Curvilinear motion problem

  • #1
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Homework Statement


A weight is suspended from a spring 50 cm long and stretches it by 1 cm. Take the other end of the spring in your hand and rotate the weight in a horizontal plane so that the spring is stretched by 10 cm. What is the velocity of the weight? (The force with which the stretched spring acts is proportional to the amount of its tension. Disregard the force of gravity when the weight is rotating.)​

Homework Equations



Fs=Fg=Fc
kx = mg = mv2/r

Where Fs is the spring force, Fg is the weight of the body, Fc is the centripetal force, k is the spring constant and x is the elongation of the spring.

The Attempt at a Solution


mv2/r = kx = mg
v=√(gr)=√(g*60cm)=2.42 m/s.

But the textbook says the right answer is 7.7 m/s. My problem is I can't determine k, the spring constant, and what to do next. Also, the problem says to ignore the force of gravity. Doesn't that mean that we assume the mass has 0 weight?[/SUP]
 
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Answers and Replies

  • #2
gneill
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My problem is I can't determine k, the spring constant. Also, the problem says to ignore the force of gravity. Doesn't that mean that we assume the weight has 0 mass?
Work symbolically until the end. An expression for k can be had from the first scenario with the weight hanging. No, ignoring gravity doesn't make the mass zero. Imagine you're doing the experiment in space.
 
  • #3
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Work symbolically until the end. An expression for k can be had from the first scenario with the weight hanging. No, ignoring gravity doesn't make the mass zero. Imagine you're doing the experiment in space.
My bad, I meant to say the weight would become zero.
 
  • #4
PKM
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My problem is I can't determine k, the spring constant, and what to do next.
Why not? Find ##k## in terms of ##m##, and proceed by substitution.
Also, the problem says to ignore the force of gravity. Doesn't that mean that we assume the mass has 0 weight?
It is asked to ignore gravity while rotating the mass in horizontal plane. Indeed, this idealization makes the calculations a lot easier. You don't need to consider any other forces other than the radial one.
 
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  • #5
collinsmark
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To elaborate on what @gneill said, don't worry about calculating the numeric value for the spring constant k. But you can use the first scenario (the hanging scenario) to find k in terms of the mass m and g.

In the spinning scenario, imagine that the spring/mass system is rotating on a horizontal, frictionless plane (like an ideal sheet of ice). (Or like gneill suggested, in space.) That way, gravity doesn't effect the force on the spring.
 
  • #6
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I have figured out how to solve the problem. Thanks to those who have helped me.

I am answering this thread so other viewers can learn:

When the body is hanging by a spring from an unyielding body, such as a ceiling, the spring is elongated by x1. The force acting on the spring equals the weight of the body, mg. Therefore the spring constant k equals their ratio, mg/x1.

When the body is swung by the same spring in a horizontal circle, the spring is elongated by x2. The spring force in this case is kx2 = (mg/x1)x2. This quantity is the centripetal force, so we may write:

Fc = mv2/r = (mg/x1)x2

Let L be the initial length of the spring. So the radius of the revolution, r, equals L + x2.

Solving for v, we get:

v = √[g(L+x2)(x2/x1)]

Convert lengths to meters and substitute known quantities g, L, x1 and x2. v equals approx. 7.7 m/s
 
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