational motion but it doesn't have linear motion. It doesn't have translational motion. It just stays in place spinning around itself. We're going to say that it has no linear kinetic energy. It does have rotational kinetic energy because it's spinning around itself and that's given by _ Iw^2 (half I omega squared). A basketball has moments of inertia, the moment of inertia of a hollow sphere. I didn't give you the equation for that. I didn't explicitly say it was a hollow sphere but you should know that a basketball is a shell and then there's air inside, so it is a hollow sphere. I for a hollow sphere, you would look it up or it would be given to you, is 2/3 mr^2 (squared). What I'm going to do is I'm going to plug that in here, 2/3 mr^2 (squared) and then w2 (omega squared) which I have. Now we can just plug in numbers. The two cancels with the two and IÕm left with one-third. The mass is 0.62, the radius is 0.12^2 (squared) and w (omega), we have it right here 15, so 15^2 (squared). If you multiply all of this, I got it here. You get a 0.67 Joules. That's it. For the last part, we want to do the total kinetic energy. Remember the total kinetic energy is just an addition of the two types, kinetic linear plus kinetic rotational. There is no kinetic linear so the total kinetic energy is just the 0.67 that's coming from the rotational kinetic energy. That's how this stuff works. Hopefully this made sense. Let me know if have any questions and let's keep going.

Practice: A flywheel is a rotating disc used to store energy. What is the maximum energy you can store on a flywheel built as a solid disc with mass 8 x 10^{4} kg and diameter 5.0 m, if it can spin at a max of 120 RPM?

Example #1: Mass of re-designed flywheel

**Transcript**

Hey guys! Here we have a rotational kinetic energy problem of the proportional reasoning type and what that means, it's one of those questions where I ask you how changing one variable affects another variable. It's one of those. Let's check out. IÕm going to show you what I think is the easiest way to solve these.

It says you're tasked with redesigning a solid disk flywheel and you want to decrease the radius by half. First things first, solid disk means that the moment of inertia is 1/2 mr^2. That's the equation for a solid disk or solid cylinder. You want the new radius, IÕm going to call this R2 to be _ R1. I want to know by how much mass must the new flywheel have. It's the new mass relative to the original mass so that you can store the same amount of energy. You want the amount of energy that you stored to be the same. The amount of energy you stored is given by KR, that's energy stored which is given by _ Iw^2. This is energy stored as rotational kinetic energy in a flywheel. You want this number not to change. Yu want this number to be constants. How do you do this? If R changes, then I is going to change and if I changes, K is going to change and that's bad news. How do we change something else so that the K doesn't change? For the K not to change, you have to make sure that the I doesn't change. For the I not to change, you have to cancel out changing R with changing M. What I'm going to do here is I'm going to expand this equation, _, I is _ MR^2 w^2. Now I see all the variables that affect my K and again the K has to remain constant. If my radius is becoming half as large, it means that it is decreasing by a factor of 2. But the R is squared which means that when I reduce R by a factor of two, I also have to square this and R is becoming half as large but then the whole thing R^2 is becoming four times smaller. What that means is that if you want to keep everything constant, my mass has to grow by a factor of 4x. My new mass has to be four times my old mass and that's the answer. Again, R decreases by the factor of two but then you have to square because there's a square here. You get a four. If one variable decreases by four, the other one has to increase by four. Notice there are no squares in the M, so it's just a four not a two. Nothing crazy like that. That's it for this one let me know you have any questions.

Practice: When solid sphere 4 m in diameter spins around its central axis at 120 RPM, it has 10,000 J in kinetic energy. Calculate the sphere’s mass.

0 of 4 completed

Concept #1: Intro to Rotational Kinetic Energy

Practice #1: Max storing energy in flywheel

Example #1: Mass of re-designed flywheel

Practice #2: Mass of spinning sphere

A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping.
What is the ratio between the rotational kinetic energy about the center of the sphere and the sphere’s total kinetic energy?
A. 3/7
B. None of these
C. 7/2
D. 2/5
E. 2/7
F. 3/5
G. 5/3

A hollow spherical shell has mass 8.20 kg and radius 0.225 m . It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.895 rad/s2 . What is the kinetic energy of the shell after it has turned through 6.25 rev ?

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 31.0 cm . Starting from rest at t = 0, the flywheel rotates with constant angular acceleration 3.00 rad/s2 about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of 8600 kg/m3, what thickness must it have to store 800 J of kinetic energy at t = 8.00 s?

A merry-go-round has a mass of 1550 kg and a radius of 7.60 m. How much net work is required to accelerate it from rest to a rotation rate of 1.00 revolution per 9.00 s ? Assume it is a solid cylinder.

