Disk and Block (Moment of Inertia)

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The discussion centers on calculating the moment of inertia for a system consisting of a uniform disk and a block on its rim. The moment of inertia for the disk is calculated using the formula I = (1/2)(M)(R^2), resulting in 0.289 kg·m². The block's moment of inertia is determined using the parallel-axis theorem, taking into account its distance from the rotation axis. The radius used for calculating the block's acceleration is established as 0.14 m, leading to the application of angular acceleration formulas. Overall, the calculations focus on accurately determining the system's rotational dynamics.
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Homework Statement


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?

Homework Equations


I_{cm}=(1/2)(mr^{2})
I=I_{cm}+Md^{2}
v=wr
a=r(alpha)

The Attempt at a Solution


M of I for disk
I=(1/2)(4.1)(.3^{2})=.289
M of I for block
Parallel-axis theorem, but with what mass??
What radius would I use in the angular velocity/acceleration formula?
 
Last edited:
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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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