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B Question about force and torque -- Coins falling off a rod...

  1. Feb 18, 2017 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    Screenshot_2017-02-19-02-38-47.png

    By my choice it seems 1 is correct can you please help me ...confused
     
    Last edited by a moderator: Mar 2, 2017
  2. jcsd
  3. Mar 2, 2017 #2
    Hello Soumobrata,
    I think the correct answer is (D).
    The rotational motion of the plate is given by: ## \theta (t) = \frac {3 g} {4 L} t^2## , ##g## being the gravity acceleration and ##L## the plate length. Hence the material point of the plate initially located at station ##x_0## will move according to the following trajectory: ##\begin{cases} x = x_0 \cos \theta(t) \\ y = x_0 \sin \theta(t) & y > 0 \text{ downward for convenience} \end{cases}## .
    From which we can find the 2nd time derivative: ## \ddot y = x_0 \frac {3g} {2L} \cos \theta -x_0 ( \frac {3g} {2L} t )^2 \sin \theta )## .
    Now, the condition for coins to remain in contact with the plate is that gravity acceleration is greater than the plate acceleration at that particular point, which can be written as: ## x_0 \frac {3g} {2L} \cos \theta -x_0 ( \frac {3g} {2L} t )^2 \sin \theta ) \le g## . We can simplify this by introducing the non dimensional notation: ## \lambda \equiv \frac {x_0} L## . The non-separation condition simply writes: ## \frac 1 {\lambda} \ge 3 ( \sin \theta - \frac {\cos \theta } 2 )##
    The right hand side term of this condition increases from ##-\frac 3 2## to ##+3## with ##\theta## increasing from 0 to 90 deg. Obviously, all coins will remain in contact with the plate during the first part of the rotation, except possibly for those which may be sliding past the edge and then free fall faster that the plate: that corresponds to sketch (D). Vertical separation will start at the plate edge then propagate inboard toward the hinge (table edge). This begins to occur for ##\lambda_{sep} = 1## that is for the angle ##\theta_{sep} = 2 \tan^{-1} ( \frac 1 3 -2 + \sqrt { \frac 1 9 +4}) \approx 39.7 \text{ deg}##
    Unless I made some error in the derivation above, of course...
    I hope this can help you. TGVF
     
  4. Mar 2, 2017 #3

    Baluncore

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    The centroid of mass of the scale will accelerate downwards due to gravity.
    The coins will rest on the rule except where the rule accelerates faster than gravity.
    A, B and C are similar models except for the break point where coins separate from the rule.
    Is the centroid at the mid-point of the rule?
    Does the energy of rotation in the rule reduce the acceleration downwards?
     
  5. Mar 2, 2017 #4
    I assumed the plate|rule has constant mass density which has two consequences:
    1. Hinge moment (about table edge) due to gravity is: ##M_g = g \frac {ML} 2 \cos \theta##
    2. Inertia about hinge is: ##I{yy} = \frac {ML^2} 3##
    I just realized my motion solution was wrong since I forget the ##\cos \theta## factor in the hinge moment expression! So, expressions are a bit more involved and results will change quantitatively and but, qualitatively and for small angles at least, the conclusion remains the same: I believe solution (D) to be the right answer.
     
  6. Mar 5, 2017 #5
    B abd C look similar
     
  7. Mar 5, 2017 #6
    Thanks
     
  8. Mar 5, 2017 #7

    Philip Wood

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    Surely this is the key to it. I should guess we're supposed to take cos θ as 1, in which case simple rotational dynamics shows that the point on the ruler which has acceleration g is 2/3 of the way along from the hinge. Therefore, following Baluncore, B is the right answer.
     
  9. Mar 13, 2017 #8
    but the acceleration on a slope is ##g*cos^2\theta##.
     
  10. Mar 13, 2017 #9

    Baluncore

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    The coins will not begin to slide until Tan(theta) exceeds the coefficient of friction of coin on meter-scale.
     
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