Problem Solving Strategies - 2024-03-28

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3:43 AM

Im thinking of finding speed from this equation you wrote, but i fall short of equations, ig we will need to find speed from angular momentum and energy conservation only.

2 hours later…

6:10 AM

@PinkAura Orbital motion is surprisingly complicated to describe in detail.

We often resort to just using conservation of energy and angular momentum as you describe. using this leads to the Vis Viva equation:

Vis-viva equation

In astrodynamics, the vis-viva equation, also referred to as orbital-energy-invariance law or Burgas formula, is one of the equations that model the motion of orbiting bodies. It is the direct result of the principle of conservation of mechanical energy which applies when the only force acting on an object is its own weight which is the gravitational force determined by the product of the mass of the object and the strength of the surrounding gravitational field. Vis viva (Latin for "living force") is a term from the history of mechanics, and it survives in this sole context. It represents the...

6:43 AM

user image

@JohnRennie Hi !

7:05 AM

Hi :-)

How far have you got with this?

@JohnRennie Hi ! Sorry I got logged out.

So the velocity:

makes an angle 60 degrees with the horizontal

at the time of projection

when it strikes the other plane perpendicularly it will make an angle 30 degrees with the horizontal

since the horizontal component does not change, I proceeded to calculate the time taken for the vertical component to change accordingly

but this didn't work

7:24 AM

I need to go I'm afraid. Sorry :-(

I'll have a look at this as soon as I'm back.

Ok :-)

1 hour later…

8:29 AM

@KavinIshwaran I can't immediately see how to do this, and what I usually do in these circumstances is t just start and see what I can see.

user image

We don't know the positions of the points, but there are some things we do know.

Suppose the equation for the parabola is y(t)

Yes

Then we know y'(t₁) = tan(60)

And we know y'(t₂) = -tan(30)

Yes

Suppose we write y(t) = at² + bt + c then y'(t) = 2at + b

Yes

8:33 AM

Then:
2at₁ + b = tan(60)
2at₂ + b = -tan(30)

Hmm, four unknowns and two equations ...

We need to find t₂ - t₁ as that's the time of flight.

@JohnRennie My method ?

So we get 2a(t₂ - t₁) = -tan(30) - tan(60)

And we are told the launch velocity is 20 m/s

Yes

Argh! :-(

It's x on the horizontal axis not t.

I'm getting confused :-)

:-)

1 hour ago, by Kavin Ishwaran

since the horizontal component does not change, I proceeded to calculate the time taken for the vertical component to change accordingly

8:39 AM

y = ax² + bx + c
y' = 2ax + b

2ax₁ + b = tan(60)
2ax₂ + b = -tan(30)

@KavinIshwaran Ah, OK, give me a moment ...

Ok :-)

OK, I see what your method is:
v₁x = v₂x = 20 cos(30)

And we know v₁y = 20 sin(30)

Yes

And we know v₂y/v₂x = tan(30)

Yes

8:48 AM

So v₂y = 20 cos(30) tan(30)

So the time must be (20 sin(30) + 20 cos(30) tan(30))/g

i.e. the time for the vertical velocity to change

t = 20/g

Is that what you got?

No

What is the answer?

I am getting something different

One minute

4/root(3) seconds

is the correct answer

initially the vertical component of velocity is 20root(3)/2

Oops, I just realised I go the wrong equation for v₁ x and y

but when it strikes it is it is 20root3/2 but downward

8:57 AM

v₁x = v₂x = 20 cos(60) = 10

v₁y = 20 sin(60) = 10 √3

Yes

v2y is also 20root(3)/2

It is correct that v₂y/v₂x = tan(30) = 1/√3

yes

So v₂y = 10/√3
Yes?

Yes

9:02 AM

So the change in vertical velocity is:
Δvy = 10 √3 + 10/√3

Yes?

Yes

Then t = Δy/g = √3 + 1/√3

Δvy = 40/root(3)

t = 4/root(3)

:-)

OK :-)

7 mins ago, by Kavin Ishwaran

but when it strikes it is it is 20root3/2 but downward

I calculated the velocity wrong here

9:04 AM

Aha!

