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## Homework Statement

## Homework Equations

## The Attempt at a Solution

Is it correct?

- Thread starter javii
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Is it correct?

- #2

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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.Is it correct?

- #3

berkeman

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- #4

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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.The movement of the liquid inside the carton would need to be accounted for in a full solution, IMO.

- #5

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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?

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So, the formula i have to use is: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.

1/3*m*(b^2+l^2)?

and when i use that one i get

which leads me to:

so the angular velocity is 9.15 rad/s ?

- #7

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Why are you dividing the 0.195 by 2 but not the 0.07?So, the formula i have to use is:

1/3*m*(b^2+l^2)?

and when i use that one i get

View attachment 196008

- #8

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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.which leads me to:

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

skewed.

So your are saying, that I also have to divide 0.07 with 2?

- #10

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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?So your are saying, that I also have to divide 0.07 with 2?

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

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

- #12

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

Thanks

- #13

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I assume you mean MdI_p= 1/12M(l^2+a^2)+M+d^2

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

What is d

- #14

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So what do you get now for the vertical displacement of the mass centre?I see, it is a mistake by me. Its just the height.

Thanks

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