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Questions 2

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?

Note by Tanishq Varshney
2 years, 5 months ago

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Answer 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 - 2 years, 5 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 - 2 years, 5 months ago

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Try this. Tension in the Ring!

Nishant Rai - 2 years, 5 months ago

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answers please @Tanishq Varshney

Nishant Rai - 2 years, 5 months ago

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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 known

Tanishq Varshney - 2 years, 5 months ago

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I got second one.

Satvik Pandey - 2 years, 5 months ago

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@Satvik Pandey

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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 - 2 years, 5 months ago

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@Satvik Pandey your answer to \(T\) is dimensionally incorrect. Correcr answer = \( T = \frac{M}{2\pi}g\)

Nishant Rai - 2 years, 5 months ago

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@Nishant Rai I missed g in last line. It is a typo. Thanks for pointing. :)

Satvik Pandey - 2 years, 5 months ago

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@Satvik Pandey if the cone is rotated about its axis with angular velocity \(\omega\) what will be the tension in the ring now??

Tanishq Varshney - 2 years, 5 months ago

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

Raghav Vaidyanathan - 2 years, 5 months ago

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@Raghav Vaidyanathan is it possible to find plz reply

Tanishq Varshney - 2 years, 5 months ago

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

Tanishq Varshney - 2 years, 5 months ago

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@Tanishq Varshney Yes, it is possible to find the answer in this case. Actually, I did not get what you meant by: "the ring remains in equilibrium".

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 - 2 years, 5 months ago

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@Raghav Vaidyanathan the friction will act upwards along the slant height??

Tanishq Varshney - 2 years, 5 months ago

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@Tanishq Varshney No the vertical components are balanced. I think it will act along the tangent opposite to the direction of angular velocity.

Satvik Pandey - 2 years, 5 months ago

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@Satvik Pandey may be both are possible.

Tanishq Varshney - 2 years, 5 months ago

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@Tanishq Varshney 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.

Raghav Vaidyanathan - 2 years, 5 months ago

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@Raghav Vaidyanathan 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?

Satvik Pandey - 2 years, 5 months ago

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@Satvik Pandey Assume you have a cylindrical rod which is smooth. It is kept vertically on a smooth table. A smooth ring which just fits on the rod is put around the rod such that it lies on the table. Now, if you rotate the rod about its axis, does the ring move?

The rotation of the ring and rod are independent.

Raghav Vaidyanathan - 2 years, 5 months ago

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@Raghav Vaidyanathan Oh! I got it. The ring is not rotating with the cone! Thanks Raghav! :)

Satvik Pandey - 2 years, 5 months ago

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I think the first problem's answer is \(21/8 s\)

Raghav Vaidyanathan - 2 years, 5 months ago

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i want the method

Tanishq Varshney - 2 years, 5 months ago

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Equation of motion of first pendulum is: \[x_1=-acos(\frac{2\pi} {3}t)\]

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 - 2 years, 5 months ago

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@satvik pandey @Raghav Vaidyanathan

Tanishq Varshney - 2 years, 5 months ago

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What are the answers?

Satvik Pandey - 2 years, 5 months ago

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