The Proof is &#x27;not-so&#x27; Trivial - Physics Edition

Problem 29**
Imagine a cylindrical barrel full of water, height

H

. Cutting a small hole in the side of the barrel causes water to emerge from the hole. Prove that the emerging water travels furthest away from the barrel when the hole is cut half way between the floor and the top of the barrel.

Reply 189

Kyx

6

Original post by Llamas123

Problem 29**
Imagine a cylindrical barrel full of water, height

H

. Cutting a small hole in the side of the barrel causes water to emerge from the hole. Prove that the emerging water travels furthest away from the barrel when the hole is cut half way between the floor and the top of the barrel.

I'm guessing we have to use the fact that the water leaves the hole with more force if the hole is lower down. But the higher the hole, the more time it has to fall, so has more time to travel away from the barrel. ???

So we have to equate water pressure and suvat with respect to H ???

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Reply 190

Initially considering the problem using forces makes it much more difficult. You have to use a different technique for the first step.

Reply 191

Original post by Arieisit

Problem 22 **

A wheel consists of a uniform disk of radius

R

, with a circle of radius

R/2

removed as shown. Show the position of the centre of mass is given by

\bar{y} =R/6

and the mass moment of the wheel about the centre of mass is

\frac{37}{72}mR^2

The wheel is released from rest in the position shown. If no slipping occurs between the disk and horizontal, determine the expression for the angular velocity reached by the disk when its kinetic energy is a maximum. Also determine the reaction force between the disc and horizontal surface at that instant.

Disc has com (0, 0), Cut-out has com (0, - R/2), Wheel has com (0, y) for y > 0

Area D = πR^2 => Mass D = pπR^2, Area C = πR^2/4 => Mass C = pπR^2/4. So Mass W = (3/4)pπR^2

Take moments about (0, - R):
(3/4)pπR^2(y + R) = pπR^2(R) - pπ(R^2/4)(R/2)
=> (3/4)(y + R) = R - R/8 = 7R/8 => y + R = 7R/6
=> y = R/6 m

Mass W = m => Mass D = 4m/3, Mass C = m/3
About their centres: I_d = (4m/3)(R^2)/2, I_c = 0.5(m/3)(R/2)^2 = (m/3)(R^2)/8

// axis theorem to find I_d, I_c about wheel's com:
I_d = (4m/3)[R^2/2 + (R/6)^2] = (4m/3)(19/36)R^2
I_c = (m/3)[R^2/8 + (2R/3)^2] = (m/3)(1/8 + 4/9)R^2 = (m/3)(41/72)R^2

So I_w = I_d - I_c = (m/3)(76/36 - 41/72)R^2
=> I_w = (m/3)(111/72)R^2
=> I_w = (37/72)mR^2 kgm^2

For KE max and no slip, all GPE max converts to KE (rot) & KE (linear). Rot dist = linear dist => v (linear) = v (rot) => v (linear) = wR.

So mg(Δh max) = (1/2)m(wR)^2 + (1/2)(I_w)w^2
=> 2mgR = (1/2)w^2[mR^2 + (37/72)mR^2]
=> 2g = (1/2)(109/72)Rw^2
=> w^2 = g(4*36*2)/(109R)
=> w = 12Sqrt[2g/(109R)] rad/s

Upward N at wheel's contact w/ ground, downward mg at wheel's com. Net centrip F at contact aligns to wheel's com, at @ to vert.

@ forms RA triangle: opp = y = R/6, adj = R, hyp = R.Sqrt[1 + (1/6)^2] = R.Sqrt(37/36) = (R/6)Sqrt37
So cos@ = R/[(R/6)Sqrt37] = 6/Sqrt37

Resolve forces to action line of centrip F:
Ncos@ - mgcos@ = mRw^2 = mRg(4*36*2)/(109R)
=> (6/Sqrt37)(N - mg) = mg(288/109)
=> N - mg = mg(48/109)Sqrt37
= > N = mg[1 + (48/109)Sqrt37] N

(edited 6 years ago)

Reply 192

Original post by Llamas123

Problem 29 **

Imagine a cylindrical barrel full of water, height

H

. Cutting a small hole in the side of the barrel causes water to emerge from the hole. Prove that the emerging water travels furthest away from the barrel when the hole is cut half way between the floor and the top of the barrel.

H = barrel height, h = top-to-hole distance

For max velocity/distance from barrel, model mass m of water doing work on m at the hole; mathematically equivalent to m moving through h, transferring its KE, equal to its GPE decrease.

