An object of mass M = 4.00 kg is attached to a spring with spring constant k = 1100 N/mN/m whose unstretched length is L = 0.150 mm, and whose far end is fixed to a shaft that is rotating with an angular speed of ω (omega) = 5.00 radians/s . Neglect gravity and assume that the mass also rotates with an angular speed of 5.00 radians/s as shown. When solving this problem use an inertial coordinate system, as drawn here. Given the angular speed of ω (omega) = 5.00 radians/s , find the radius R (omega) at which the mass rotates without moving toward or away from the origin.

The radius at which the mass doesn't move towards or away from the origin is 165 mm. To get to the solution, all we have to find is the radius at which the centripetal force due to the circular motion is equal to the elastic force from the spring. Just use the known formulae for the elastic force and the centripetal force and substitute in the values. When you equate both of them, you can find the R value.

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The problem of finding the radius as a function of omega such that the mass rotates without moving toward or away from the origin can be translated to the problem of finding the correct radius. We need to find the radius at which...

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The problem of finding the radius as a function of omega such that the mass rotates without moving toward or away from the origin can be translated to the problem of finding the correct radius. We need to find the radius at which the centripetal force is equal to the spring force, such that there is no radial acceleration relative to the origin.

This means that the elastic force generated by the spring will be responsible for the circular motion at radius R!

So, all we have to do is use the fact that R must be such that `F_(cp) = F_s` , where `F_s` is the force caused by the spring and `F_(cp)` is the centripetal force due to the circular motion.

Now, the centripetal force `F_(cp)` is given by:

`F_(cp)=mw^2R`

`F_(cp) = (4 kg)(5)^2R`

`F_(cp) = 100R`

And the elastic force (when at a distance R from the shaft where the spring is attached to) is given by:

`F_s = k(R-L)`

`F_s = (1100 N/m)(R-0.00015 m)`

Note that we converted 0.15 millimeters to meters in the equation above.

Now setting `F_(cp) = F_s` give us:

`(1100 N/m)(R - 0.00015) = 100 R `
`1100R - 0.165 = 100 R`
`1000R = 0.165`
`R = 0.000165 m`

So, the answer is that the radius R must be 165 mm.

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