Thart is why I was getting wrong answer :-)

A very simple error in rearrangement !

Phew :-)

I must admit I had not spotted your method so you did better than me!

:-)

@JohnRennie A conceptual doubt. In calculating velocity of efflux why we take the pressure just outside the orifice ?

which is the atmospheric pressure

if we take it at the orifice, the pressure due to liquid column will also influence right

What matters is the pressure change, because the change in KE is related to the pressure change.

potential energy ?

9:14 AM

Well the heights are the same so there is no PE change.

We just have a velocity immediately before the orifice and a velocity immediately after it.

No, like if we take the a bucket fully filled with water, if we drill an hole at the very bottom of the bucket

Right, but what I'm saying is we are considering what happens at the orifice.

Yes

The flow is probably complicated exactly at the orifice, but we know that just inside the orifice the pressure is ρgh and just outside it is zero. Yes?

Yes

9:18 AM

(add 1 atm to both if we're using absolute pressure instead of gauge pressure)

Zero in the sense atmospheric pressure

Yes

@JohnRennie Yes

So the pressure change at the orifice is ρgh and that has to be equal to the KE change at the orifice.

There isn't any PE change as we assume the height doesn't change (much) from just inside the orifice to just outside it.

Yes

Actually I was thought differently. the Bernoulli's theorem was applied on top of the bucket and just outside orifice

9:22 AM

You can apply it between any two points.

Yes

But the point should be within the system right ?

I guess it doesn't make sense to take points that are not inside the liquid.

Yeah, that is where I am getting confused. I guess they are taking just outside the orifice to avoid liquid pressure, since the flow is normal to liquid column

I'm not sure I get what you mean.

We don't know the pressure at the orifice because the flow is complicated there.

Yes

9:27 AM

But we know the pressure outside is zero (1 atm) because the liquid is in equilibrium with the atmosphere.

And we know the pressure inside is ρgh.

So as we move from inside to outside we have a pressure change of ρgh.

Yes?

Yes

26 mins ago, by Kavin Ishwaran

@JohnRennie A conceptual doubt. In calculating velocity of efflux why we take the pressure just outside the orifice ?

That's why we take the pressure just outside.

Yes

I get it :-)

OK :-)

I was confused as I was taught by taking two different points one at top and one just outside the orifice and apply Bernoulli's theorem

9:44 AM

@KavinIshwaran You can do that as well if you want.

The point of all these calculations is they tell us what the net change must be if we start from point A and end at point B. What happens in between can involve complicated flows that may be hard to calculate. But that doesn't matter as all we care about is the net change between our two points.

Yes

I thought it was conceptually simpler to consider two points just inside and just outside the orifice as it seems to me that's easier to understand.

But you can start at the top end end just outside as well, and the calculation still works.

Either way we can't take a point at the orifice because the flow is complicated and unknown exactly at the orifice.

I need to go now. I'll be back in about an hour.

Ok :-)

2 hours later…

12:01 PM

@JohnRennie i see sir. Thanks :)

2 hours later…

2:06 PM

user image

@JohnRennie Can you please look into this problem when you are free ?

3:03 PM

@KavinIshwaran is the answer 2π/√(k/7m)

Yes

Do you want the solution?

Yes :-)

1 hour later…

4:15 PM

@KavinIshwaran sorry i went offline

user image

Thanks :-)

4:48 PM

@KavinIshwaran :)

@JohnRennie Sir my teacher told that horizontal hydrostatic pressure can be simply calculated by pressure at COM × horizontal projected area, which I am able to derive for projected area for rectangle,etc. But im unable to get this if projected area is semicircular arc (say for a half cone). Can you pls help me do it?

Sorry it is pressure at COM × vertical projected area= horizontal pressure force

Also,i searched it on google and it says, the formula is actually Avg Pressure × Vertical projected area, which one is correct?

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