Calculate the mass of the disk?

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    Disk Mass
The moment of inertia of the disk-clay system about the central axis of the disk is I=1.5kgm2.The mass of the disk is M.The mass of the clay is 0.12kg.The disk's moment of inertia is I=1.5kgm2.The clay's moment of inertia is I=0.12kgm2.The total moment of inertia is I=1.8kgm2.
  • #1
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Homework Statement


The moment of inertia of the disk-clay system about the central axis of the disk is I = 1.5 kg m2
Disk's mass = M
The clay's mass = 0.12 kg
Calculate the mass of the disk?

Homework Equations


I = m r^2

The Attempt at a Solution


I thought that the moment of inertia may just be the sum of the individual moment of inertia's of the two objects. I know that I for a disk is equal to 1/4 MR^2, but I do not know the one for the clay.[/B]
 
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  • #2
I thought that the moment of inertia may just be the sum of the individual moment of inertia's of the two objects.
Correct.

I know that I for a disk is equal to 1/4 MR^2, but I do not know the one for the clay.

You sure you've interpreted the axis correctly?
https://en.wikipedia.org/wiki/List_of_moments_of_inertia

Was there a diagram? As it is, we don't know r disk or rclay so can't find mclay.
 
  • #3
billy_joule said:
Correct.



You sure you've interpreted the axis correctly?
https://en.wikipedia.org/wiki/List_of_moments_of_inertia

Was there a diagram? As it is, we don't know r disk or rclay so can't find mclay.
Yes. The diagram showed a piece of clay is sitting at the rim of the disk and not sliding on the disk.
 
  • #4
Post the diagram and the entire question statement word for word, without it we're just guessing/relying on your interpretation.
 
  • #5
upload_2016-6-9_1-36-57.png


A horizontal uniform disk with a mass M and a radius R=1.5m is free to rotate about its central axis perpendicular to the disk’s flat surface. A piece of clay with a mass m=0.12kg is sitting at the rim of the disk and does not slide on the disk. The moment of inertia of the disk-clay system about the central axis of the disk is I=1.5kgm2.At time t=0s, a constant external force, F, is applied tangentially at the rim of the disk causing the disk-clay system, initially at rest, to start rotating with a constant angular acceleration ofα=0.25rad/s2.
 
  • #6
There's doesn't appear to be a question?
Can you follow the link in post #2 and identify the relevant MOI equations for the disk and clay?
 
  • #7
alaa amed said:
perpendicular to the disk’s flat surface
Note that part of the axis specification.
 
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