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

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Homework Help Overview

The problem involves calculating the moment of inertia for two welded metal disks and determining the speed of a block suspended from a string wrapped around one of the disks. The subject area includes rotational dynamics and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the moment of inertia formula and the conservation of energy principles. There are attempts to clarify the correct formula for the moment of inertia of a solid disk and how to account for kinetic energy in the calculations.

Discussion Status

Some participants have provided guidance on the correct formula for the moment of inertia and the need to include kinetic energy in the energy conservation equation. Others are exploring different approaches to the problem, but there is no explicit consensus on the final calculations.

Contextual Notes

Participants note discrepancies between their calculations and the answers provided in the textbook, indicating potential misunderstandings of the formulas or assumptions involved.

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?
 
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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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