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An object of mass = 5.00 is attached to a spring with spring constant = 495 whose unstretched length is = 0.150 , and whose far end is fixed to a shaft that is rotating with an angular speed of = 3.00 . Neglect gravity and assume that the mass also rotates with an angular speed of 3.00

2007-06-23 14:34:20 · 2 answers · asked by amp 1 in Science & Mathematics Physics

I found that if I set the 2 equations equal to each other, they cancel out the radius, which baffels me as to how that could then give me the correct answer, even though the hints I was given with the question basically said the same thing. Any additional help would be greatly appriciated.

2007-06-23 15:39:00 · update #1

2 answers

Let the displacement of the spring be x, then r = (x + 0.15) so x = r - 0.15

Force F = kx and also equals mrω^2

k (r - 0.15) = mrω^2

(r - 0.15) /r = mω^2/k

1 – 0.15 /r = mω^2/k

1 – mω^2/k = 0.15 /r

r = 0.15 / [1 – mω^2/k] = 0.165 m

2007-06-23 22:15:40 · answer #1 · answered by Pearlsawme 7 · 0 0

This was a little hard to picture (not worded well), but I think what it means is that the whole apparatus is rotating, with one end of the spring at the center of rotation, and the spring extending radially with the mass attached at the other end, swinging in a circle. Okay.

Since the mass is moving in a circle with constant speed, its acceleration is:
a = ω²r
where ω is 3.00 (radians per second, presumably); and we don't know how much r is (that's what we're trying to find out).

So the force acting on the mass must be:

F = ma = mω²r

But the force is supplied entirely by the tension in the spring. That means the force is also equal to this:

F = (amount of stretch)•(spring constant)

Where (amount of stretch) is how far the spring is stretched beyond its equilibrium (unstretched) length. That is:

(amount of stretch) = (current length of spring) – (unstretched length)

But we know that (current length of spring) is surely "r". So:

(amount of stretch) = r – (unstretched length)
= r – 0.150

Plugging that into the previous "F" equation:

F = (r – 0.150) • (spring constant)
= (r – 0.150) • 495

So, now we have two expressions for "F". Set them equal to each other, then solve for "r", and you're done.

*

2007-06-23 15:00:01 · answer #2 · answered by RickB 7 · 0 0

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