Mass sliding on rough and smooth surfaces

[Karol]

Homework Statement

[/B]
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 x_i?
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?

Homework Equations

Centripetal acceleration: $a=\frac{v^2}{R}$
Kinematics: $v^2=2ax$

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:
$$mgx_0\sin\alpha=F_f x_{tot}\;\rightarrow\;mg\frac{R\cos\alpha\sin\alpha}{\sin\alpha-\mu\cos\alpha}=mg\mu\sin\alpha x_{tot}\;\rightarrow\;x_{tot}=\frac{R\cos\alpha}{\mu(\sin\alpha-\mu\cos\alpha)}$$

 

[haruspex]

You lost me here:

The loss of energy each time m is on AB again:
$\frac{W_f}{E_0}=\frac{F_f}{E_0}=\frac{mg\mu\cos\alpha\cdot x_0}{mgx_0\sin\alpha}=\frac{\mu}{\tan\alpha}\triangleq \beta$

Why is it not just the ratio of x₁ to x₀ that's of interest?

 

[Nathanael]

The force m applies before B is the component of the weight $mg\cos\alpha$

What about friction?

The loss of energy each time m is on AB again:
$$\frac{W_f}{E_0}=\frac{F_f}{E_0}=\frac{mg\mu\cos\alpha\cdot x_0}{mgx_0\sin\alpha}=\frac{\mu}{\tan\alpha}\triangleq \beta$$

Say it starts at x_n and ends at x_n+1 (it is stationary at both x_n and x_n+1). Your equation is saying the distance traveled against friction is x_n, but it also travels a distance x_n+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.

You lost me here:

The loss of energy each time m is on AB again:
$$\frac{W_f}{E_0}=\frac{F_f}{E_0}=\frac{mg\mu\cos\alpha\cdot x_0}{mgx_0\sin\alpha}=\frac{\mu}{\tan\alpha}\triangleq \beta$$

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 E_n (w.r.t. point B) is proportional to x_n (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 W_f, he thought that $\frac{W_f}{E_n}\triangleq\beta$ was a constant (but it's not).

 

[haruspex]

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

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 x_i then up x_i+1 it starts with energy c₁x_i and finishes with energy c₁x_i - c₂(x_i+x_i+1) = c₁x_i+1. So x_i+1/x_i=(c₁-c₂)/(c₁+c₂).

 

[Nathanael]

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 x_i then up x_i+1 it starts with energy c₁x_i and finishes with energy c₁x_i - c₂(x_i+x_i+1) = c₁x_i+1. So x_i+1/x_i=(c₁-c₂)/(c₁+c₂).

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 - E_n+1/E_n = 1 - x_n+1/x_n) which is not constant.

 

[haruspex]

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 - E_n+1/E_n = 1 - x_n+1/x_n) which is not constant.

I don't understand. According to my post, with which you agree, 1 - x_n+1/x_n is constant.

 

[Nathanael]

I don't understand. According to my post, with which you agree, 1 - x_n+1/x_n is constant.

Right sorry. I was seeing that the x_n+1/x_n does not cancel out from β and I was getting mixed up. Brain fart sorry.
All I meant to say then is that β is not immediately known, but somehow I convinced myself it's not constant

 

[Karol]

I consider an arbitrary point, say for x_n, and the velocity at B for that starting point is v_B:
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 x_n+1:
$$\frac{1}{2}mv_B^2=mgh_1+F_fx_1=mg(\sin\alpha+\mu\cos\mu)x_1 \;\rightarrow\; x_1=\frac{2v_B^2}{g(\sin\alpha+\mu\cos\mu)}$$
$$\beta\triangleq \frac{x_1}{x_0}=\frac{4(\sin\alpha-\mu\cos\alpha)}{\sin\alpha+\mu\cos\mu}$$

 

[haruspex]

I consider an arbitrary point, say for x_n, and the velocity at B for that starting point is v_B:
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 x_n+1:
$$\frac{1}{2}mv_B^2=mgh_1+F_fx_1=mg(\sin\alpha+\mu\cos\mu)x_1 \;\rightarrow\; x_1=\frac{2v_B^2}{g(\sin\alpha+\mu\cos\mu)}$$
$$\beta\triangleq \frac{x_1}{x_0}=\frac{4(\sin\alpha-\mu\cos\alpha)}{\sin\alpha+\mu\cos\mu}$$

Nearly right. Check the factors of 2, and you have a typo in the Greek.

 

[Karol]

$$\frac{1}{2}mv_B^2=mgh_1+F_fx_1=mg(\sin\alpha+\mu\cos\mu)x_1 \;\rightarrow\; x_1=\frac{v_B^2}{2g(\sin\alpha+\mu\cos\mu)}$$
$$\beta\triangleq \frac{x_1}{x_0}=\frac{\sin\alpha-\mu\cos\alpha}{\sin\alpha+\mu\cos\alpha}$$

 

[haruspex]

$$\frac{1}{2}mv_B^2=mgh_1+F_fx_1=mg(\sin\alpha+\mu\cos\mu)x_1 \;\rightarrow\; x_1=\frac{v_B^2}{2g(\sin\alpha+\mu\cos\mu)}$$
$$\beta\triangleq \frac{x_1}{x_0}=\frac{\sin\alpha-\mu\cos\alpha}{\sin\alpha+\mu\cos\alpha}$$

That's it.

 

[Karol]

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

 

[haruspex]

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

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.

 

[Karol]

$$x_1=x_0\beta,\quad x_2=x_1\beta=x_0\beta^2\quad \rightarrow\quad x_n=x_0\beta^n$$
$$\ln x_n=n\ln\beta+\ln x_0\quad \rightarrow\quad n>\frac{\ln x_n-\ln x_0}{\ln\beta}=\frac{\ln\left( \frac{x_n}{x_0} \right)}{\ln\beta}$$
$$n>\frac{\ln\left( \frac{0.21}{0.6} \right)}{\ln 0.875}=7.8$$
Verification: $x_8=0.6\cdot 0.875^8=0.2<0.21$