# Calculate the angular velocity of the milk carton

1. Apr 19, 2017

### javii

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Is it correct?

2. Apr 19, 2017

### kuruman

No. You need to use the moment of inertia about the corner of the carton and you need to find the correct height of the center of mass when tipping occurs.

3. Apr 19, 2017

### Staff: Mentor

Maybe this problem does not intend to involve the full complexity of this situation, but milk cartons are not "full" of liquid, even when new. The movement of the liquid inside the carton would need to be accounted for in a full solution, IMO.

4. Apr 19, 2017

### kuruman

That is true, but since no mention is made in the problem about liquid in the carton, we may assume that it is empty. It could an empty carton on a cafeteria conveyor belt on its way to the trash disposal area.

On edit: Of course, the moment of inertia of an empty carton is more complicated to calculate.

5. Apr 19, 2017

### haruspex

It's also an issue for cartons that are full. The milk would tend to rotate internally in the opposite direction.
But an earlier part of the question says to ignore any sloshing around https://www.physicsforums.com/threads/calculate-the-maximum-acceleration-of-a-milk-carton.911818/.

There is another complication. Is it easily shown that the carton does not lose contact with the ground?

Last edited: Apr 19, 2017
6. Apr 20, 2017

### javii

So, the formula i have to use is:
1/3*m*(b^2+l^2)?

and when i use that one i get

so the angular velocity is 9.15 rad/s ?

7. Apr 20, 2017

### haruspex

Why are you dividing the 0.195 by 2 but not the 0.07?

8. Apr 20, 2017

### kuruman

And why do you think the CM is at height $\frac{0.195}{2}~m$ just before the carton tips over? That's its height when the carton sits flat on the conveyor belt.

9. Apr 20, 2017

### javii

For me, it is obvious that I have to divide 0.195 with 2, but I am not sure if 0.07 also have to be divided with 2 since the carton is
skewed.
So your are saying, that I also have to divide 0.07 with 2?

10. Apr 20, 2017

### kuruman

Yes, but it is important to understand why. Start with $I_{CM}=\frac{1}{12}M(l^2+a^2)$ and apply the parallel axes theorem. Try using symbols, not numbers. What do you get?

11. Apr 20, 2017

### javii

I_p= 1/12M(l^2+a^2)+M+d^2

I_p=1/12M*l^2+1/12*a^2+M+d^2 ?

12. Apr 20, 2017

### javii

I see, it is a mistake by me. Its just the height.
Thanks

13. Apr 20, 2017

### haruspex

I assume you mean Md2, not M+d2.
What is d2 in terms of l and a?

14. Apr 20, 2017

### haruspex

So what do you get now for the vertical displacement of the mass centre?