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A string is wrapped several times around the rim of a small hoop with a radius of 0.0800 $\mathrm{m}$ and a mass of 0.180 $\mathrm{kg}$ . If the free end of the string is held in place and the hoop is released from rest (see Figure 9.30), calculate the angular speed of the rotating hoop after it has descended 0.750 $\mathrm{m} .$

$[33.9 \mathrm{rad} / \mathrm{s}]$

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 Washington

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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A string is wrapped severa…

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in this problem, we're going to use the conservation of energy, which tells us that the initial kinetic energy plus the initial potential is equal to the final kinetic plus the final potential. Now, if we let the final position be defined as zero, then that tells us that the final potential is equal to zero. Jules. So why a fecal zero implies you f is equal to zero. This means that the initial height is equal to 0.75 meters. Now. That means this term was away since you f zero. But also the initial kinetic energy is zero since it starts arrest. And so the final relationship is that the initial potential is equal to the final Kinetic. The final kinetic is equal to that 1/2 don't be square, which is the linear, kinetic energy plus 1/2 Iomega square, which is the rotational connect energy. Since there's no slipping involved, we know V is equal to our omega. This will let us solve for B here in place of Omega. Also, we know the moment inertia. This object is equal to M R square. And so plugging these two things into this expression gives 1/2 um, our Omega squared plus 1/2 Hilmar sward Omega squared. So we have 1/2 film R squared. Omega's Word plus 1/2 film are swear Omega squared that he was a fool. R squared Omega's word. So it's the final connect energy. The initial potential energy is equal to mass times, gravity times the initial height, which is equal to 0.75 So tickle a mass G 0.75 Equating these two things, What's us? Cancel the mass and sulfur the Omega. Why do that? I get Omega is equal to the square root of G times the initial height, which I substitute in the last expression. But now I'm just going to keep it as why I over r squared. Now I can plug everything in at once and get a value of 33.9 radiance per second, and that's the final answer

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