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What side does the bread land?

  1. Oct 11, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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.

    3. 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...
     
  2. jcsd
  3. Oct 11, 2016 #2

    mfb

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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?
     
  4. Oct 11, 2016 #3
    You are saying that there is no torque due to gravity after it falls off the table, but it still rotates?
     
  5. Oct 11, 2016 #4

    mfb

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    Gravity never leads to torque around the center of gravity - by definition.
    It still rotates, sure.
     
  6. Oct 11, 2016 #5
    Would it have an initial linear velocity if it has an initial rotational velocity? I know the equations and physics say so but Im 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.
     
  7. Oct 11, 2016 #6
    Sorry for all of the questions - this is the last one.
     
  8. Oct 11, 2016 #7

    mfb

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    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.
     
  9. Oct 11, 2016 #8

    haruspex

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    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.
     
  10. Oct 11, 2016 #9
    I solved this problem earlier today. Thank you for the help.
     
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