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$\bullet$ The kinetic energy of running. Using the previous prob-lem as a guide, apply it to a person running at $12 \mathrm{km} / \mathrm{h},$ withhis arms and legs each swinging through $\pm 30^{\circ}$ in $\frac{1}{2} \mathrm{s}$ . Asbefore, assume that the arms and legs are kept straight.

a) $=2.1 \mathrm{rad} / \mathrm{s}$b) $20 \mathrm{J} $c) $ 436 \mathrm{J}$d) 4.58$\%$

Physics 101 Mechanics

Chapter 9

Rotational Motion

Physics Basics

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

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.

02:21

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.

07:54

The Kinetic Energy of Runn…

02:12

Using the previous problem…

00:39

Calculate the kinetic ener…

the two runners have the s…

01:03

If a person of mass $M$ si…

A running man has half the…

02:47

03:00

Determine the total energy…

03:36

The kinetic energy possess…

02:33

Some Typical Kinetic Energ…

this problem should be done directly after the previous problem because they share a lot of the same strategies. So party is asking for thie averaging your velocity. And so, in order to get what we need, the angle gone through and the time it takes to go through that angle. And so the problem tells us it goes through six degrees and so the Final three radiance and does so in half a second. So 0.5 seconds. This is equal to 2.1 radiance per second. Assoc answer to party for part B. We need the moment, Hersholt, which is going to be the same moment inertia is used in the previous problems. And so I total is equal to nine 0.8 kilograms meter squared. See the previous problem if you want to know how to get this because it was Dr there now to find the rotational connect energy, we need 1/2. I only x squared. Well, this is I and here's omega. So plugging that in because a rotational connect energy of twin jewels, it's Los answer to part B. Before we do perceive, we need to convert the velocity which has given us 12 corners an hour and a meters per second. So when you do that just to simple conversions, we get 3.33 meters per second and this is what we'll actually get. And so the translational connect energy is equal to 1/2 don't be squared. So this is equal to 1/2 times the mass, which is 35 times 3.33 which is the velocity squared and I get 400 and 16. Jules, this means that the total kinetic energy rotational plus translational is he going for 16 plus 20 which is 436 Jules and party to find the ratio. All we do is we take the rotational connect energy and divided by the total connect energy. So you're 20 over for 36 which is equal to 0.46 or 4.6%. So 4.6% of the total Connecticut city is rotational, and that completes the problem

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