- #1
Karol
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Homework Statement
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?
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:
$$mgx_0\sin\alpha=F_f x_{tot}\;\rightarrow\;mg\frac{R\cos\alpha\sin\alpha}{\sin\alpha-\mu\cos\alpha}=mg\mu\sin\alpha x_{tot}\;\rightarrow\;x_{tot}=\frac{R\cos\alpha}{\mu(\sin\alpha-\mu\cos\alpha)}$$