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Mass sliding on rough and smooth surfaces

  1. Nov 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Snap1.jpg

    Mass m starts sliding down on a rough surface with coefficient of friction μ. it reaches point B and starts sliding frictionlessly till it reaches point D without velocity, i.e. without escaping the arc.
    What is the maximum length AB=x0 not to escape the arc.
    What is the force the mass applies on the track just before and after point B at the first pass?
    At what distance x1 the box stops on the line AB the first time after it was released?
    After how many instantaneous stops on the line AB will the box stop at a distance not bigger than xi?
    What is the force the mass applies at point C after many instantaneous stops on AB?
    What's the total distance the box travels on AB after many instantaneous stops?

    2. Relevant equations
    Centripetal acceleration: ##a=\frac{v^2}{R}##
    Kinematics: ##v^2=2ax##

    3. The attempt at a solution
    $$F=ma\;\rightarrow\;mg\sin\alpha-\mu mg\cos\alpha=ma\;\rightarrow\;a=g(\sin\alpha-\mu\cos\alpha)$$
    What is the velocity at B so that at C it stops: ##mgh=\frac{1}{2}mv_B^2##
    $$mgR\cos\alpha=\frac{1}{2}mv_B^2\;\rightarrow\;v_B^2=2gR\cos\alpha$$
    Kinematics: ##v_B^2=2ax\;\rightarrow\;2gR\cos\alpha=2\cdot g(\sin\alpha-\mu\cos\alpha)\;\rightarrow\;x_{0}=\frac{R\cos\alpha}{\sin\alpha-\mu\cos\alpha}##
    The force m applies before B is the component of the weight ##mg\cos\alpha##, while after it the centripetal force is added: ##F=mg\cos\alpha+m\frac{v_B^2}{R}##
    When m returns the first time to B it has kinetic energy, which is transformed to potential energy+the friction work:
    $$\frac{1}{2}mv_B^2=mgh_1+F_fx_1\;\rightarrow\;\frac{1}{2}\cdot 2gR\cos\alpha=gx_1\sin\alpha+g\mu\cos\alpha\;\rightarrow\;x_1=\frac{R\cos\alpha}{\sin\alpha+\mu\cos\alpha}$$
    The loss of energy each time m is on AB again:
    $$\frac{W_f}{E_0}=\frac{F_f}{E_0}=\frac{mg\mu\cos\alpha\cdot x_0}{mgx_0\sin\alpha}=\frac{\mu}{\tan\alpha}\triangleq \beta$$
    $$x_1=x_0(1-\beta),\; x_2=x_1(1-\beta)=x_0(1-\beta)^2\;\rightarrow\;x_n=x_0(1-\beta)^n$$
    $$n>\frac{\ln x_n-\ln x_0}{\ln(1-\beta)}=\frac{\ln\left( \frac{x_n}{x_0} \right)}{\ln(1-\beta)}$$
    After many oscillations m will swing between B and the same height on the right of the arc. the height difference between B and C is (R-h). this potential difference will be translated to kinetic energy at C:
    $$mgR(1-\cos\alpha)=\frac{1}{2}mv_B^2\;\rightarrow\;v_B^2=2gR(1-\cos\alpha)$$
    $$F_C=\frac{mv_B^2}{R}=2mg(1-\cos\alpha)$$
    The total distance x_{tot} m travels is calculated from equating the friction work along x_{tot} and the initial potential energy from the vertical height:
    $$mgx_0\sin\alpha=F_f x_{tot}\;\rightarrow\;mg\frac{R\cos\alpha\sin\alpha}{\sin\alpha-\mu\cos\alpha}=mg\mu\sin\alpha x_{tot}\;\rightarrow\;x_{tot}=\frac{R\cos\alpha}{\mu(\sin\alpha-\mu\cos\alpha)}$$
     
  2. jcsd
  3. Nov 4, 2015 #2

    haruspex

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    You lost me here:
    Why is it not just the ratio of x1 to x0 that's of interest?
     
  4. Nov 4, 2015 #3

    Nathanael

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    What about friction?
    Say it starts at xn and ends at xn+1 (it is stationary at both xn and xn+1). Your equation is saying the distance travelled against friction is xn, but it also travels a distance xn+1 against friction on the way up, right?

    Considering the fractional change in energy β is not helpful because β is not the same between each stop.

    It also confused me why he wanted to use the fractional change in energy, but what he did in the next line is logically correct because the total energy En (w.r.t. point B) is proportional to xn (and the proportionality constant, mgsinα, is the same for all n).

