What is the maximum speed an object can spin without breaking the light string?

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Homework Help Overview

The problem involves determining the maximum speed at which an object can rotate while attached to a light string that can support a stationary load of 27.0 kg. The object has a mass of 2.81 kg and rotates in a circle of radius 0.809 m on a frictionless table.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between centripetal force and the tension in the string, questioning how to calculate the force that the string can tolerate. There are attempts to substitute known values into the equations governing the system.

Discussion Status

Some participants have provided calculations for the force the string can support and have suggested solving for the speed using the derived equations. Others raise considerations about the physical properties of the string and its natural frequency, indicating that the maximum speed may be affected by these factors.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the string's properties, including its mass and stiffness, which may influence the maximum speed before breaking. The gravitational constant is also mentioned as a necessary component for calculating the force on the string.

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Homework Statement



A light string can support a stationary hanging load of 27.0 kg before breaking. An object of mass m = 2.81 kg attached to the string rotates on a frictionless, horizontal table in a circle of radius r = 0.809 m, and the other end of the string is held fixed as in the figure below. What range of speeds can the object have before the string breaks?

Homework Equations



ƩF=ma
ac=v2/r

The Attempt at a Solution



The light string can suppport a stationary hanging load of 27.0 kg before breaking.
Object's mass = 2.81 kg
r = 0.809 m

What range of speeds can the object have before the string breaks?

Substitute the formula for centripetal acceleration into Newton's second law and you get:

ƩF = m v2/r

Then you plug in the known values into the equation, and you get:

ƩF = (2.81)v2/(0.809). How would you calculate force in this case?
 
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If the string can suppport a stationary hanging load of 27.0 kg before breaking, then what is the force the string can tolerate before breaking? Might involve the gravitational constant 'g'.
 
Dick said:
If the string can suppport a stationary hanging load of 27.0 kg before breaking, then what is the force the string can tolerate before breaking? Might involve the gravitational constant 'g'.

ƩF=mg
ƩF=(27.0 kg)(9.8 m/s2)
ƩF=264.6 Newtons.

264.6 = (2.81)v^2/(0.809), then solve for V?
 
AryRezvani said:
ƩF=mg
ƩF=(27.0 kg)(9.8 m/s2)
ƩF=264.6 Newtons.

264.6 = (2.81)v^2/(0.809), then solve for V?

Pretty much, yes!
 
Dick said:
Pretty much, yes!

Thanks, brotha. :-p
 
The light string still has some mass and stiffness, and it has a mass hanging from it. All this means that it has a natural frequency. If the mass spins at that frequency long enough to approach equilibrium, the the string will break at a much lower speed than you calculated here. But if the system quickly passes thru that frequency, and all modes of it, then there is no limit to how fast the mass could spin as far as the string is concerned.
 

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