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

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The discussion focuses on determining the maximum speed an object can rotate without breaking a light string that supports a load of 27.0 kg. The object has a mass of 2.81 kg and rotates in a circle with a radius of 0.809 m. The force the string can tolerate before breaking is calculated to be 264.6 Newtons, derived from the equation ƩF=mg. It is noted that the string's mass and stiffness, along with the hanging load, introduce a natural frequency that could affect the breaking point. If the object spins quickly enough to bypass this frequency, the string may withstand higher speeds than initially calculated.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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