Rope Sliding on Double Ramp

Let $x$ be the length of rope (in meters) on the right side of the ramp and $1.5 - x$ be the length of rope on the left side. As the entire rope has mass $1$ kg, the mass of the portion of the rope on the right side is $\dfrac{x}{1.5}$ kg, and on the left side $\dfrac{1.5 - x}{1.5}$ kg.

The force pulling the rope down the right side is then $\dfrac{x}{1.5}g\cos(30^{\circ})$ , i.e., the component of the weight of the rope on this side that is in line with the ramp, and the force restraining it is $\dfrac{1.5 - x}{1.5} g\cos(30^{\circ})$ , the component of the weight of the rope on the left side that is in line with the left side of the ramp. Thus the acceleration of the rope down the right side of the ramp is

$a = \dfrac{x}{1.5}g\cos(30^{\circ}) - \dfrac{1.5 - x}{1.5} g\cos(30^{\circ}) = \dfrac{2x - 1.5}{1.5} \times 10 \times \dfrac{\sqrt{3}}{2} = \dfrac{5\sqrt{3}}{3}(4x - 3)$ .

Now by the chain rule $a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \dfrac{dx}{dt} = v\dfrac{dv}{dx}$ , so we have that

$v\dfrac{dv}{dx} = \dfrac{5\sqrt{3}}{3}(4x - 3) \Longrightarrow vdv = \dfrac{5\sqrt{3}}{3}(4x - 3) dx \Longrightarrow$

$\dfrac{v^{2}}{2} = \dfrac{5\sqrt{3}}{24}(4x - 3)^{2} + C \Longrightarrow v = \dfrac{dx}{dt} = \sqrt{\dfrac{5\sqrt{3}}{12}(4x - 3)^{2} + C}$ .

Now $v = 0$ when $x = 1$ , so we see that $C = -\dfrac{5\sqrt{3}}{12}$ , and so

$\dfrac{dx}{dt} = K \times \sqrt{(4x - 3)^{2} - 1}$ where $K = \dfrac{1}{2}\sqrt{\dfrac{5}{\sqrt{3}}}$ .

Separating and integrating, we then have that

$\displaystyle \int \dfrac{dx}{\sqrt{(4x - 3)^{2} - 1}} = Kt + D \Longrightarrow \int \sec(\theta) d\theta = 4Kt + D$ ,

where the substitution $4x - 3 = \sec(\theta)$ was made. This last integral is standard, so we have

$\ln(4x - 3 + \sqrt{(4x - 3)^{2} - 1}) = 4Kt + D$ , with the initial condition $x(0) = 1$ giving us $D = 0$ .

Solving for $t$ , we have that $t = \dfrac{1}{2}\sqrt{\dfrac{\sqrt{3}}{5}} \times \ln(4x - 3 + \sqrt{(4x - 3)^{2} - 1})$ .

Plugging in $x = 1.5$ gives us a desired time of $\dfrac{1}{2}\sqrt{\dfrac{\sqrt{3}}{5}} \times \ln(3 + 2\sqrt{2}) = \boxed{0.519}$ seconds to 3 decimal places.

Let the length of the rope be $L$ , and $L_1,L_2$ the length of rope on either side. Define $u$ such that $L_1,L_2 = \frac L 2 \pm u.$ The advantage of this definition is its symmetry: at $u = 0$ the rope is in (unstable) equilibrium. Note that for the given values, $L = 1.5$ and the initial and final values of $u$ are $u_0 = L/6, u_f = L/2$ .

Let $\mu$ be the linear mass density of the rope. I don't expect it to matter for the answer; it will cancel out in the calculations.

We determine the potential energy of each part of the rope relative to the top of the triangle. We multiply its mass $\mu L_i$ by $g$ and by the vertical distance of its center of mass below the top, which is $-\tfrac12cL$ . Here, $c$ describes the geometry of the ramp: $c = \cos 30^\circ = \tfrac12\sqrt 3$ . Thus $U_1, U_2 = mgh = \mu L_i g (-\tfrac12 cL_i) = -\tfrac12 \mu c g L_i^2.$ Expressing the $L_i$ in our variable $u$ we find for the total potential energy $U = U_1 + U_2 = \tfrac12 \mu c g \left(\left(\frac L 2 + u\right)^2 + \left(\frac L 2 - u\right)^2\right) \\ \tfrac12 \mu c g(2 u^2 + \cdots) = \mu c g u^2.$ The expression on the dots does not depend on $u$ . We can ignore it, because that simply means choosing a new zero for potential energy.

Conservation of energy tells us that the kinetic energy during any moment of the motion is $K = U_0 - U = \mu c g (u^2 - u_0^2).$ Using $K = \tfrac12 m v^2 = \tfrac 12 \mu L v^2$ we find $v = \sqrt{\frac {2cg}L} \sqrt{u^2 - u_0^2}.$ The rate of change of $u$ is equal to the speed of each part of the rope. Therefore we can write for any infinitesimal movement of the rope $dt = \frac{du}v.$ Integrate this: $t = \int_{u_0}^{u_f} dt = \int_{u_0}^{u_f} \frac{du}v \\ = \sqrt{\frac{L}{2cg}} \int_{u_0}^{u_f} \frac{du}{\sqrt{u^2 - u_0^2}} \\ = \sqrt{\frac{L}{2cg}} \left(\ln(u + \sqrt{u^2 - u_0^2}) - \ln u_0\right) \\ = \sqrt{\frac{L}{2cg}} \ln\left(\frac{u}{u_0} + \sqrt{\left(\frac{u}{u_0}\right)^2 - 1}\right).$ In the given situation, $u/u_0 = 3$ ; $L = 1.5$ ; $g = 10$ ; and $c = \tfrac12\sqrt 3$ . Thus $t = \sqrt{\frac{1.5}{10\sqrt{3}}}\ln (3 + \sqrt 8) \approx \boxed{0.518747}.$

