Bead and a Rope

A massive homogenous rope of length l l is suspended between two nails A A and B B driven in the same horizontal level. The rope makes an angle θ \theta with the horizontal at the nails and it's lowest point at a depth d d below the nails.
A bead threaded in the rope released from the nail A A slides down the rope without friction.
If the bead is so light that the shape of the rope remains unaffected find acceleration of the bead when it passes the lowest point of the rope .
Answer comes in the form of a b e a d = α g d tan θ l a_{bead}=\frac{\alpha gd \tan \theta}{l}
Find α = ? \alpha=?

Gravity is acting downward and it is g g .


The answer is 4.

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2 solutions

Karan Chatrath
Sep 9, 2020

This problem is a very good one.

Solving this requires some background knowledge. When a rope hangs between two points at the same level, the shape of the rope is a catenary. The equation of a catenary is:

y = a cosh x a y = a \cosh{\frac{x}{a}}

Where a a is constant. Let the lowermost point of the rope be the origin. The Y-axis points upwards while the X-axis points to the right. Let the coordinates of the points A A and B B be:

A = ( x o , d ) A = (-x_o,d) B = ( x o , d ) B = (x_o,d)

Now, the length of the rope is L L which can be found using the following relation:

L = x o x o 1 + ( d y d x ) 2 d x L = \int_{-x_o}^{x_o} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx

Plugging in the derivative and solving the integral gives the result:

L = 2 a sinh x o a ( 1 ) L = 2a \sinh{\frac{x_o}{a}} \dots (1)

Now, the derivative at point B B is tan θ \tan{\theta} . This gives the second equation as follows:

tan θ = sinh x o a ( 2 ) \tan{\theta} = \sinh{\frac{x_o}{a}} \dots (2)

From (1) and (2), we get:

L = 2 a tan θ ( 3 ) L = 2a \tan{\theta} \dots (3)

Now, consider the motion of the bead. By applying the energy conservation principle, the velocity acquired by the bead at the lowermost point is:

v 2 = 2 g d ( 4 ) v^2 = 2gd \dots(4)

At the lowermost point, the bead moves horizontally. the net acceleration of the bead is it's centripetal acceleration which can be found as such:

a b e a d = v 2 R a_{bead} = \frac{v^2}{R} a b e a d = 2 g d R a_{bead} = \frac{2gd}{R}

Now, to find R R , the following formula can be applied at x = 0 x=0 :

R = ( 1 + ( d y d x ) 2 ) 3 / 2 d 2 y d x 2 R = \frac{\left(1 + \left(\frac{dy}{dx}\right)^2\right)^{3/2}}{\frac{d^2y}{dx^2}}

Simplifying gives:

R = a R = a

a b e a d = 2 g d a \implies a_{bead} = \frac{2gd}{a}

Using (3) and eliminating a a gives

a b e a d = 4 g d tan θ L \implies a_{bead} = \frac{4gd\tan{\theta}}{L}

I wonder how they expect 17 year olds to do this on a timed test... lol

Nice solution btw.

Krishna Karthik - 9 months ago

@Karan Chatrath yeah it is a good one., thanks for the solution Specially posted for Steven sir , once he said, he had a love affair with this types of problem.

Talulah Riley - 9 months ago

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@Lil Doug

Yeah; he did post a lot of those "bead on parabola"; "bead on sin"; "bead on every function" type problems and kept solving them using Lagrangian mechanics.

My bead problems have a bit of a twist:

Friction and springs. That's what makes them require numerical solutions

Btw how'd your date go? XD

Krishna Karthik - 9 months ago

Btw what formula did you use for R R ?

Krishna Karthik - 9 months ago

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It is the general formula for calculating the radius of curvature of a 2D curve at any point.

Karan Chatrath - 9 months ago

@Karan Chatrath @Krishna Karthik I solved this problem without calculating the length of rope.

Talulah Riley - 9 months ago

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Tbh I'm never going to attempt a bead-curve problem ever again. I'm done with that kind of thing lol

Krishna Karthik - 9 months ago

I will post my solution after 10 minutes

Talulah Riley - 9 months ago

@Krishna Karthik You are not able to solve a bead curve problem and saying I'm done with that kind of thing.
What the fuck you want to say bro?

Talulah Riley - 9 months ago

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? I didn't solve this particular problem ; I went ahead and clicked on the "discuss solutions" button. Instead, I solved all of Steven Chase's bead curve problems.

Krishna Karthik - 9 months ago

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@Krishna Karthik who said you to do that? You have to solve bro.
Karan sir can also do that thing, but he didn't.

Talulah Riley - 9 months ago

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@Talulah Riley Why do I have to solve? Why can't I just appreciate a good solution when it comes?

Krishna Karthik - 9 months ago

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@Krishna Karthik @Krishna Karthik ok leave it bye :)

Talulah Riley - 9 months ago
Talulah Riley
Sep 9, 2020

Feel free to ask anything regarding my solution.

Talulah Riley - 9 months ago

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Nice. Yeah; I didn't think you had to calculate the value of l l anyway.

Krishna Karthik - 9 months ago

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@Krishna Karthik which earphone/headphone do you have bro?

Talulah Riley - 9 months ago

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@Talulah Riley I have a HyperX Cloud II. Gaming headset, although the audio mixing options can be adjusted for every purpose.

Krishna Karthik - 9 months ago

@Lil Doug

Btw what does the quote in your bio mean?

Krishna Karthik - 9 months ago

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@Krishna Karthik Ask your Parents, it is simple english bro.

Talulah Riley - 9 months ago

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@Lil Doug

I know, I just want to know what you make of it. Here's what I make of it:

"Some people who you think are your friends are actually just deceptive dirtbag scavengers that want to profit off of your loss."

Would that be correct? So, who's it by?

Krishna Karthik - 9 months ago

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@Krishna Karthik @Krishna Karthik little bit correct you are. I will talk some other day if it, not now.

Talulah Riley - 9 months ago

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@Talulah Riley @Lil Doug

Tell me. Why this quote? I'd like to know out of curiosity.

Krishna Karthik - 9 months ago

@Lil Doug

How did you solve Spaceship Mechanics? I've tried solving countless differential equations; my brain's swarming with maths... help pls!!!!!!!

Krishna Karthik - 9 months ago

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@Krishna Karthik Today i will not be able to help.
I will help you tomorrow. In the mean time you can ask Steven and Karan sir.
Today i have date with my girlfriend.

Talulah Riley - 9 months ago

@Lil Doug What's your bio mean this time? What's this bio? Lmao

Krishna Karthik - 8 months, 4 weeks ago

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