Rope through hole

Classical Mechanics Level 3

A rope of length $\SI{90}{\centi\meter}$ lies in a straight line on a frictionless table, except for a very small piece at one end which hangs down through a hole in the table.

This piece is released, and the rope slides down through the hole. What is the speed $($ in $\text{m/s})$ of the rope (to 2 decimal places) at the instant it loses contact with the table?

Details : $g = \SI[per-mode=symbol]{9.81}{\meter\per\second\squared}.$

The answer is 2.97.

4 solutions

Denton Young
Mar 31, 2017

Relevant wiki: Conservation of Mechanical Energy

Let $σ$ be the mass density of the rope. From conservation of energy, we know that the rope’s final kinetic energy, which is $((σL)v^2)/2$ , equals the loss in potential energy. This loss equals $(σL)(L/2)g$ , because the center of mass falls a distance L/2.

Therefore, v = $\sqrt{gL}$ .

After substituting the values $g = 9.81 \text{ m/s}^2$ and $L = 0.9 \text{ m}$ , we get $v = 2.97 \text{ m/s}$ .

Nice solution. For some reason I tend to first go with the kinematics approach and often regret the effort. If $x(t)$ is the length of the rope that has slipped off the table after time $t$ and $x_{0}$ the initial length of rope hanging down through the hole then we end up with the differential equation

$\dfrac{d^{2}x}{dt^{2}} = \dfrac{g}{L}x$ with initial conditions $x(0) = x_{0}$ and $x'(0) = 0$ .

This has solution $x(t) = x_{0}\cosh\left(\sqrt{\dfrac{g}{L}}t\right)$ .

Letting $T$ be the time when the all the rope has slipped off the table, we have $x(T) = L$ , which when solved for $T$ gives us

$T = \sqrt{\dfrac{L}{g}}\cosh^{-1}\left(\dfrac{L - x_{0}}{x_{0}}\right)$ .

Then $x'(T) = x_{0}\sqrt{\dfrac{g}{L}}\sinh\left(\cosh^{-1}\left(\dfrac{L - x_{0}}{x_{0}}\right)\right)$ .

Now in general $\sinh(\cosh^{-1}(y)) = \sqrt{y^{2} - 1}$ . Applying this identity and simplifying gives us that

$x'(T) = \sqrt{gL} \times \sqrt{1 - \dfrac{2x_{0}}{L}}$ , which goes to $\sqrt{gL}$ as $x_{0} \to 0$ .

This approach gives us more information in the end, but is overkill for the purpose of solving this problem.

Brian Charlesworth - 4 years, 2 months ago

umm from the first equation can't we do the following?

$a_{avg} = \dfrac{\lim_{x \to 0} \frac{g}{L}x + \lim_{x \to L} \frac{g}{L}x }{2} = \dfrac{g}{2}$

and we plug it into this equation $v_{f}^2 = v_{i} ^ 2 + 2 * a * d = 0 + 2 * \frac{g}{2} * L = g * L$ .

and can you please explain how did you get the equation $\dfrac{d^{2}x}{dt^{2}} = \dfrac{g}{L}x$ ?

Mehdi K. - 4 years, 1 month ago

As for how I got the equation, note that the force on the rope is the weight of the portion $x$ of the rope that hangs down through the hole. If the mass of the entire rope is $M$ , then the mass of the portion $x$ is $\dfrac{x}{L} \times M$ . Now the acceleration of the entire rope is the same as the acceleration of the portion $x$ , so by Newton's Second Law

$M\dfrac{d^{2}x}{dt^{2}} = \left(\dfrac{x}{L} \times M\right)g \Longrightarrow \dfrac{d^{2}x}{dt^{2}} = \dfrac{gx}{L}$ .

Your first point is interesting. As you note the acceleration is linear $x$ , so the average acceleration with respect to x can be calculated as you have done. However, it is not linear with time, which is how we normally express it, and the formula you use for $v_{f}^{2}$ can generally only be used in the case where acceleration is constant with respect to time. So it would appear that your approach yielded the correct answer by good fortune. :)

Brian Charlesworth - 4 years, 1 month ago

@Brian Charlesworth – Thank you for clearing that up.

Mehdi K. - 4 years, 1 month ago

Puts in 90 mm... facepalm

Jerry McKenzie - 4 years, 1 month ago

Actually finding velocity was quite easy, why dont lets find acceleration when it has fallen through a distance x and also the impulse provided on the foor due to floor due to falling rope and equation of rope left on table as a function of time...

Aarsh Verdhan - 4 years, 2 months ago

Tables are typically about 72cm high. Are we to assume that this table is also, or unusually more than 90cm high?