This solution was a little bit off because I do not know a lot of physics formulas, but this is what I did. Let's assume that the rope has an equal mass distribution throughout. We see that the acceleration down the slope on both sides of the triangle is $\frac{10}{\cos(30)} = \frac{20}{\sqrt{3}} = \frac{20 \cdot \sqrt{3}}{3} \frac{m}{s^2}$ Using the force equation $F = m \cdot a$ , we see that the force on the right side is $F_{right} = \frac{20 \cdot \sqrt{3}}{3} \frac{m}{s^2} \cdot \frac{2}{3} kg = \frac{40 \cdot \sqrt{3}}{9} N$ and the force on the left side is $F_{left} = \frac{20 \cdot \sqrt{3}}{3} \frac{m}{s^2}\cdot \frac{1}{3} kg = \frac{20 \cdot \sqrt{3}}{9} N$ Since the force on the left side is opposing the force on the right side, we subtract the force on the left side from the force on the right side. $F_{right} - F_{left} = \frac{40 \cdot \sqrt{3}}{9} - \frac{20 \cdot \sqrt{3}}{9} = \frac{20 \cdot \sqrt{3}}{9} N$ Converting this back to an acceleration constant, we see the acceleration of the entire rope towards the right side is $\frac{20 \cdot \sqrt{3}}{9} \frac{m}{s^2}$ In order to find the time it takes for the rope to move all the way to the right side, we take the integral of this to find the velocity in respect to time. $\int \frac{20 \cdot \sqrt{3}}{9} dt = \frac{20t \cdot \sqrt{3}}{9} \frac{m}{s}$ Now we want to find when the distance traveled is $0.5$ meters Therefore, we find the value $x$ such that $\int_{0}^{x} \frac{20t \cdot \sqrt{3}}{9} dt = 0.5$ $\int_{0}^{x} \frac{20t \cdot \sqrt{3}}{9} dt = 0.5 \\ \frac{10x^2 \cdot \sqrt{3}}{9} - \frac{10 \cdot 0^2 \cdot \sqrt{3}}{9} = 0.5 \\ \frac{10x^2 \cdot \sqrt{3}}{9} = 0.5 \\ 10x^2 \cdot \sqrt{3} = 4.5 \\ x^2 = \frac{3 \cdot \sqrt{3}}{20} \\ x = \sqrt{\frac{3 \cdot \sqrt{3}}{20}} \approx \boxed{0.510}$ The actual answer was $\boxed{0.519}$ . If someone can explain to me what I did wrong, that would be great.

Consider at time $t$ when the rope has slid a length of $x$ on the right side. The force on the rope is due to the weight on the right side minus that on the left side, that is:

$\begin{aligned} F & = {\color{#3D99F6}\lambda} (1+x - (0.5-x)) g \cos 30^\circ & \small \color{#3D99F6} \text{where }\lambda \text{ is the unit mass of the rope.} \\ {\color{#D61F06}m\ddot x} & = {\color{#3D99F6}\frac 23} \left(\frac 12+2x \right) \cdot 10 \cdot \frac {\sqrt 3}2 & \small \color{#D61F06} \text{where } m \text{ and } \ddot x \text{ are the mass and acceration of the rope.} \\ \ddot x & = \frac 5{\sqrt 3} (1+4x) \\ \sqrt 3 \ddot x & = 20 x + 5 & \small \color{#3D99F6} \text{Solving the 2nd order linear ODE.} \\ x(t) & = c_1 e^{\frac {2\sqrt 5}{\sqrt[4]3}t} + c_2 e^{-\frac {2\sqrt 5}{\sqrt[4]3}t} - \frac 14 & \small \color{#3D99F6} \text{Putting }t = 0 \implies c_1 + c_2 = \frac 14 \\ \dot x(t) & = \frac {2\sqrt 5}{\sqrt[4]3} \left(c_1 e^{\frac {2\sqrt 5}{\sqrt[4]3}t} - c_2 e^{-\frac {2\sqrt 5}{\sqrt[4]3}t}\right) & \small \color{#3D99F6} \text{As velocity }\dot x(0) = 0 \implies c_1=c_2 = \frac 18 \\ x(t) & = \frac 18 \left(c_1 e^{\frac {2\sqrt 5}{\sqrt[4]3}t} - c_2 e^{-\frac {2\sqrt 5}{\sqrt[4]3}t}\right) - \frac 14 \\ & = \frac 14\cosh \left(\frac {2\sqrt 5}{\sqrt[4]3}t\right) -\frac 14 \end{aligned}$

When all the rope is just on the right side, $x = \frac 12$ , then:

$\begin{aligned} \frac 14\cosh \left(\frac {2\sqrt 5}{\sqrt[4]3}t\right) -\frac 14 & = \frac 12 \\ \cosh \left(\frac {2\sqrt 5}{\sqrt[4]3}t\right) & = 3 \\ \frac {2\sqrt 5}{\sqrt[4]3}t & = \ln(3+2\sqrt 2) \\ \implies t & = \ln(3+2\sqrt 2) \cdot \frac {\sqrt[4]3}{2\sqrt 5} \\ \approx \boxed{0.519} \end{aligned}$