An experimental flywheel, used to store energy and replace...

An experimental flywheel, used to store energy and replace an automobile engine, is a solid disk of mass $200.0 \mathrm{kg}$ and radius $0.40 \mathrm{m} .$ (a) What is its rotational inertia? (b) When driving at $22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph}),$ the fully energized flywheel is rotating at an angular speed of $3160 \mathrm{rad} / \mathrm{s} .$ What is the initial rotational kinetic energy of the flywheel? (c) If the total mass of the car is $1000.0 \mathrm{kg},$ find the ratio of the initial rotational kinetic energy of the flywheel to the translational kinetic energy of the car. (d) If the force of air resistance on the car is $670.0 \mathrm{N},$ how far can the car travel at a speed of $22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph})$ with the initial stored energy? Ignore losses of mechanical energy due to means other than air resistance.

Key Concepts

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion about an axis. It depends on both the mass of the object and how that mass is distributed relative to the axis of rotation. For various shapes, such as a solid disk, standard formulas exist that relate its mass and radius to its moment of inertia.

Rotational Kinetic Energy

Rotational kinetic energy represents the energy stored in a rotating object due to its motion. It is calculated using the formula (1/2)I?², where I is the moment of inertia and ? is the angular speed. This concept is crucial when analyzing systems where energy is stored in rotational motion, as it accounts for the energy that can be released when the object decelerates.

Translational Kinetic Energy

Translational kinetic energy is the energy that an object possesses due to its movement through space and is given by the formula (1/2)mv², where m is the mass and v is the velocity. This concept is fundamental in comparing different forms of energy in physical systems, especially when analyzing the relationship between rotational energy storage and linear motion.

Work and Energy Conversion

Work and energy conversion involve the relationship between energy stored in a system and the work that can be done by or against forces, such as friction or air resistance. By equating stored energy to the work done over a certain distance (using the relation work = force × distance), one can determine outcomes like how far a system can move given its energy reserves. This is essential in connecting energy storage to practical applications like vehicular range.

Transcript

00:01 Okay, so in this problem, we're dealing with a flywheel in a car, and we're supposed to figure out a whole list of different quantities about it.

00:11 So, you know, there's four parts of this problem, so there's a bunch of different things we're supposed to find.

00:16 So we're given, let's start with what we're given.

00:18 We are told that the mass of the flywheel is equal to 200 .0 kilograms, and we're told that this radius is 0 .40.

00:31 Meters.

00:33 And in the first part, part a, we are asked to find what the moment of inertia of the wheel is.

00:40 So since we're assuming that it's, you know, a solid disk or solid cylinder, we can use the formula that i is equal to one half times m r squared, which is exactly what we're given.

00:55 So we just go ahead and plug it in.

00:57 I is equal to one half times a two, two, 200 .0 times 0 .40 times, or squared.

01:10 And if you plug that into your calculator, you will get that the moment of inertia of the wheel is 16 kilograms meters squared.

01:21 And it's important to note that since the radius is just 0 .40, we only have two significant figures throughout this whole problem.

01:30 That's going to carry on throughout the rest of it because as you'll see.

01:36 So we only have two significant figures here and throughout the rest of the problem.

01:39 We're only going to have two significant figures.

01:42 So moving on to part b, we are told that when it is at when it's fully energized, the flywheel has an omega of 3 ,160 radians per second.

02:00 And with that information and what we found from the previous problem we are supposed to figure out what the kinetic energy the flywheel has when it's fully energized so we just take the formula that rotational kinetic energy k -e -rote as i like to just write down is equal to one -half times i times omega squared and again we you know we have all of that information available to us now.

02:34 So we can just say that k .e.

02:37 Rotational is equal to one -half times 16 times 3 ,160 squared.

02:50 And that is equal to, didn't write, here we go, 8 .0 times 10 to the 7 joules...

Please give Ace some feedback