The massless rod in a simple pendulum is replaced by a massless spring such that you have a spring-mass and a simple pendulum system at the same time. You pulled the mass at some distance from its equilibrium position and displaced the entire spring-mass system at some small angle with respect to the vertical. If the spring has a relaxed length of 1 m and the attached mass is 10 kg, what must be the spring constant in order for the object to move in a circular path?

Question

anonymous · Answer

hi morningrush,
  first of all you think that what is the value of spring constant for which massless spring behaves as a massless rod.
        here relaxed length of spring is 1m.for circular motion this relaxed length of spring must be constant i.e it will be radius of the circle.Imagine a situation at the lowest point of the circle.The total vertical force must be zero i.e kl=mg=>k=mg/l.
where l is relaxed length(radius of the circle),k is spring constant,and m is the mass of body.....if any query ,talk to me...

anonymous · Answer

first of all the spring does not oscillate or go in a circular path when you put only the weight mg
so there should be a force that you give ( this shud have  a vertical component which will provide a velocity of minimum root of 5gl if u want it to go thru a circular path)
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