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Kavin Ishwaran
Kavin Ishwaran
05:32
@JohnRennie Hi !
John Rennie
John Rennie
Hi :-)
Kavin Ishwaran
Kavin Ishwaran
3
Q: How ring expansion is possible here?

Kavin IshwaranRecently I have came across this question. I went with the 3rd option as ring expansion cannot occur due to the formed stable 3 degree carbocation. But the answer key claims 4th option. How can product through the ring expansion be the major product here ? As the 3 degree carbocation is stable,...

reaction-mechanism stability carbocation
John Rennie
John Rennie
I have no idea. Sorry :-(
Kavin Ishwaran
Kavin Ishwaran
Its Ok :-)
Kavin Ishwaran
Kavin Ishwaran
06:28
@JohnRennie A simple pendulum of length 1 m is oscillating with an angular frequency of 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10^−2m. The relative change in the angular frequency of the pendulum is best given by:
John Rennie
John Rennie
I would guess the acceleration up and down causes the effective value of 𝑔 to change i.e. it's 9.81 ± the acceleration of the suppiort.
Kavin Ishwaran
Kavin Ishwaran
Yes ! I also think the same, but I am having trouble finding the g eff
John Rennie
John Rennie
For the support we have:
y = 0.01 sin(t)
so:
d²y/dt² = -0.01 sin(t)
We get g = 10 ± 0.01 i.e. a change of 0.1%
Kavin Ishwaran
Kavin Ishwaran
the effective value of g will vary throughout the oscillation right as the support of pendulum undergoes shm, i.e a is proportional to y
John Rennie
John Rennie
Yes, but we just need the max and min values of 𝑔 as these will give the max and min values of the frequency of the pendulum.
And the max and min values are g = 10 ± 0.01 i.e. a change of 0.1%
Yes?
Kavin Ishwaran
Kavin Ishwaran
06:37
@JohnRennie "d²y/dt² = -0.01 sin(t)"
John Rennie
John Rennie
Yes ...
Kavin Ishwaran
Kavin Ishwaran
a = -0.01y
How it gives the change in g ?
Oh wait
John Rennie
John Rennie
y = 0.01 sin(t) so a = -y not -0.01y
Kavin Ishwaran
Kavin Ishwaran
Yes I get it :-)
John Rennie
John Rennie
OK :-)
Kavin Ishwaran
Kavin Ishwaran
06:45
So we can use error method to find the relative change in angular frequency right
John Rennie
John Rennie
@KavinIshwaran Yes
Δg = 0.1% and f ∝ √g so Δf = 0.05%
Kavin Ishwaran
Kavin Ishwaran
Yes
Kavin Ishwaran
Kavin Ishwaran
07:14
@JohnRennie They have given the relative change in angular frequency as 10^-3 rad/s
John Rennie
John Rennie
That's not a relative change. That's an absolute change.
So that's a change of 0.01% ...
I don't understand how they got that answer.
Kavin Ishwaran
Kavin Ishwaran
@JohnRennie I have found a solution online
byjus.com/question-answer/…
But i can't understand what they are doing..
John Rennie
John Rennie
The solution says:
For the vertical oscillations of the support, wₛ = √g/A
But that doesn't make any sense at all.
Kavin Ishwaran
Kavin Ishwaran
If we rearrange that term, we get Aw^2 = g
John Rennie
John Rennie
07:29
But then they calculate dg/dĪ‰ and arbitrarily set that equal to dg. Huh?
Kavin Ishwaran
Kavin Ishwaran
It doesn't make any sense to me
John Rennie
John Rennie
Me neither.
But we got 0.05% and that's ¹â„â‚‚ × 10âģ³
So A is the closest value
Kavin Ishwaran
Kavin Ishwaran
I see
Kavin Ishwaran
Kavin Ishwaran
07:51
@JohnRennie Aw^2 is maximum acceleration of SHM
Are they saying g is the maximum acceleration of vertical ?
John Rennie
John Rennie
I think that Byjus article is rubbish and gets the right answer by pure luck.
Our method is correct and it does get the correct answer (within a factor of 2)
Kavin Ishwaran
Kavin Ishwaran
Yeah
 
3 hours later…
Kavin Ishwaran
Kavin Ishwaran
10:54
@JohnRennie I found why the answer is given twice that of the result we obtained. I think the question expects us to take change in g to be difference between minimum g_eff - maximum g_eff which is 2a. (a = 0.01)
John Rennie
John Rennie
Hi :-)
@KavinIshwaran Yes, possibly.
Kavin Ishwaran
Kavin Ishwaran
:-)
@JohnRennie Are you free ?
John Rennie
John Rennie
Yes :-)
Kavin Ishwaran
Kavin Ishwaran
suppose there is a heavy uniform rope which hangs in a support, how to find the elongation due to its own weight?
My approach was, I took the COM and which will be l/2 distance from bottom, and calculated the elongation for the rest half of the rope.
F =Mg/2
L - l/2
A is cross sectional area (constant)
Y = young's modulus
And we apply FL/AY = e
John Rennie
John Rennie
You do it by dividing the rope into infinetesimal lengths dx, finding the elongation due to the weight of the rope below the segment, then integrating.
Kavin Ishwaran
Kavin Ishwaran
11:05
Oh
John Rennie
John Rennie
Suppose we have a dx at a distance x from the bottom end (we'll integrate upwards).
The the elongation is k (x/𝓁) m dx
where 𝓁 is the length of the rope and m is the mass. 𝑘 is some force constant that will be related to the Young's modulus and area as you say.
Kavin Ishwaran
Kavin Ishwaran
@JohnRennie I framed the equation like de = k(M/l)g(l-x)dx
John Rennie
John Rennie
Yes, that's taking x = 0 at the top rather than the bottom.
Kavin Ishwaran
Kavin Ishwaran
@JohnRennie Got the answer :-)
John Rennie
John Rennie
11:21
OK :-)
Kavin Ishwaran
Kavin Ishwaran
It turns out to be the same result as we do with finding COM
MLg/2AY
John Rennie
John Rennie
I guess the expansion is linear with position, so it will average out to be the same as the expansion at the centre of the rope.
Kavin Ishwaran
Kavin Ishwaran
where 1/AY is the k in our equation
@JohnRennie Yes
 
3 hours later…
Bumblebee
Bumblebee
14:22
Hi @JohnRennie :)

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