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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 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. · 2 years ago

$$N$$ 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 $$v_t = \sqrt{\frac{T}{\lambda}}$$ · 2 years ago

Try this. Tension in the Ring! · 2 years 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 known · 2 years ago

I got second one. · 2 years ago

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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 $$T=\frac { M }{ R2\pi } R$$ · 2 years ago

your answer to $$T$$ is dimensionally incorrect. Correcr answer = $$T = \frac{M}{2\pi}g$$ · 2 years ago

I missed g in last line. It is a typo. Thanks for pointing. :) · 2 years ago

if the cone is rotated about its axis with angular velocity $$\omega$$ what will be the tension in the ring now?? · 2 years 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. · 2 years ago

is it possible to find plz reply · 2 years ago

what if the cone is rough?? and we have to calculate minimum coefficient of friction so that the ring remains in equilibrium. · 2 years ago

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 $$\mu$$. · 2 years ago

the friction will act upwards along the slant height?? · 2 years ago

No the vertical components are balanced. I think it will act along the tangent opposite to the direction of angular velocity. · 2 years ago

may be both are possible. · 2 years 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. · 2 years 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? · 2 years ago

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. · 2 years ago

Oh! I got it. The ring is not rotating with the cone! Thanks Raghav! :) · 2 years ago

I think the first problem's answer is $$21/8 s$$ · 2 years ago

i want the method · 2 years ago

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. · 2 years ago

What are the answers? · 2 years ago