Lab experiment on moment of inertia dynamic

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experiment set up.png

Homework Statement


That is the set up for experiment, and I attached the data I found in the lab.
The lab given me the equation to find the moment of inertia of the cylinder

Homework Equations


Icyl = (2hgM_cylinder) * (1/(W_withcylinder)^2 - 1/(W_nocylinder)^2)
The height is 0.9m

The Attempt at a Solution


Me and my lab partner work separately and then we compare the Icyl value, in the attached file column J is his value, column K is my value and the two graph. I'm not sure which one is right, please explain to me what we did wrong?
And my graph (I cyl) doesn't look right too, please explain. Thank you very much
 

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Answers and Replies

  • #2
haruspex
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Your diagram shows two masses and a rod. Where does the cylinder come in?
What is the purpose of the lab? What is the procedure?
Are the distances in your diagram to the centres of the two masses? It looks like they're to the outside edges, but the mass widths are not given.
Icyl = (2hgM_cylinder) * (1/(W_withcylinder)^2 - 1/(W_nocylinder))
Comparing with the column K formula, you've left out a power of 2 in there.
The column J formula is different from K's in two ways:
- there's an extra constant factor 0.0797 in col K
- col J has a division instead of a multiplication: Icyl = (2hgM_cylinder) / (1/(W_withcylinder)^2 - 1/(W_nocylinder)^2)
 
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  • #3
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Your diagram shows two masses and a rod. Where does the cylinder come in?
What is the purpose of the lab? What is the procedure?
Are the distances in your diagram to the centres of the two masses? It looks like they're to the outside edges, but the mass sizes are not given.
I_Cyl is I_sys - I_rod.
I think by the moment of inertia of the cylinder they mean the moment of inertia of the two masses.
Can you please look at the excel sheet at column J that is my partner data, I don't know where he got that equation to calculate the I_cyl
 
  • #4
haruspex
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I_Cyl is I_sys - I_rod.
I think by the moment of inertia of the cylinder they mean the moment of inertia of the two masses.
Can you please look at the excel sheet at column J that is my partner data, I don't know where he got that equation to calculate the I_cyl
You might have posted that before I updated my post.
You'll need to explain all the variables in the equation.
Where did you get the equation from ?
 
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This equation is given in the lab manual to find I_cyl (I mis-typed in the post sorry)
Icyl = (2hgM_cylinder) * (1/(W_withcylinder)^2 - 1/(W_nocylinder)^2)
where h is the height = 0.9 m (height from hanging mass to the floor)
M_cylinder mass of the cylinders = 0.0757 kg
W_withcylinder and W_nocylinder recorded by rotary sensor
I don't understand why the moment of inertia get smaller when the mass get bigger (the hanging mass act as a torque to make the rod rotating)
My partner (column J) has an entirely different number, and I have no idea where he got that equation from
 
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haruspex
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W_withcylinder and W_nocylinder recorded by rotary sensor
Ok, but what are they? At the least, what sort of entity: distances, times, speeds..?
 
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Ok, but what are they? At the least, what sort of entity: distances, times, speeds..?
The Angular velocity when the hanging mass hit the floor, I attached a better sketch
 

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  • #8
haruspex
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Sorry, I made a mistake in characterising the differences between the two equations. I missed a leading "1/".
The differences are:
- the constant 0.0797 I mentioned (M)
- col J is the inverse of col K
That is, col J is M/Icyl
 
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