Suppose that some time in the future we decide to tap the moons rotational energy for use on earth. In additional to the astronomical data in Appendix F in the textbook, you may need to know that the moon spins on its axis once every 27.3 days. Assume that the moon is uniform throughout.a) How much total energy could we get from the moons rotation?b) The world presently uses about 4.0
1020 J of energy per year. If in the future the world uses five times as much energy yearly, for how many years would the moons rotation provide us energy?c) In light of your answer, does this seem like a cost-effective energy source in which to invest?

The flywheel of a gasoline engine is required to give up 600 J of kinetic energy while its angular velocity decreases from 780 rev/min to 510 rev/min. What moment of inertia is required?

A square, with a side length of 40 cm, has a 5 kg mass at each of its corners. If the square rotates about an axis through its center, perpendicular to the plane of the square, with an angular speed of 5 rad/s, how much kinetic energy does the square have?

If we multiply all design dimensions of an object by a scaling factor f, its volume and mass will be multiplied by f3.a) By what factor will its moment of inertia be multiplied?b) If a (1/48)large{{frac{1}{48}}}-scale model has a rotational kinetic energy of 2.5 J, what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

Biomedical measurements show that the arms and hands together typically make up 13.0 % of a persons mass, while the legs and feet together account for 37.0 % . For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. Let us consider a 76.0 kg person having arms 68.0 cm long and legs 93.0 cm long. The person is running at 12.0 km/h , with his arms and legs each swinging through 30° in 1/2 slarge{frac{1}{2};{
m s}}. Assume that the arms and legs are kept straight.a) What is the average angular velocity of his arms and legs?b) Calculate the amount of rotational kinetic energy in this persons arms and legs as he walks.c) What is the total kinetic energy due to both his forward motion and his rotation?d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

A grinding wheel in the shape of a solid disk is 0.200 m in diameter and has a mass of 3.0 kg. The wheel is rotating at 2200 rpm about an axis through its center.
a) What is its kinetic energy?
b) How far would it have to drop in free fall to acquire the same amount of kinetic energy?

The three 210 g masses in the figure are connected by massless, rigid rods.a) What is the triangle’s moment of
inertia about the axis through the
center?b) What is the triangle’s kinetic energy if it rotates about the axis at 5.2 rev/s ?

The L-shaped object shown in the figure below right consists of the masses connected by light rods. How much work must be done to accelerate the object from rest to an angular speed of 3.25 rad/s about the x-axis?
a) 62.3 J
b) 23.7 J
c) 17.6 J
d) 47.4 J
e) 34.2 J

Consider a particle of mass m that is revolving with angular speed ω around an axis. The perpendicular distance from the particle to the axis is r. (Figure 1) 1.) Find the kinetic energy K of the rotating particle. Express your answer in terms of m, r, and ω.2.) Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing through its center. Express your answer in terms of m and r.

A typical ten-pound car wheel has a moment of inertia of about 0.35kg *m2. The wheel rotates about the axle at a constant angular speed making 70.0 full revolutions in a time interval of 5.00 seconds. What is the rotational kinetic energy K of the rotating wheel? Express answer in Joules.

A softball pitcher throws an underhanded pitch with the arm fully extended (straight at the elbow). The ball leaves the pitcher's hand with a speed of 25 m/s. Assume that the pitching motion originates with the arm straight back and terminates with the arm straight down. Do not ignore gravity. Treat the pitcher's arm as a rod rotating about one end.The ball has a mass of 0.755 kg and the moment of inertia of the pitcher's arm is 0.72 kg · m2. Find the rotational kinetic energy of the pitcher's arm (including the ball) if the ball leaves the hand at a distance of 0.585 m from the pivot in the shoulder.

Calculate the moment of inertia and the rotational kinetic energy for the following objects spinning about a central axis (in units of Joules): a. a solid cylinder with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. b. a hoop with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec.

Calculate the moment of inertia and the rotational kinetic energy for the following objects spinning about a central axis (in units of Joules): a. a solid sphere with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. b. a thin spherical shell with a mass of 200 grams and a radius of cm rotating with an 5.0 angular speed of 2.5 rad/sec.

If a rotating skater triples her rate of rotation by decreasing her moment of inertia by 1/3, by what factor does her rotational kinetic energy change?

This problem illustrates the two contributions to the kinetic energy of an extended object: rotational kinetic energy and translational kinetic energy. You are to find the total kinetic energy of a dumbbell of mass when it is rotating with angular speed and its center of mass is moving translationally with speed . Denote the dumbbell's moment of inertia about its center of mass by . Note that if you approximate the spheres as point masses of mass each located a distance from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by , but this fact will not be necessary for this problem. Find the total kinetic energy of the dumbbell. Express your answer in terms of m, v, , and .

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