Will Irvine - 4 years, 1 month ago

I think yes, we need to assume that the table is high enough. The diagram on the right also shows the same.

Rohit Gupta - 4 years, 1 month ago

Nour Kfoury
Apr 17, 2017

Logically, for the rope to slide down half of its length should be hanging down through the hole so the center of gravity loses its balance but if we want to take it that way we can solve it with the law of conservation of mechanical energy:

Em( before sliding) =Em (after sliding down)

If we take the reference level 90 cm down from the table

$\frac{1}{2}mV^2 + mgh = \frac{1}{2}mv^2 + mgH$

( V=0 since the rope was with no speed at the beginning )

$\frac{1}{2}mv^2 = mg(h-H)$ ; ( m is different from 0 so we can ÷ with m)

$\frac{1}{2}v^2 = g ( h-H)$ ;

$v^2 = 2 g ( h-H )$ ;

$v^2 = 8.829$ ;

$v = 2.97 \text{m/s}$

but I still insist that this small piece that is hanging down can't pull the rope down through the hole

I do not think this solution is correct. Because we have an (imaginary) frictionless table, and the problem asserts that the rope slides through the hole, like a rope through a pulley, this is no different from a weight falling vertically 90cm - how long does that take and what speed is it falling when the rope clears the hole. center of mass does not matter here. and half the length of the rope does not factor in. you need ot use the whole length to calculate the problem as worded.

in fact, I believe it is a mass independent problem.

how fast is a weight going if it starts at zero velocity and falls 90cm? Vf, which we want, is Vi (which is zero) + at, so we need t

the distance is 90cm. 0.90 = ViT + 1/2 a t ^2. the first part is zero, so 1/2 a T^2 = .90 m

t = .428 sec.

vf = 9.8 * .428 = 4.2m/s

william wrightson - 4 years, 1 month ago

My answer also. Frictionless pulley and frictionless table are equivalent. Find the time for an object to fall .9m and then find the instantaneous velocity at that time. Answer is 4.2 m/s

Raymond Peterson - 4 years, 1 month ago

I dont know how you are considering frictionless pulley and table equivalent in case of pulley com is not important while it becomes so in table

Aarsh Verdhan - 4 years, 1 month ago

This was also my thinking. I was surprised to get this question wrong.

Sridev Humphreys - 4 years, 1 month ago

This is the answer I submitted, but it was rejected by Brilliant

Patrick Tawil - 4 years, 1 month ago

Did you take the acceleration of the rope equal to acceleration due to gravity? If yes, then that is not correct, it will vary with the fall of the rope.

Rohit Gupta - 4 years, 1 month ago

Guys you are not considering the center of mass which is here at the center of the rope because every point is of equal mass so the distance is 45 cm and not 90cm

Nour Kfoury - 4 years, 1 month ago

Why is the acceleration is equal to the acceleration due to gravity? Won't the table exert a force on the rope?

Rohit Gupta - 4 years, 1 month ago

Consider the end of the rope that slides on the table and falls through the hole last. The force that is pulling that end of the rope is a tension force. By Newton's Third Law, there must also be a tension force pulling upwards against gravity on the end of the rope that falls through the hole first. Thus, the acceleration that the rope experiences is not a constant 9.81 m/s^2, which is what you assumed, and where you went wrong.

maxalabama . - 4 years, 1 month ago

Hey I'm new to this, can someone please explain to me why kinematics won't work? I tried ${V_f}^2 = {V_i}^2 + 2ad$ , so ${V_f}^2 = 17.658$ and rooted both sides to get 4.20. Also isn't it impossible for the rope to move like that without a big weight difference at the tip of the rope where the hole is?

John Smith - 4 years, 1 month ago

You haven't considered the center of gravity which is at the center of the rope because every point is of equal mass so the distance will be 45 cm and not 90cm that's if you take the reference level 90cm down the table

Nour Kfoury - 4 years, 1 month ago

This eq is only valid for motion with constant acceleration but u can see the acceleration will be function of x in this case

Aarsh Verdhan - 4 years, 1 month ago

You can use the law of conservation of mechanical energy whenever the movement is frictioneless

Nour Kfoury - 4 years, 1 month ago

"Logically, for the rope to slide down half of its length should be hanging down through the hole so the center of gravity loses its balance" I don't get this point. The rope should slide if the table is smooth even if a small fraction of it is overhanging, shouldn't it?

Rohit Gupta - 4 years, 1 month ago

If you're sitting on the edge your legs are hanging down and the surface is smooth would you slide down? No because your legs are a small part of your mass and the bigger part is holding you in your sitting position on the top of the building ( for exemple) so in order for the rope to slide down the force that is pulling it down should be bigger than the force that is keeping it up ( the reaction of the table)

Nour Kfoury - 4 years, 1 month ago

Well, the upward force is the normal reaction and the downward force is the gravitational pull. The normal reaction will be smaller than $mg$ because some part of the rope is hanging which won't add up in the weight on the table.

