1) Two pendulums of same amplitude but time period \(3s\) and \(7s\) start oscillating simultaneously from two opposite extreme positions. After how much time they will be in phase?

2) A ring of mass \(m\) and radius \(R\) rests in equilibrium on a smooth cone of semi vertical angle \(45^{o}\),The radius of the cone is \(2R\). the radius of circular cross section of the ring is \(r(r<<R)\).

a) What will be the tension in the ring ?

b) What will be the speed of transverse wave on the ring?

## Comments

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TopNewestAnswer to second problem @Tanishq Varshney

Upward component of normal force balances weight of the chain. The horizontal component causes tension in ring. We can find tension in chain using the value of horizontal component. – Raghav Vaidyanathan · 1 year, 10 months ago

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\(N\) in the normal force acting on a small part of the ring.

\(dm*g\) is the weight of that part and Tension acts in the ring along the tangent.

So \(\frac{N}{\sqrt{2}}=dmg ...... (i)\)

and \( 2Tsin(\frac{d \theta}{2}) =\frac{N}{\sqrt{2}} .........(ii)\)

Using \((i)\) & \((ii)\) we get,

\( 2Tsin(\frac{d \theta}{2})=\lambda Rd\theta g \) where \( \lambda \) is the linear mass density of the ring.

So \(Td \theta=\lambda Rd\theta g\)

and \(T=\frac { M }{ 2\pi } g\)

For finding the speed of the transverse wave , use \( v_t = \sqrt{\frac{T}{\lambda}}\) – Nishant Rai · 1 year, 10 months ago

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Try this. Tension in the Ring! – Nishant Rai · 1 year, 10 months ago

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answers please @Tanishq Varshney – Nishant Rai · 1 year, 10 months ago

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– Tanishq Varshney · 1 year, 10 months ago

for the first one \(\frac{21}{8}\) and second a) \(\frac{mg}{2\pi}\) b) \(\sqrt{gr}\). b) part is easy once a) part is knownLog in to reply

– Satvik Pandey · 1 year, 10 months ago

I got second one.Log in to reply

img

In the figure \(N\) in the normal force acting on a small part of the string \((dm)g\) is the weight of that part. And Tension acts in the string along the tangent.

So \(N/ \sqrt{2}=dmg\)....(1)

and \( 2Tsin(\frac{d \theta}{2}) =N/ \sqrt{2}\)....(2)

Using (1) and (2) we get

\( 2Tsin(\frac{d \theta}{2})=\lambda Rd\theta g\)........................ (\(\lambda=\frac { M }{ R2\pi } )\)

So \(Td \theta=\lambda Rd\theta g\)

and \(T=\frac { M }{ R2\pi } R\) – Satvik Pandey · 1 year, 10 months ago

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– Nishant Rai · 1 year, 10 months ago

your answer to \(T\) is dimensionally incorrect. Correcr answer = \( T = \frac{M}{2\pi}g\)Log in to reply

– Satvik Pandey · 1 year, 10 months ago

I missed g in last line. It is a typo. Thanks for pointing. :)Log in to reply

– Tanishq Varshney · 1 year, 10 months ago

if the cone is rotated about its axis with angular velocity \(\omega\) what will be the tension in the ring now??Log in to reply

– Raghav Vaidyanathan · 1 year, 10 months ago

It will not change. The cone is supposed to be smooth. Hence the answer doesn't get affected by the cone spinning. Troll FIITJEE.Log in to reply

– Tanishq Varshney · 1 year, 10 months ago

is it possible to find plz replyLog in to reply

– Tanishq Varshney · 1 year, 10 months ago

what if the cone is rough?? and we have to calculate minimum coefficient of friction so that the ring remains in equilibrium.Log in to reply

We can find the maximum angular velocity that the rough cone can have so that the ring doesn't slip on it for a given value of \(\mu\). – Raghav Vaidyanathan · 1 year, 10 months ago

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– Tanishq Varshney · 1 year, 10 months ago

the friction will act upwards along the slant height??Log in to reply

– Satvik Pandey · 1 year, 10 months ago

No the vertical components are balanced. I think it will act along the tangent opposite to the direction of angular velocity.Log in to reply

– Tanishq Varshney · 1 year, 10 months ago

may be both are possible.Log in to reply

– Raghav Vaidyanathan · 1 year, 10 months ago

Oh, it is more complicated than I thought. Yes, it will act upwards along slant height. Also, if the cone rotates, it will also create a torque on the ring which makes it start rotating too.Log in to reply

– Satvik Pandey · 1 year, 10 months ago

I have a confusion. When the cone would be rotating then if we consider a reference frame moving with cone then we have to include centrifugal force on that small part while writing equations. Then how the tension would remain unaffected?Log in to reply

The rotation of the ring and rod are independent. – Raghav Vaidyanathan · 1 year, 10 months ago

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– Satvik Pandey · 1 year, 10 months ago

Oh! I got it. The ring is not rotating with the cone! Thanks Raghav! :)Log in to reply

I think the first problem's answer is \(21/8 s\) – Raghav Vaidyanathan · 1 year, 10 months ago

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– Tanishq Varshney · 1 year, 10 months ago

i want the methodLog in to reply

Equation of motion of the second pendulum is: \[x_2=acos(\frac{2\pi} {7}t)\]

Just find \(t\) for which \(x_1=x_2\) and also velocities in same direction. – Raghav Vaidyanathan · 1 year, 10 months ago

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@satvik pandey @Raghav Vaidyanathan – Tanishq Varshney · 1 year, 10 months ago

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– Satvik Pandey · 1 year, 10 months ago

What are the answers?Log in to reply

@Kushal Patankar @Nishant Rai @Rohit Shah @Abhineet Nayyar – Tanishq Varshney · 1 year, 10 months ago

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