@JohnRennie hi

@AshishAhuja hi :-)

give me a second, I'll upload a picture.

How far have you got with this?

Not very far at all. If you could point me in a certain direction I can go try that out, I'm not looking for a complete solution.

If the oscillations are small the rod oscillates in the horizontal plane

So the rod oscillates about the point A in the horizontal plane..

The moment of inertia of the rod about A is m(2L)² where m is thw mass if the ball at the end of the rod.

OK so far?

Yup. I think I can go try it out now. Thank you, I will come back if I don't get it.

OK :-)

Yup, it worked.

I was missing the simplification that the rod oscillates only in the horizontal plane, and was trying to consider the string as well.

:-)

wait though

@JohnRennie I am having a hard time trying to think how the vertical string maintains it length

hmm yeah that is true, it is written inextensible?

@satan29 for small oscillations $\cos\theta \approx 1$

So the vertical position of the rod remains constant.

@JohnRennie yes, but

okay consier point b

take a horizontal line parallel to B

tsay line L

then the distance between

line

L and the rod remains the same, right?

I guess this is only possible in the theoretical situation where theta approaches 0, so in that case the distance would remain same, or at least the difference approaches 0?

but i mean the inextensibility is a constraint , isnt it

yeah that is there, not sure...

the system simply cant behave in a way that changes the length of the string

hang on, i am uploading an image

(sorry but I gtg for lunch now will be back in ~30 mins)

@satan29 the real motion is complicated, but as long as the angle is small the bob moves approximately along a line normal to the screen. That's the limit the question is asking about.

sir have a look at this

is this how you have treated the situation?

I don't want to do the question without using the approximation that the motion is in a straight line. Life is too short.

so the conclusion is that the motion is **approximately ** straight, in light of which, BX and BY are roughly equal?

hmmm i get it, in reality, upon displacing the rod will be slightly raised upwards upon displacing (to maintain BX=BY), but we are neglecting that height and assuming approximately straight l. motion

@JohnRennie right?

Yes

@AshishAhuja what answer are you getting?

i am getting w=sqrt(5g/4)

@satan29 I'm getting $\omega = \sqrt{\frac{g}{2l}} = \sqrt{\frac{5g}{2}}$

That is the given answer as well.

actually instead of doing it using MOI we can just directly plug in $2L$ in the time period of pendulum formula

@AshishAhuja i got that initially, but through a calc error

what did you do?

nvm, i made a mistake

@JohnRennie Hi!

@Wolgwang hi :-)

I am getting it around 4.5

The trajectory is symmetric. The time spent travelling up is the same as the time spent travelling down again.

Why is there a .5 difference?

The question doesn't make sense.

Uhm?

If the object took 5 seconds to reach its maximum height then the initital velocity must have been 50 m/s.

That's because g = -10 m/s² so it takes 5 seconds to slow from 50 m/s to zero at the max height.

OK so far?

Yes

Then the distance is given by v² = u² + 2as

Yes

so we get 0 = (50)² - 20s

s = 125m

O_o

But the question says s = 100m, not 125m

So the time the question gives is not consistent with the height the question gives.

@JohnRennie Yes

So there must be an error in the question.

Maximum speed=Average speed?

@Wolgwang which question are we discussing now?

@Wolgwang @JohnRennie This

Graph one

acceleration is a = dv/dt, so to get v(t) we have to integrate:

v(t) = ∫ a(t) dt

Yes?

Yes

Now we could write down the equation for a(t), but there is a shortcut. If we have a graph of a(t) then the integral is just the area under the line. Yes?

Yes

This area. Yes?

Yes

And the area of that triangle is ½ base times height.

Yes 55

Yes :-)

@JohnRennie But is this the maximum speed?

That integral is the change in velocity during the 11 seconds. Yes?

Yes

And the question tells us that the particle started from rest i.e. from v = 0.

Ohk got it

Thanks :-)

@JohnRennie Are you getting this? >_<

Yes, it did that to me as well. I'm guessing it's a false alarm as that extension hasn't been updated since 2015 so it can't have suddenly got malware.

But chrome isn't allowing me to enable it.

But there is an updated version of it you can install.

This is the updated version

:-)

Install this instead. You'll need to reimport the definitions but I can post mine here if you want.

Unfortunately I don't know how to get back any new definitions you made in the old version.

@JohnRennie Done ;-)

{
  "+-": "±",
  "-->": "⟶",
  "1/2 ": "½",
  "1/3 ": "⅓",
  "1/4 ": "¼",
  "2/3 ": "⅔",
  "3/2 ": "³⁄₂",
  "3/4 ": "¾",
  "<--": "⟵",
  "D_ ": "Δ",
  "F_ ": "Φ",
  "G_ ": "Γ",
  "L_ ": "Λ",
  "O_ ": "Ω",
  "S_ ": "Σ",
  "Y_ ": "Ψ",
  "^0 ": "°",
  "^1 ": "¹",
  "^2 ": "²",
  "^3 ": "³",
  "^4 ": "⁴",
  "^5 ": "⁵",
  "^6 ": "⁶",
  "^7 ": "⁷",
  "^8 ": "⁸",
  "^9 ": "⁹",
  "_0 ": "₀",
  "_1 ": "₁",
  "_2 ": "₂",
  "_3 ": "₃",
  "_4 ": "₄",
  "_5 ": "₅",
  "_6 ": "₆",
  "_7 ": "₇",
  "_8 ": "₈",
  "_9 ": "₉",