Spring Extension in Circular Motion

[dayum_samantha]

Hi so the question is as follows and I am not really good at physics, but I really want to improve and get better, so please DO NOT give me the answer, just push me in the right direction, thanks

 

{ The natural length of a spring is 10cm and its force constant is 5N/m. Now one side of the spring is fixed and the other side is connected to a ball, which has a mass of 6g. If the ball is moving in a horizontal circle with a uniform speed 10m/s, what is the total length of the spring when it is extended?}

 

 

I tried using Hooke's Law Force, the Centripetal Force equation and the Force equation (f=ma) but they don't seem to get me anywhere? Is there anything I am missing out on, or am i doing something wrong?

 

Please reply with useful advice.

 

 

[swansont]

Can you show how you tried applying these equations? Because using them will be how to solve the problem.

[studiot]

swansont is correct, but perhaps to help you produce something closer to an appropriate analysis, here is a hint.

 

Beginners often think that centripetal force is due to the circular motion and is somehow additional to the other forces that are acting.

 

This is not so, it is the other way round,

 

The circular motion is due to some force, in this case the one in the spring, and we call it the centripetal force.

But this is the same force that is described by Hooke's Law, not an additional or different one.

 

Does this help?

[swansont]

Also, the distance used in Hooke's law is not the same one used for the centripetal force.

[Sriman Dutta]

As far as I can think - we first need to calculate outward centrifugal force F and use Hooke's Law- F=-kx

[swansont]

As far as I can think - we first need to calculate outward centrifugal force F and use Hooke's Law- F=-kx

There is no outward force. As F =-kx implies, the force would be inward.