Need Help - Regarding the mixture of circular motion and hooke's law

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Homework Help Overview

The problem involves a mass attached to a spring on a frictionless surface, exploring the interaction between circular motion and Hooke's Law. The specific question is about determining the stretch of the spring when the mass moves in a circular path at a given rate.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the mass, including the spring force and gravitational force, and question the assumptions about the surface's orientation. There is an exploration of the equations relating circular motion to spring extension.

Discussion Status

The discussion includes attempts to clarify the setup of the problem, particularly regarding the horizontal nature of the surface. Some participants provide equations and reasoning related to the forces involved, while others seek confirmation of the assumptions made.

Contextual Notes

Participants note that the problem does not specify the angle of inclination of the surface, leading to assumptions about it being horizontal. There is also mention of the need to consider acceleration in the context of circular motion.

Lolagoeslala
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Homework Statement



A 1.01 kg mass is attached to a spring of force constant 9.5 N/cm and placed on a frictionless surface. By how much will the spring stretch if the mass moves along a circular path of radius 0.485 m at a rate of 2.14 revolutions per second?

What i was thinking was using the following equation:
Fc= Kx - mg
Which would turn into
a=kx-g
(4pi^2(R)(F)^2 +g)/K=x
 
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Lolagoeslala said:

Homework Statement



A 1.01 kg mass is attached to a spring of force constant 9.5 N/cm and placed on a frictionless surface. By how much will the spring stretch if the mass moves along a circular path of radius 0.485 m at a rate of 2.14 revolutions per second?

What i was thinking was using the following equation:
Fc= Kx - mg
Which would turn into
a=kx-g
(4pi^2(R)(F)^2 +g)/K=x

Is it a horizontal surface the mass moves on?

ehild
 
ehild said:
Is it a horizontal surface the mass moves on?

ehild

It does not state...
 
I guess the surfaceis horizontal, the angle of inclination would be stated otherwise. So the only force in the plane of motion is the force of the spring. Gravity cancels with the normal force of the surface.

ehild
 
ehild said:
I guess the surfaceis horizontal, the angle of inclination would be stated otherwise. So the only force in the plane of motion is the force of the spring. Gravity cancels with the normal force of the surface.

ehild

Would u like to show me that using the equation...
Kx=Mg somewhere you also need to include the acceleration i guess
 
Mg is a vertical force. The object moves horizontally, along a circle. What force is needed to make it move along that circle of radius 0.485 m with 2.14 revolutions per second?

ehild
 
ehild said:
Mg is a vertical force. The object moves horizontally, along a circle. What force is needed to make it move along that circle of radius 0.485 m with 2.14 revolutions per second?

ehild

m= 1.01 kg
K= 9.5 N/cm = 950 N/m
r= 0.485 m
f= 2.14 Hz

Kx= ma
kx= m x 4pi^2 x r x f^2
950 N/mx = 88.472 N
x= 0.0931 m
 
Your solution is correct.

ehild
 

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