Questions 2

1) Two pendulums of same amplitude but time period 3s3s and 7s7s start oscillating simultaneously from two opposite extreme positions. After how much time they will be in phase?

2) A ring of mass mm and radius RR rests in equilibrium on a smooth cone of semi vertical angle 45o45^{o},The radius of the cone is 2R2R. the radius of circular cross section of the ring is r(r<<R)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
4 years, 1 month 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 - 4 years, 1 month ago

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NN in the normal force acting on a small part of the ring.

dmgdm*g is the weight of that part and Tension acts in the ring along the tangent.

So N2=dmg......(i)\frac{N}{\sqrt{2}}=dmg ...... (i)

and 2Tsin(dθ2)=N2.........(ii) 2Tsin(\frac{d \theta}{2}) =\frac{N}{\sqrt{2}} .........(ii)

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

2Tsin(dθ2)=λRdθg 2Tsin(\frac{d \theta}{2})=\lambda Rd\theta g where λ \lambda is the linear mass density of the ring.

So Tdθ=λRdθgTd \theta=\lambda Rd\theta g

and T=M2πgT=\frac { M }{ 2\pi } g

For finding the speed of the transverse wave , use vt=Tλ v_t = \sqrt{\frac{T}{\lambda}}

Nishant Rai - 4 years, 1 month ago

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@satvik pandey @Raghav Vaidyanathan

Tanishq Varshney - 4 years, 1 month ago

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What are the answers?

satvik pandey - 4 years, 1 month ago

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I think the first problem's answer is 21/8s21/8 s

Raghav Vaidyanathan - 4 years, 1 month ago

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i want the method

Tanishq Varshney - 4 years, 1 month ago

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Equation of motion of first pendulum is: x1=acos(2π3t)x_1=-acos(\frac{2\pi} {3}t)

Equation of motion of the second pendulum is: x2=acos(2π7t)x_2=acos(\frac{2\pi} {7}t)

Just find tt for which x1=x2x_1=x_2 and also velocities in same direction.

Raghav Vaidyanathan - 4 years, 1 month ago

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answers please @Tanishq Varshney

Nishant Rai - 4 years, 1 month ago

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for the first one 218\frac{21}{8} and second a) mg2π\frac{mg}{2\pi} b) gr\sqrt{gr}. b) part is easy once a) part is known

Tanishq Varshney - 4 years, 1 month ago

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I got second one.

satvik pandey - 4 years, 1 month ago

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@Satvik Pandey img img

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

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

and 2Tsin(dθ2)=N/2 2Tsin(\frac{d \theta}{2}) =N/ \sqrt{2}....(2)

Using (1) and (2) we get

2Tsin(dθ2)=λRdθg 2Tsin(\frac{d \theta}{2})=\lambda Rd\theta g........................ (λ=MR2π)\lambda=\frac { M }{ R2\pi } )

So Tdθ=λRdθgTd \theta=\lambda Rd\theta g

and T=MR2πRT=\frac { M }{ R2\pi } R

satvik pandey - 4 years, 1 month ago

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

Tanishq Varshney - 4 years, 1 month 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 - 4 years, 1 month 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 - 4 years, 1 month 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 - 4 years, 1 month ago

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

satvik pandey - 4 years, 1 month 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 - 4 years, 1 month 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 μ\mu.

Raghav Vaidyanathan - 4 years, 1 month ago

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

Tanishq Varshney - 4 years, 1 month 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 - 4 years, 1 month 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 - 4 years, 1 month ago

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@Satvik Pandey may be both are possible.

Tanishq Varshney - 4 years, 1 month ago

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@Raghav Vaidyanathan is it possible to find plz reply

Tanishq Varshney - 4 years, 1 month ago

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@Satvik Pandey your answer to TT is dimensionally incorrect. Correcr answer = T=M2πg T = \frac{M}{2\pi}g

Nishant Rai - 4 years, 1 month ago

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

satvik pandey - 4 years, 1 month ago

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Try this. Tension in the Ring!

Nishant Rai - 4 years, 1 month ago

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