Rohit Gupta - 4 years, 1 month ago

@Rohit Gupta – I feel that the ever so slight pull on the end of the rope by gravity will cause a sideways motion on the rest of the rope on the table. In a normal world with friction the mass of the rest of the rope would be enough to counteract this sideways force-- but in a frictionless environment it would not be. The rope would have to travel sideways, and in doing so more of the rope falls through the hole, causing greater forces pulling on the rest of the rope. Basically you do not want to be sitting on a the edge of a frictionless surface with legs hanging over-- you will fall off. ;-) Speaking from personal experience sitting on icy surfaces...

D H - 4 years, 1 month ago

Agree, when there is no friction, the rope should fall down and dont stop, because there is no friction that hold the rope

John Hendrick Halim - 4 years, 1 month ago

@Nour Kfoury h - H = 0.9 ? or 0.45 ?

Bahout Hani - 4 years, 1 month ago

For the energy conservations, the height of the center of mass should be considered. Thus, $H-h = 0.45 \text{ m}$ .

Rohit Gupta - 4 years, 1 month ago

I think you over complicated this. All we need to consider is the center of mass's location when the rope leaves the table since it is frictionless. The center of mass is 45 cm down. so as v= root 2gh where h = 0.45m =Root (8.829) =2.971 m/s

David Turner - 4 years, 1 month ago

Alex Li
Apr 19, 2017

By conservation of energy:

$\Delta$ Potential Energy = $\Delta$ Kinetic Energy

$mg \Delta h = 1/2mv^2$

$g \Delta h = 1/2v^2$

The change in height is more formally stated as the change in the height of the center of mass. Therefore, $\Delta h = 9/2$ .

We then solve the resulting equation to get 2.97

Exactly. Simplest and thus best answer so far.

David Turner - 4 years, 1 month ago

Wow, I really did this the hard way now that I see how simply it works with conservation of energy. I had to use differential equations. Let x be the distance the rope falls in time t. By Newton's 2nd Law: (x/L)mg=ma. This can be rewritten as x''-(g/L)x=0. From there, solve for x in terms of t and take a derivative to get v in terms of t. Then, solve for t in terms of x and substitute in to the velocity equation plugging L in for x. You won't need to solve for the constant that pops out of the solution to the differential equation which seems to be an infinitesimal anyway.

Ben Bruemmer - 4 years, 1 month ago

Wouldn't it depend on how tall the table is? If it is only, say, 45 cm high,. then I think it would lose some momentum as contact with the floor would create a (possibly slight, but not non-existent) upward force on the falling column of rope?

Roger Cline - 4 years, 1 month ago

It would, but, though it isn't really stated in the problem, we're assuming that the rope won't hit the floor. Perhaps that bit should be added.

Alex Li - 4 years, 1 month ago

J M
Apr 17, 2017

Neglecting friction:

$a=v/t$ and

$v=at$ , but $t=d/v$

therefore: $v=a(d/v)$

Solve for $v$ :

$v\times v = v^2 = ad$

Square root of both sides

$v= \sqrt {ad} = \sqrt{9.81 \times 0.9} = \sqrt{8.829} = \boxed{2.97 \text{ m/s}}.$

the v in the first and second equation is $v_f$ (final velocity) and the one in the third equation is $v_{avg}$ (average velocity) = $v_{avg} = \frac{v_i + v_f}{2} = \frac{0 + a . v_i}{2} = \frac{a . v_i}{2}$ , so the final equation would be $v_f = \sqrt{ 2 . a . d}$ . Am I wrong here?

Mehdi K. - 4 years, 1 month ago

Why the acceleration of the rope is equal to the acceleration due to gravity? Won't the table exert any force on the rope?

Rohit Gupta - 4 years, 1 month ago

$a = v/t$ is correct only if the acceleration is constant. Why would the acceleration be constant in this case? It seems to me that the acceleration of the rope will increase as the rope falls through the hole.

Pranshu Gaba - 4 years, 1 month ago

The acceleration is constant because only gravity is acting on the rope and there is no friction. I think you are thinking that velocity would not be constant.

Jesse Drevelton - 4 years, 1 month ago

Please look at Brian's comment in Denton's solution. The acceleration is not constant with time. As more and more rope falls through the rope, the normal reaction decreases, so the acceleration increases with time.

Pranshu Gaba - 4 years, 1 month ago

Rope through hole

The answer is 2.97.

4 solutions

0 pending reports