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A bicycle racer is going downhill at when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when itreaches the foot of the hill if it rolled without slipping all the way down? (b) How much total kinetic energy does the wheel have when it reaches the bottom of the hill?

29.25 $\mathrm{m} / \mathrm{s}$1925 $\mathrm{J}$

Physics 101 Mechanics

Chapter 9

Rotational Motion

Physics Basics

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

McMaster University

Lectures

04:16

In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

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In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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in this case, the moment of inertia of the wheel is a disk, So it's m a r squared. Where ILM is the mass that's 2.25 kilograms and are the radius is 0.45 years, so rolling without slipping, which is what the wheel is doing in this problem, rolling without slipping implies that there's a relationship between the angular velocity and the linear velocity. This right here. Now I know the initial linear velocity is 11 years for second. We're going to let the positive Y direction be upwards. We're gonna let why is equal to zero at the bottom of the hill. And this means that the initial high is 75 meters in the final height. His era Mears. So now we're ready to actually start problem. We're going to solve this by using conservation of energy. Show me the labels that conservation of energy implies that the initial kinetic energy, plus the initial potential energy is equal to the final connect energy, plus the final potential energy now any arbitrary, kinetic energy Not necessarily the initial final, but just in general, the kinetic energy is 1/2 empty square, which is the Lanier kinetic Energy plus the rotational connect purity, which is 1/2 Iomega squared. This means that this is equal to 1/2 only squared. And then now I'm going to play in what I is for a problem. And it's no more square. I stated it on this page right here and then Now I'm going to use the relationship between the linear and your velocities, which is right here. I'm going to solve for Omega T V over R, and I'm gonna plug this in for Omega. So here I will make a square. I'm going to be over r squared and I'll see this are square. The denominator cancels this r squared. So I get 1/2 A T squared, which is what this is, and so they add to give me a full m p squared. And any kinetic energy of this process is equal to this. And so that means the initial kinetic energy is equal to him. Initial velocity squared, these were both given from, so the initial kinetic energy is equal to 272 jewels. The final connect energy is equal to l'm V f squared and we don't know what the office. So we just have to even in this form, the initial potential energy is mass times, gravity times the initial position in the white direction. And then you plug in these values here they're all given to us, and I get 16 50 for Jules, and then the final potential energy is equal to mg. Why have, But why F zero? So this is your jewels. Supplying these values into the conservation of energy gives us that. Yes, that was the final velocity squared, which is the final. Kinetic energy is equal to 1926 Jules and then solving for final velocity. I get a value of 29.3 years per second like this. And that's the answer. Party Party is asking for the final kinetic energy which we saw was just equal to the mass times the final glossy square. Or we just found the final velocity so you can usually find us to be 1932 Jules. And that's the problem.

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