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$\bullet$ A flywheel with a radius of 0.300 $\mathrm{m}$ starts from rest and accelerates with a constant angular acceleration of 0.600 $\mathrm{rad} / \mathrm{s}^{2} .$ Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration ofa point on its rim (a) at the start, (b) after it has turned through $60.0^{\circ},$ and $(\mathrm{c})$ after it has turned through $120.0^{\circ} .$

a) $=0.18 \mathrm{m} / \mathrm{s}^{2}$b) $=0.418 \mathrm{m} / \mathrm{s}^{2}$c) $=0.775 \mathrm{m} / \mathrm{s}^{2}.$

Physics 101 Mechanics

Chapter 9

Rotational Motion

Physics Basics

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Hope College

University of Winnipeg

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.

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.

10:25

II A flywheel with a radiu…

05:59

A flywheel with a radius o…

06:32

04:44

03:05

A flywheel with radius $0.…

01:41

A A flywheel having consta…

01:39

The flywheel has a diamete…

01:54

$\cdot \mathrm{A}$ wheel r…

02:44

A gyroscope flywheel of ra…

03:44

02:06

07:37

A flywheel with a diameter…

03:29

A flywheel has angular acc…

07:31

03:21

A wheel of diameter 40.0 c…

let's begin this problem by stating the various forms of the accelerations that we're looking for. So the radio acceleration has a form our omega squared, and so this is non constant, since omega is going to be changing throughout this process. So it's the one constant. The tangential acceleration is you two are alfa. Instance are an Alfa rough, constant process. This will be constant and then the results in acceleration. I'm just called by a and it's equal to the square root of a tangential square, plus a radio squared. This is true because a tangential and a radio go on perpendicular directions, so this could be a tangential and this is a radio. And since they go and perpendicular directions, the resultant acceleration, like Pythagorean theorem, is the square root of the sum of the squares since the perpendicular. And so now that we have the forms for the different accelerations now we just applied them in the three situations that we have, So her eh Omega's equals zero, since we haven't really started yet and this implies that the radio acceleration zero, the tangential acceleration is not zero because the angular acceleration is not zero in fact, it's equal to the radius 0.3 times angle acceleration 016 and so is equal to 0.180 meters per second. Scoring. This will be true when data zero when they did 61 days at 1 20 Throughout this whole process, this will be the tangential acceleration because again it's constant now since the radio acceleration zero the results of exploration is just the tangential acceleration. And so this is also equal to 0.180 This will not be constant, though, because the radio acceleration will not be constant. And this depends on. But this is the final answer party and part B. We rotated 60 degrees, which is the same saying we rotated pie over three radiance. Now we're going to use the Kinnah Matic equation. We'll make a squared is equal to omega, not square plus two Alfa data mined Stay tonight and when we want to do is solve for the final omega here. No, maybe not. Zero. We know what alphas and we know what itemize they did not is It's this right here. So playing this in gives us an omega and we take that omega. We plug it in for the radio acceleration, which is equal to our omega squared. And so when we do that, we get the radio. Acceleration is equal to 0.3. She's the Radius times too, which is us to hear times acceleration, which is 0.6 times the distance. Travel was just power three. This is equal to 0.377 meters per second squared. Yeah. Now get in. The tangential acceleration is the same as it was last time. So 0.180 meters per second squared. And that means the resultant acceleration is a squared of 0.180 squared plus 0.377 squared. And then spirit of all that. And this is, uh, being zero plane for 18 meters for six. Word in that completes part B. Park SI is asking for 120 degrees rotation, which is equal to two pie or three radiance. Now, we're going to use the same exact method as we used last time. And so I'm not gonna repeat the whole process. I'll just simply quote the answer. The radio acceleration ends up being to your 0.754 meters per cents worth the tangential explorations again 0.180 squared and so the results and acceleration is 0.180 squared plus 0.754 squared and this gives 0.775 meters per second squared.

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