Question

A circular rope of weight W and radius r = $\frac{R}{2}$ is resting on a smooth sphere of radius R. The tension in the rope is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

So what do we know

r = $\frac{R}{2}$ --------------------(I)

where R is the radius of sphere

r is the radius of circular rope

To find tension in the rope let's look at the smaller section of the rope from top view.

Tension in the string is same T.

Here we have taken a small dl length of the rope and zoomed in to see. Let's assume that the angle that this section forms at the centre is d$\theta$ and I have also divided this symmetrically just for convenience.

That's why$\frac{d\theta }{2}$.

I guess you all recall now we find plane angles

eg:- a circle of radius r. find the angle of a quarter arc at the centre.

Find the angle made by arc AB at centre. I know it looks ${90}^{\circ }$ or $\frac{\pi }{2}$ but how it actually works is that we take the arc length in question and divide it by the radius to get the angle made by it at the centre.

So since its quarter arc

So its length is $\frac{2\pi r}{4}=\frac{\pi r}{2}$

So angle at center is $\frac{\frac{\pi r}{2}}{r}=\frac{\pi }{2}$

Similarly here I have arc length as dl radius = r

angle at center = $d\theta$

$\Rightarrow \frac{dl}{r}=d\theta$

dl = rd$\theta$ --------------------(ii)

now weight of the entire rope is w then what will be the weight of this segment of rope. for $2\pi r$ lenght w weight

1 unit = $\frac{w}{2\pi r}$

dl unit =$\frac{w}{2\pi r}dl=\frac{w}{2\pi r}rd\theta$

= $\frac{wd\theta }{2\pi }$ weight ------------------(iii)

Now tension as we see in diagram will be tangential to the rope ends.Resolving it along x and y we get

$\sum T_y=2Tsin\left(\frac{d\theta }{2}\right)$ $\sum T_x=Tcos\frac{d\theta }{2}-Tcos\frac{d\theta }{2}=0$

$\sum T_y=2Tsin\left(\frac{d\theta }{2}\right)$

sin$\theta \approx \theta$ if $\theta$ is very very small

$\Rightarrow \sum T_y=2T\left(\frac{d\theta }{2}\right)=td\theta$ ---------------(iv)

Now lets look at side view of the same section.

=$\frac{wd\theta }{2\pi }$

sin$\theta =\frac{R}{2R}=\frac{1}{2}$

$\Rightarrow \theta ={30}^{\circ }$

Since equilibrium condition Rope isn't moving so net force on it will be 0

$\Rightarrow Td\theta =Nsin\theta$ ---------------(v)

$\frac{wd\theta }{2\pi }=Ncos\theta$ -----------------(vi)

Dividing (v) & (vi)

$T=\frac{w}{2\pi }tan\theta$ $(\theta ={30}^{\circ })$

T = $\frac{w}{2\pi \sqrt{3}}=\frac{w}{\pi \sqrt{12}}$