Challenges in Statics

Classical Mechanics Level 5

Imagine a rope, attached to two points and whirling about the line joining those points at a constant angular velocity ω \omega .

Now you start to view the rope in a rotating frame as well so that it appears something as in the figure, and you wish to calculate the area under the curve that describes the rope in your frame.

All you have is a protractor and hence you know the angle the rope makes at the two points (A and B) with the axis.

You also know the minimum value of tension in the rope (don't ask me how), and the mass per unit length of the rope is λ \lambda .

So if the area can be expressed as:

l n ( A s e c θ + B t a n θ A s e c θ − B t a n θ ) C ( T ) D ( λ ω 2 ) ln(\frac { Asec\theta +Btan\theta }{ Asec\theta -Btan\theta } )\frac { C(T) }{ D(\lambda { \omega }^{ 2 }) }

and λ \lambda is the linear density, T T is the minimum tension in the rope and ω \omega is the angular velocity of rotation,

then find A+B+C+D.

Details and Assumptions

1) There is no gravity.

2) Assume ideal conditions;

3) It might help to think about why the rope possesses a different shape in the first place;

In the spirit of JEE, I am making it an integer type question;
As usual, this is original.
2 3 6 5 7 8 4 1

Mvs Saketh by Mvs Saketh , 24, India

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1 solution

Mvs Saketh
Feb 17, 2015

Ok consider a small rope element,

The vertical gradient of Tension at that point should support the centripetal force and the horizontal gradient should be 0 as no external force is there along it,

so let the angle of the tangent at that point be θ \theta ,

Then we have

d ( T s i n θ ) = − λ w 2 y d l d ( T c o s θ ) = 0 d(Tsin\theta )=-\lambda { w }^{ 2 }ydl\\ \\ d(Tcos\theta )=0\\ \\

Now let tension at θ = 0 \\ \theta =0 be T,

Then we have the horizontal component tension at any angle as

T = T 0 s e c θ T={ T }_{ 0 }sec\theta

(dont worry that it may be negative, it is just that it refers to the direction of horizontal component, right or left)

Now , put in the first equation to get

d ( T s i n ( θ ) ) = T 0 ( s e c θ ) 2 d θ = T 0 ( 1 + ( y ′ ) 2 ) 1 + ( y ′ ) 2 y ′ ′ d x = T 0 y ′ ′ d x = − λ w 2 y d l = − λ w 2 y 1 + ( y ′ ) 2 d x s o u s i n g d y ′ / d x = y ′ ′ T 0 λ w 2 ∫ t a n θ t a n ( − θ ) d ( y ′ ) 1 + ( y ′ ) 2 = − ∫ 0 X y d x = − ( a r e a u n d e r c u r v e ) = T 0 λ w 2 l n ( s e c − t a n s e c + t a n ) s o A = T 0 λ w 2 l n ( s e c + t a n s e c − t a n ) A = B = C = D = 1 A + B + C + D = 4 d(Tsin(\theta ))={ T }_{ 0 }(sec\theta )^{ 2 }d\theta \quad =\quad { T }_{ 0 }\frac { (1+(y')^{ 2 }) }{ 1+(y')^{ 2 } } y''\quad dx\quad =\quad { T }_{ 0 }y''dx\quad \\ \\ =\quad -\lambda { w }^{ 2 }ydl\quad =-\lambda { w }^{ 2 }y\sqrt { 1+(y')^{ 2 } } dx\quad \\ \\ \\ so\quad using\quad dy'/dx\quad =\quad y''\\ \\ \frac { { T }_{ 0 } }{ \lambda { w }^{ 2 } } \int _{ tan\theta }^{ tan(-\theta ) }{ \frac { d(y') }{ \sqrt { 1+(y')^{ 2 } } } } \quad =\quad -\int _{ 0 }^{ X }{ ydx\quad =\quad -\quad (area\quad under\quad curve)\quad =\quad } \frac { { T }_{ 0 } }{ \lambda { w }^{ 2 } } ln(\frac { sec\quad -\quad tan }{ sec\quad +\quad tan } )\quad \\ \\ so\quad A=\frac { { T }_{ 0 } }{ \lambda { w }^{ 2 } } ln(\frac { sec\quad +\quad tan }{ sec\quad -\quad tan } )\\ \\ A=B=C=D=1\\ \\ A+B+C+D=4\\

3 Helpful 0 Interesting 0 Brilliant 0 Confused

This is a really nice question. Thanks for sharing it :).

Essentially the same math/logic as that of the catenary works here.

I'm guessing you asked for the area to make the calculations tractable?

Also, do you think something like this can come in JEE?!(I ask because you mention 'in the spirit of JEE') I really don't think all of this can be 'figured out' in three minutes.

Shashwat Shukla - 6 years, 3 months ago

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No, i do not think so, usually JEE physics is usually easier and yes on an average, one can only allot 2-3 minutes on a question and this one definitely looks weird even though it is simple once figured out, i just wanted to make it integer type

yes , it is infact that the differential equation was to complicated to figure out y as a function of x itself which was originally my thought, but i would appreciate if you can do that and share it, because i failed,

But it is much easier to solve in a uniform gravitational field, it is amazing how a change as small as the dependence of 'y' on the force complicates it so much that we can find the area easily but not so simply the function itself,

Mvs Saketh - 6 years, 3 months ago

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Yep, the extra y does make a good deal of difference. I was able to find y in terms of x but it gets messy towards the end.

Let λ ω 2 T 0 = K \frac{\lambda \omega ^2}{T_{0}} =K .

Then we have:

y ′ ′ 1 + ( y ′ ) 2 = K y \frac{y''}{\sqrt {1 + (y')^2}}=Ky

Multiplying both sides by y ′ y' and integrating: ∫ y ′ y ′ ′ 1 + ( y ′ ) 2 = ∫ K y y ′ \int \frac{y'y''}{\sqrt {1 + (y')^2}}=\int Kyy'

Thus, 2 1 + ( y ′ ) 2 = K y 2 + C 2\sqrt {1 + (y')^2}=Ky^2+C .

At x=0 , we have y = 0 , y ′ = t a n θ y=0,y'=tan\theta . This gives C = 2 s e c θ C=2sec\theta .

Squaring, rearranging and taking the square root again we get: 2 ( y ′ ) = ( K y 2 + 2 s e c θ − 4 ) 2(y')=\sqrt{(Ky^2+2sec\theta-4)}

This is easily integrated. What you get in the end is a quadratic in y, containing a term of the form A e B x Ae^{Bx} in one of the coefficients.

Solving the quadratic gives the required function. But it's nowhere close to being as appealing as the equation of the catenary.

Shashwat Shukla - 6 years, 3 months ago

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@Shashwat Shukla – indeed nothing as awesome as

k ( e x + e − x − 2 ) = y k({ e }^{ x }+{ e }^{ -x }-2)\quad =\quad y

Mvs Saketh - 6 years, 3 months ago

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@Mvs Saketh – Precisely :)

Shashwat Shukla - 6 years, 3 months ago

Can You Please State your Bubble Problem more precisly . Answer to my query in your disput Box . Thank's a lot!

Karan Shekhawat - 6 years, 3 months ago

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