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$\bullet$ $\bullet$ A 15.0 kg mass fastened to the end of a steel wire with an unstretched length of 0.50 $\mathrm{m}$ is whirled in a vertical circle with angular velocity 2.00 $\mathrm{rev} / \mathrm{s}$ at the bottom of the circle. The cross-sectional area of the wire is 0.010 $\mathrm{cm}^{2} .$ Calculate theelongation of the wire when the mass is at the lowest point of the path. (See Table $11.1 . )$

$3.328 \times 10^{-3} \mathrm{m}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

Lectures

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in this problem. We're dealing with a steel wire, so we're going to need Young's module. ISS were that wire. So it's 2.0 times 10 to the 11 which is the Youngs modules for steel. We're also given the angular velocity and revolutions per second, so we need to convert that into radiance for second. I multiplied by two by we do that, we get 12.6 write ins per second and we're also told the initial length of the wire 0.5 meters. And the first step to prop this problem is the fear it with the radio acceleration ists while the radio acceleration Cammi right in terms of the angular velocity as radius times angular, velocity squared, multiplying the Senate is full 0.6 and adding in radius gets 79.4 meters per second squared. You know, the second step is to figure out what f purpose the force being applied on the bottom end of the wire. Not to do this, we need to actually draw forced diagram for the object. So if purpose pulling it up, gravity mg point down, there's acceleration upward of a rad and my corner system will be. Plus why upward and so Newton's second law in the white direction gives f perp minus mg is equal to mass times a rat. Teo. Now we know what a rat is. We know what the masses and we know GS so we can solve for f perp. When we do that, we get 1.34 times 10 to the third noon. So now we need to like this into the form for the Youngs modules. So we know the Youngs module a CE is Ellen on times per over cross sectional area time. Still tell solving this for Delta L Ezel non times after over a why. And now we're complying and everything we know in the right hand side. So Ellen, on a 0.5 f purpose 1.34 time stand to third. That's what we just found out earlier. The cross sectional area is 0.1 times 10 to the minus. Four is given in the problem, and Young's mom Angelus for steel is 2.0 times 2.0 10 to 11. And so all of this equals 3.35 times 10 to the minus 30 meters. And this is the amount of the steel wire stretches under this rotation

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