    Basically Karol was doing this:

    ##E_{n+1}=E_n(1-\frac{W_f}{E_n})##
    ##kx_{n+1}=kx_n(1-\frac{W_f}{E_n})## // k = mgsinα,
    ##x_{n+1}=x_n(1-\frac{W_f}{E_n})##

    Since he miscalculated Wf, he thought that ## \frac{W_f}{E_n}\triangleq\beta## was a constant (but it's not).
     
  5. Nov 4, 2015 #4

    haruspex

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    Are you sure it isn't? The energy lost by each slide, up or down, is a constant multiple of the distance. If it slides down xi then up xi+1 it starts with energy c1xi and finishes with energy c1xi - c2(xi+xi+1) = c1xi+1. So xi+1/xi=(c1-c2)/(c1+c2).
     
  6. Nov 4, 2015 #5

    Nathanael

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    Yes, I got the same result, but that is not what I meant by "the fractional change in energy." In hindsight, it would have been more clear if I had called it "the fraction of the energy lost." At any rate, what I was referring to is what Karol called "β," (which is 1 - En+1/En = 1 - xn+1/xn) which is not constant.
     
  7. Nov 4, 2015 #6

    haruspex

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    I don't understand. According to my post, with which you agree, 1 - xn+1/xn is constant.
     
  8. Nov 4, 2015 #7

    Nathanael

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    Right o:) sorry. I was seeing that the xn+1/xn does not cancel out from β and I was getting mixed up. Brain fart :blushing: sorry.
    All I meant to say then is that β is not immediately known, but somehow I convinced myself it's not constant o_O
     
  9. Nov 6, 2015 #8
    I consider an arbitrary point, say for xn, and the velocity at B for that starting point is vB:
    Kinematics:
    $$v_B^2=2ax\;\rightarrow\; v_B^2=2\cdot g(\sin\alpha-\mu\cos\alpha)x_0\;\rightarrow\; x_0=\frac{v_B^2}{2g(\sin\alpha-\mu\cos\alpha)}$$
    Finding xn+1:
    $$\frac{1}{2}mv_B^2=mgh_1+F_fx_1=mg(\sin\alpha+\mu\cos\mu)x_1 \;\rightarrow\; x_1=\frac{2v_B^2}{g(\sin\alpha+\mu\cos\mu)}$$
    $$\beta\triangleq \frac{x_1}{x_0}=\frac{4(\sin\alpha-\mu\cos\alpha)}{\sin\alpha+\mu\cos\mu}$$
     
  10. Nov 6, 2015 #9

    haruspex

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    Nearly right. Check the factors of 2, and you have a typo in the Greek.
     
  11. Nov 6, 2015 #10
    $$\frac{1}{2}mv_B^2=mgh_1+F_fx_1=mg(\sin\alpha+\mu\cos\mu)x_1 \;\rightarrow\; x_1=\frac{v_B^2}{2g(\sin\alpha+\mu\cos\mu)}$$
    $$\beta\triangleq \frac{x_1}{x_0}=\frac{\sin\alpha-\mu\cos\alpha}{\sin\alpha+\mu\cos\alpha}$$
     
  12. Nov 6, 2015 #11

    haruspex

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    That's it.
     
  13. Nov 7, 2015 #12
    I had a mistake in the OP.
    $$x_1=x_0(1-\beta),\; x_2=x_1(1-\beta)=x_0(1-\beta)^2\;\rightarrow\;x_n=x_0(1-\beta)^n$$
    $$n>\frac{\ln x_n}{\ln[x_0(1-\beta)]}$$
    If i substitute: R=42[cm], α=370, μ=0.05 then:
    $$\beta=\frac{0.6-0.05\cdot 0.8}{0.6+0.05\cdot 0.8}=0.875,\quad x_0=\frac{0.42\cdot 0.8}{0.6-0.05\cdot 0.8}=0.6,\quad n>\frac{\ln 0.21}{\ln[0.6(1-0.875)]}=0.6$$
    Wrong
     
  14. Nov 7, 2015 #13

    haruspex

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    How did ##\beta## in post #10 turn into ##1-\beta## in post #12?
    You also ended up with the n on the wrong side when you took logs.
     
  15. Nov 7, 2015 #14
    $$x_1=x_0\beta,\quad x_2=x_1\beta=x_0\beta^2\quad \rightarrow\quad x_n=x_0\beta^n$$
    $$\ln x_n=n\ln\beta+\ln x_0\quad \rightarrow\quad n>\frac{\ln x_n-\ln x_0}{\ln\beta}=\frac{\ln\left( \frac{x_n}{x_0} \right)}{\ln\beta}$$
    $$n>\frac{\ln\left( \frac{0.21}{0.6} \right)}{\ln 0.875}=7.8$$
    Verification: ##x_8=0.6\cdot 0.875^8=0.2<0.21##
     
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