GPE -> KE: mgh = 0.5mv^2 => 2gh = v^2
=> v = Sqrt(2gh)

Projectile: x = Sqrt(2gh)t, y = (H - h) - 0.5gt^2
At y = 0: 0.5gt^2 = H - h => t^2 = 2(H - h)/g
=> t = Sqrt[2(H - h)/g]

So x = Sqrt(2gh).Sqrt[2(H - h)/g] = 2Sqrt[h(H - h)]

Differentiate x wrt h (or complete square): dx/dh = 2[Sqrt(H - h)0.5h^(-1/2) - Sqrt(h)0.5(H - h)^(-1/2)]

For max x: 0 = Sqrt[(H - h)/h] - Sqrt[h/(H - h)]
=> h/(H - h) = (H - h)/h
=> h^2 = (H - h)^2 = H^2 - 2hH + h^2
=> 0 = H(H - 2h)

H > h > 0 => only solution from H - 2h = 0
=> h = H/2

(edited 4 years ago)

Reply 193

Nice solution Enemy. You could have also completed the square inside the square root, line 8, to prove h=H/2 is the solution without requiring the use of calculus.

Also you do not need to write water molecules fall from the top for max horizontal velocity. You can consider the Potential energy loss and Kinetic energy gain as a small unit mass

\Delta m

of water emerges from the barrel.
Using conservation of energy, KE=0.5mv^2 and PE=mgh

\frac 1 2 \Delta m v^2 = \rho A h g \Delta x

Where

\rho

is the density of water,

A

is the mean cross sectional area of the barrel above the hole,
and

\Delta x

is the distance through which the centre of mass of the water above the hole is displaced.

This is because

\rho A h

is the mass of water displaced a over a distance

\Delta x

.

Luckily,

\Delta m = \rho A \Delta x

as the volume/mass of water must be conserved.
Therefore they cancel and we get,

\frac 1 2 v^2=gh

, which is the same result you got.

But you do not need to assume water molecules fall from the top.

(edited 6 years ago)

Reply 194

Original post by Physics Enemy

But it helps (me) since these molecules are the best case (max gpe, ke, v, x) for any hole. Or are you saying those near the hole eject as far as those from the top? I don't think your reply suggests that.

Only water molecules right next to the hole are ejected. But we can treat it has water molecules from the top of the barrel falling a distance of h and being ejected, even though this is not actually happening.

Reply 195

Why can't molecules move from the top to the hole and be ejected? Also as water ejects, water level falls and the top layer gets ejected. And surely molecules are moving around everywhere?

(edited 6 years ago)

Reply 196

Original post by Physics Enemy

Why can't water molecules move from the top to near the hole and be ejected? Also as water ejects, the water level falls and the remaining top layer of water will have to be ejected. And surely water molecules are moving around everywhere?

Water molecules can move from the top to the hole but they are not in free fall so you cannot use a simple 0.5mv^2=mgh. The speed they are ejected depends on force on them due to the weight of water above them, and it just so happens, as shown above, that 0.5v^2=gh
Also the hole is "small" so the water level never actually falls. It remains the same height as only an infinitesimally small amount of water is being ejected from the hole.
You are correct that water molecules are moving around everywhere, and this is too complicated for me to describe mathematically. But the maths and physics we have used together is still correct as we have not used any false assumptions. Therefore we have proved h=H/2.

Reply 197

Thanks. But I don't think you need to be in free-fall to lose mgh, as long as you move through mgh (under presence of gravity)? Can one consider modelling a molecule at the top losing gpe / transferring ke to a molecule at the hole?

(edited 6 years ago)

Reply 198

Original post by Physics Enemy

Thanks. But I don't think you need to be in free-fall to lose mgh, as long as you move through mgh under the presence of gravity? I could have said 'pushed through' or 'moves through' rather than 'fell' to be clearer.

Also I guess one could consider modelling the problem as a molecule from the top losing gpe and transferring ke to a molecule near the hole?

I'll edit my post to replace/clarify 'fell'.

I do not think this is correct. It implies water molecules ejected from the hole could have a range of speeds depending on how high above the hole they "moved through". But I just tried to do the experiment with a milk bottle and clearly all water molecules are ejected with the same speed ie the square root of 2gh.

I think (but I am not certain) the problem with the model is that by looking at the molecular level, you cannot ignore the continual random motion of individual molecules. Molecules at the top of the barrel still have kinetic energy.

Therefore you have to consider not the change in energy of a molecule, but that of a unit mass of water ejected from the barrel and that of the water above the hole.

Reply 199