Moment of Inertia of the disk

  • Thread starter tsnikpoh11
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  • #1

Homework Statement



A uniform disk of radius 0.12 m is mounted on a frictionless, horizontal axis. A light cord wrapped around the disk supports a 1 kg object, as shown in the figure from rest the object falls with a downward acceleration of 2.3 m/s^2.The acceleration of gravity is 9.8 m/s^2 When released What is the moment of inertia of the disk? in units of kg m^2


Homework Equations



Moment of Inertia of a disk by itself= 1/2MR^2
I= Torque/Angular Acceleration



The Attempt at a Solution




1/2*(M)*(.12^2) = I

If there is downward acceleration of 2.3 m/s^2 is that added to the 9.8 m/s^2 from gravity? 12.1 m/s^2

Im having a hard time knowing where to start with how this 1kg object affects the moment of inertia of the disk, and how I find M in the disk.
 

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Answers and Replies

  • #2
Doc Al
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You'll need to separately analyze the forces on the hanging mass and on the disk.

Start with the hanging mass. Apply Newton's 2nd law to find the tension in the cord.
 

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