How Do You Calculate the Moment of Inertia for Welded Disks?

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SUMMARY

The discussion focuses on calculating the total moment of inertia for two welded disks with given dimensions and masses, specifically using the formula I = Σ mi * ri^2. The first disk has a radius of 2.5 cm and a mass of 0.8 kg, while the second disk has a radius of 5.0 cm and a mass of 1.6 kg, leading to a calculated moment of inertia of 0.0045 kg*m. The conversation also addresses the calculation of the speed of a falling block using conservation of energy principles, highlighting the need to account for both the kinetic energy of the block and the rotational energy of the disks.

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  • Understanding of moment of inertia calculations for solid cylindrical disks
  • Familiarity with conservation of energy principles in physics
  • Knowledge of basic kinematics and dynamics
  • Ability to manipulate equations involving rotational motion
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  • Study the formula for the moment of inertia of solid cylindrical disks: I = 0.5 * M * R^2
  • Learn about energy conservation in mechanical systems, particularly in rotational dynamics
  • Explore the relationship between linear and angular velocity in rotating systems
  • Practice solving problems involving combined translational and rotational motion
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jhwatts
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Homework Statement



2 metal disks, one with radius R1= 2.5cm and mass M1=.8kg and the other with radius R2=5.00cm and mass M2=1.60kg, are welded together and mounted on a frictionless axis through their common center.

a)What is the total moment of inertia of the two disks?
b)A light string is wrapped around the edge of the small disk, and a 1.5kg block, suspended from the free end of the string. If the blook is 2.00m above the floor what is the speed of just before it strikes the floor?


Homework Equations


I = Σ mi *ri^2
K = 1/2 *I*ω^2
W(g)= m*g*h


The Attempt at a Solution


a) I = 0.8kg*(2.5cm)^2 +1.6kg(5.00cm)^2 = 45kg*cm = .0045kg*m

b) m*g*h = 1/2 *I*ω^2
i) 1.5kg*9.8m/s^2*2m = .5*(.0045kg*m) * ω^2
ii) ω = sqrt[(1.5kg*9.8m/s^2*2m ) / (.5*(.0045kg*m))]

This is what i thought i should do, but looking at the solution in the back of the book, i doesn't seem to be correct. Any help is appericated, thanks.
 
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If you are going to use energy methods, I think you need to account for the ke of the falling block as well. In other words some of its potential energy of the block goes into rotating the flywheel and some into its own motion. Also I think you dropped a factor of 2m in "1)"
 
Ture, that was just a copy error though from my paper to the computer.
In part a. my book says the answser should be have of what i calculated it to be, do you know why that might be?
 
Last edited:
jhwatts said:
In part a. my book says the answser should be have of what i calculated it to be, do you know why that might be?
You are using the wrong formula for the moment of inertia of a solid cylindrical disk. It is 0.5MR^2, not MR^2, which is what you are using.
 
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Thanks i figured it out when i was reading my book, and i figured out why my velocity calcuation was coming out wrong i forgot to add the KE of the weight. Thanks for your help.
 
I am doing this question, and I am using Conservation of Energy and the following equation:

mgh = 0.5*I*(v^2/r^2)+0.5*m*v^2, which I then arrange:

2mgh = v^2((I/(r^2))+m)

v^2 = (2mgh)/((I/r^2)+m)

but cannot get the right answer when I put in the variables. Any Ideas why?



TFM
 
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