Does Bread Land Peanut Butter Side Down?

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Homework Help Overview

The problem involves analyzing the motion of a piece of bread as it falls off a table, specifically determining whether it lands peanut butter side down. The scenario includes initial conditions such as the angle of the bread and its angular velocity at the moment it leaves the table, along with relevant equations of motion and moment of inertia considerations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of the bread's angular velocity and the absence of torque once it leaves the table. Questions arise regarding the relationship between linear and angular velocity at the moment of free fall, and the relevance of moment of inertia in this context.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have provided guidance on neglecting linear velocity and focusing on the conditions at the moment the bread leaves the table. Others are questioning the assumptions made about torque and angular acceleration.

Contextual Notes

Participants note that the problem may involve complexities related to the initial conditions of the bread's motion, including its angular velocity and the effects of gravity on its rotation. There is also mention of the moment of inertia being potentially irrelevant after the bread leaves the table.

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Homework Statement


Problem is to verify that bread lands peanut butter side down.The bread makes an angle of 30 deg below the horizontal before if falls off table. At that instant it has angular velocity of 0.956*the square root of g/l where g is gravity and l is the length of the toast at 0.10m. We assume the toast is square and so it's moment of inertia is (ML^2)/3. The height of the table is 0.500m. Assuming the bread is peanut butter side up initially I basically just need to find the change in rotation from the moment it falls off the table until it hits the floor so I can verify that it lands butter side down.

Homework Equations


Theta final = theta initial + omega(t) +(0.5)alpha(t^2)
Torque = (ML^2)/3 *alpha
Y=y initial + vt + 0.5at^2
Omega = initial omega + alpha(t)
A number of other possible kinematic/CoE/force rotational motion equations.

The Attempt at a Solution


I've gotten loads of different answers. I'm not sure how I'm supposed to go about approaching the angular acceleration due to gravity, the time (I've got like 4 times all differently) etc...
 
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You are overthinking the problem. You are given the point where it leaves the table. Where do you see torque if the bread is not connected to the table any more?
How long does it take in free fall to hit the ground?
 
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mfb said:
You are overthinking the problem. You are given the point where it leaves the table. Where do you see torque if the bread is not connected to the table any more?
How long does it take in free fall to hit the ground?
You are saying that there is no torque due to gravity after it falls off the table, but it still rotates?
 
Gravity never leads to torque around the center of gravity - by definition.
It still rotates, sure.
 
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mfb said:
Gravity never leads to torque around the center of gravity - by definition.
It still rotates, sure.
Would it have an initial linear velocity if it has an initial rotational velocity? I know the equations and physics say so but I am having a hard time with this because I don't understand how a piece of bread could have an initial velocity if it falls after tilting, where it is stationary but rotates before the fall. At the initial time of free fall linear velocity must be zero...?I understand that it has initial angular velocity.
 
Sorry for all of the questions - this is the last one.
 
There is no linear velocity given, I guess you can neglect it. Alternatively, assume that the edge of the bread is leaving the table at this point, which gives a condition relating angular and linear velocity.
 
Vitani11 said:
the toast is square and so it's moment of inertia is (ML^2)/3.
As mfb has explained, there is no torque, so no angular acceleration, after the initial conditions given. Thus, the moment of inertia is irrelevant. But I will point out that the formula you quote is for a square rotating about a central axis normal to its plane, not an axis in its plane.
 
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I solved this problem earlier today. Thank you for the help.
 

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