Disk and Block (Moment of Inertia)

[Trentonx]

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?

 

[LowlyPion]

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.

 

[nlightner]

The moment of inertia for the block about the rotation axis can be calculated using the parallel-axis theorem, which states that the moment of inertia of a body about an axis parallel to its center of mass is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the two axes.

In this case, the moment of inertia for the block about the rotation axis would be:

Iblock = Icm + mblock(d^2)

Where Icm is the moment of inertia of the block about its center of mass, which can be calculated using the formula I = (1/2)mr^2, where m is the mass of the block and r is the distance from the center of mass to the rotation axis.

For the block, the radius would be the distance from its center of mass to the rotation axis, which is the sum of the radius of the disk (0.3 m) and the distance between the center of the block and the rotation axis (d = 0.16 m). So the radius for the block would be 0.46 m.

Plugging in the values, we get:

Iblock = (1/2)(2.9)(0.46^2) + 2.9(0.16^2) = 0.63 kgm^2

For part b, we can use the formula a = r(alpha), where a is the acceleration of the block, r is the distance from the rotation axis to the block (0.46 m), and alpha is the angular acceleration of the system (8.5 rad/s^2). So the magnitude of the acceleration of the block would be:

a = (0.46)(8.5) = 3.91 m/s^2