# Inertia of Disk and Block

1. Jun 14, 2010

1. The problem statement, all variables and given/known data
A uniform disk of mass Mdisk = 4.9 kg and radius R = 0.2 m has a small block of mass mblock = 2 kg on its rim. It rotates about an axis a distance d = 0.17 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) What is the moment of inertia of the disk about the rotation axis?

c) When the system is rotating about the axis with an angular velocity of 4.2 rad/s, what is its energy?

d) If while the system is rotating with angular velocity 4.2 rad/s it has an angular acceleration of 8.1 rad/s2, what is the magnitude of the acceleration of the block?

2. Relevant equations
I = mR^2

3. The attempt at a solution
I tried I = mR^2 for m = 4.9 and R = 0.2 for part b and also m = (4.9 + 2) but that was also not right.

For the block i tried I = (2)*(.17^2) but that didn't work either, so I'm completely lost.

Last edited: Jun 14, 2010
2. Jun 14, 2010

### Staff: Mentor

Last edited by a moderator: Apr 25, 2017
3. Jun 14, 2010

I found I = .5*m*R^2, and i tried both of the following:

I = .5*4.9*(.2^2) and
I = .5*4.9*(.2^2) + (2)(.17^2)

and neither worked.

4. Jun 14, 2010

Any other suggestions?

5. Jun 14, 2010

### Staff: Mentor

Good. That's the moment of inertia of a disk about its center. But its not rotating about its center, it's rotating about the specified point. So you'll need to make use of the parallel axis theorem to find the moment of inertia of the disk about the given axis.

Neither of those is quite right. The first is the moment of inertia of the disk about its center. The second is almost right--Look up the parallel axis theorem.

Note that you're working on (b). What about (a)? What's the moment of inertia of the block about the axis? (Treat the block as being a small mass.)

6. Jun 14, 2010

### dulrich

You also need the parallel axis theorem since the disk is not rotating about its center.

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax

I had to draw this out in order to really see it. The block is actually rotating around in a circle with a radius of 0.03 m -- isn't it?

7. Jun 14, 2010

So are you saying I should be using 0.03 as the radius? Which formula would I use in this case?

8. Jun 14, 2010

I also tried I = .5*4.9*(.2^2) + (.2)(.17^2), but that didn't work either.
The reason I used a 2 instead of .2 before was because I figured the mass wasn't evenly distributed due to the block being added to it.

9. Jun 14, 2010

### Staff: Mentor

Forget about the block for a moment until you've finished figuring out the moment of inertia of the disk. When you look up the parallel axis theorem, you'll see why I highlighted your use of 2 (which is the mass of the block) in your earlier post.

Figure out the moment of inertia of the disk. (Which is part (b).)

Figure out the moment of inertia of the block. (Which is part (a).)

You'll need to add them in the later parts of the problem, but keep them separate for now.

10. Jun 14, 2010

I understand part a, because it's just a circle with radius 0.03 m.

So Inertia of block = 2*(0.03)^2

But for the disk with the block on it, I'm confused,
From the ||Axis Theorum, I'm getting the following:

Intertia of disk = (1/2)*4.9*(.2^2) + .2*(1.7)^2
Since there's a block on it, would I do something like...

.5*4.9*(.2^2) + .2*1.7^2 + 2*(0.03)^2?

11. Jun 14, 2010

### Staff: Mentor

Good! So now you can forget about the block--you've done part (a).

Where did the .2 come from? The 1.7? What does the parallel axis theorem actually say?
Forget the block!

(Overall hint: The total moment of inertia of a disk with a block on it is the sum of the moment of inertia of the block plus the moment of inertia of the disk. That's why they are asking for each separately.)

12. Jun 14, 2010

I of || Axis = I of cm + Md^2

The .2 is the mass of the disk, (assuming that's what M is). Just realized I made a mistake with that one. Should be 4.9.
The .17 is the distance, (assuming that's what d is).

So with the corrected mass,
.5*4.9*(.2^2) + 4.9*1.7^2?

13. Jun 14, 2010

### Staff: Mentor

Now you've got it. (But correct the distance: 0.17, not 1.7)

14. Jun 14, 2010

Oh, right...
I guess I've been staring at a computer screen for too long...

So I added the two (using the correct distances and masses) and multiplied the sum of Inertia's by (1/2)(4.2^2) for part c and got the right answer.

for part d, do i just sum the angular acceleration and the centripetal acceleration? or am i over-complicating it again?

15. Jun 14, 2010

### Staff: Mentor

Almost. Take the vector sum of the tangential acceleration and the centripetal acceleration.

16. Jun 14, 2010

right, that's the idea.

so aT = alpha*R
aC = R*w^2

and since their directions are perpendicular, sqrt(aT^2 + aC^2), right?

17. Jun 14, 2010

### Staff: Mentor

Sounds good to me!