Disk and Block (Moment of Inertia)


by Trentonx
Tags: block, disk, inertia, moment
Trentonx
Trentonx is offline
#1
Feb21-09, 09:50 PM
P: 38
1. The problem statement, all variables and given/known data
A uniform disk of mass Mdisk = 4.1 kg and radius R = 0.3 m has a small block of mass mblock = 2.9 kg on its rim. It rotates about an axis a distance d = 0.16 m from its center intersecting the disk along the radius on which the block is situated.
a) What is the moment of inertia of the block about the rotation axis?
b) If while the system is rotating with angular velocity 4.6 rad/s it has an angular acceleration of 8.5 rad/s2, what is the magnitude of the acceleration of the block?

2. Relevant equations
I[tex]_{cm}[/tex]=(1/2)(mr[tex]^{2}[/tex])
I=I[tex]_{cm}[/tex]+Md[tex]^{2}[/tex]
v=wr
a=r(alpha)


3. The attempt at a solution
M of I for disk
I=(1/2)(4.1)(.3[tex]^{2}[/tex])=.289
M of I for block
Parallel-axis theorem, but with what mass??
What radius would I use in the angular velocity/acceleration formula?
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LowlyPion
LowlyPion is offline
#2
Feb21-09, 10:37 PM
HW Helper
P: 5,346
The axis of rotation for the || axis theorem is .16m for the disk.

That means your 1/2mr2 + md2 = I

where r is .3 and d is .16.

The block then adds 2.9*(.3 - .16)2

Since you have established the center of rotation and they are asking the acceleration on the block, then just use that .14 radius that the block is away.


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