NCU Physics Demonstration Lab 國立中央大學物理演示實驗

Physics on the Dining Table: Murphy’s Law

Toast always lands jam side down when it falls?

What?

Does toast always land on the jam side when it falls from table? Is it merely a matter of luck or reasonable physics that lies behind Murphy’s Law?

How?

Materials:

  1. A slice of toast
  2. A dining table

Experiment:

  1. Mark one side of the toast with a marker representing jam spread.
  2. Place the marked toast at the brim of a dining table. Push the toast slowly and make it fall. Observe on which side the toast lands. Repeat and record the results 10 times.

Experiments demonstration

×Experiments demonstration

Why?

What makes the toast land on the marked side?

Think...

×Think...

  1. Murphy’s Law is typically known by “if something can go wrong, it will”. When applying the Law to the case of toast fall, it denotes that, with the tendency of turning bad to worse, the tumbling toast always lands jam side down. As shown in the demonstration video, the tumbling toast lands on the side with a cross (or the side of jam). However, unsurprisingly, the landing dynamics has nothing to do with the Law; it is a complicated question involving factors like the size of toast, the onset tumbling angle, the torque of toast rotating off the table, and the moment of inertia of the toast.
  2. Neglecting the frictional force, the free body diagram of the toast at the initial state is shown in the figure below:

  3. Based on the law of Conservation of Energy, the Potential Energy gained from table height is transformed to Kinetic Energy of center of mass and Energy of rotation during the fall. In the figure above, {\omega} is the angular velocity of toast rotation; {\theta} is the angle included by the toast and the surface of table; L is the length of toast; v , the velocity of center of mass, can be divided to vertical to the toast v_a and parallel to the toast; v_b),m is the mass of toast; I is the moment of inertia. Equate the Potential Energy and the transformed Kinetic Energy, we obtain
    {\frac{mgL}{2}}sin{\theta}={\frac{1}{2}}mv^2+{\frac{1}{2}}I{\omega}^2…………(1)
    Since
    v_b={\frac{{\omega}L}{2}}
    and the toast’s moment of inertia is
    I={\frac{mL^2}{12}}………………………………..(2)
    (1) can be rewritten as
    mg\left(\frac{L}{2}\right)sin{\theta}={\frac{mv^2_a}{2}}+mv^2_b+m\left( \frac{L^2}{12} \right){\left( \frac{2v_b}{L} \right)}^2
    and further simplified to
    gL sin{\theta}=v^2_a+{\frac{3}{4}v^2_b}………………………………..(3)
    From the Impulse-Momentum Theorem, (3) can be represented as
    mg (sin{\theta})T=mv_a
    where is the time for the toast’s disconnecting from the table, which is considerably short.
    The torque formed by the distance r, from the toast’s center of mass to the point of contact on the table, is:
    mgr (cos{\theta})……………………………………………(4)
    r The r here is a function of time, r=at^2, r leading to the average of r:
    \int_{0}^{T} \left(\frac{r}{T}\right)\, dt=a\int_{0}^{T} \left(\frac{t^2}{T}\right)\, dt=\frac{aT^2}{3}……………………(5)
    In the very beginning of the toast’s disconnecting from the table, r=\frac{L}{2}. Based on {\tau}T={\Delta}I{\omega} , fit the result of (5) into (4):
    I{\omega}=mg\left( \frac{L}{6} \right)cos{\theta}=mv_a \left( \frac{L}{6} \right) cot{\theta}………………………(6)
    From the Parallel Axis Theorem, the toast’s moment of inertia is obtained:
    I=\int_{0}^{T} \frac{I}{T}\, dt=I_o+\left( \frac{m}{T} \right)\int_{0}^{T} r^2\, dt=I_o+ma\left( \int_{0}^{T} \frac{t^2}{T} \right)\, dt
    =I_o+\frac{ma^2T^5}{5T}=I_o+\frac{m\left( \frac{L}{2} \right)^2}{5}=I_o+\frac{mL^2}{20}……………………………..(7)
    Fit the result of (2) and (7) into (6), we have:
    m\left( \frac{2L^2}{15} \right)\left( \frac{2v_b}{L} \right)=mv_a\left( \frac{L}{6} \right)cot{\theta}
    simplified to
    \frac{v_a}{v_b}=\frac{8}{5} tan{\theta}………………………(8)
    Insert (8) in (1), we obtain
    \frac{4}{3v^2_b}+\left| \frac{8(tan^2{\theta})v_b}{5} \right|^2=gL sin{\theta}
    and derive after calculation
    v^2_b=gL sin{\theta}\left| \frac{3}{4}+64\left( \frac{tan^2{\theta}}{25} \right) \right|
    In our demonstration, given the included angle {\theta}=30^o , g=981\frac{cm}{sec^2} , the length of the toast L=10cm, v_b=47 \frac{cm}{sec}{\omega}=9.47\frac{red}{sec}
    and the height of the table 76cm,it takes only 0.4 seconds for the toast to fall and land. The overall angle of rotation, containing the initial tilt angle 0.4\times 9.47=3.79 rad, 217^o , and the tumbling angle 30^o, 247^o,an angle between 180^o and 360^o) In other words, the toast turns upside down and lands on the side with a cross (jam).

Questions

  1. If the toast is jammed on one side and has its center of mass slightly changed, will it still land on the jam side?
  2. Push the toast with an initial horizontal velocity along the surface of table. How will the toast land?
  3. If we change the height of table, will the result be any different?

Reference

  1. Robert, E. (1997). Jelly-Side Down. Why Toast Lands Jelly-Side Down: Zen and the Art of Physics Demonstrations (pp. 75-77). West Sussex: Princeton University Press.
  2. Edge, R. D. (1988). Murphy’s Law or Jelly-Side Down. The Physics Teacher, 26, 392.
  3. Knight, R. D. (2003). Physics for Scientists and engineers with Modern Physics. (International ed., pp.367-406). Boston, MA: Addison Wesley.

Producer

Ching-Chi Chu (朱慶琪)

Advisor

Ching-Chi Chu (朱慶琪)

Written by

Tsung-Cean Tu (杜宗勳), Ching-Chi Chu (朱慶琪)

Translator

Hsiao-Ching Su (蘇筱晴)