Pebble falling into water; drag force

[unscientific]

Homework Statement

The drag force experienced by the spherical pebble in water is given by: (0.5)Cp_wAv²

mass of pebble = 83.8g
radius of pebble = 2 cm
C = 0.7


(a) Write down the differential equation governing pebble's descent. (done)

(b) Derive an expression for terminal velocity and evaluate it. (done)

(c) The pebble is dropped 5m above surface of lake, which is also 5m deep. Show that the pebble travels about 10cm in water before reaching it's terminal velocity.

The Attempt at a Solution

Part (a)

Part (b)

Part (c)

 

[mfb]

In your exact analysis, the ball will never reach its terminal velocity - but it will approach it with a typical timescale, and you can determine this and show what it does not travel more than 10cm before it reaches a velocity very close to the terminal velocity.

 

[unscientific]

In your exact analysis, the ball will never reach its terminal velocity - but it will approach it with a typical timescale, and you can determine this and show what it does not travel more than 10cm before it reaches a velocity very close to the terminal velocity.

Is there anything wrong with my working in part (b)?? I substituted for v² using the differential equation..

 

[mfb]

I don't see any result in c, and I don't understand what you are doing there.

 

[unscientific]

I don't see any result in c, and I don't understand what you are doing there.

sorry i meant part (b)!

 

[SteamKing]

1. Are you sure that C is the drag coefficient which uses the entire surface area of the pebble or just the cross-sectional area of the pebble?

2. You recognize that the acceleration of the pebble is x double dot. What about the velocity of the pebble? Can't the velocity also be represented in terms of x?

 

[haruspex]

1. Are you sure that C is the drag coefficient which uses the entire surface area of the pebble or just the cross-sectional area of the pebble?
The area is known from the radius, and appears as A in the equation, so it's just an arbitrary coefficient.

 

[haruspex]

1. Are you sure that C is the drag coefficient which uses the entire surface area of the pebble or just the cross-sectional area of the pebble?

The area is known from the radius, and appears as A in the equation, so it's just an arbitrary coefficient.

Sorry, misunderstood the point you were making (and for some reason it's not letting me edit the post).

 

[Chestermiller]

You need to substitute x double dot = dv/dt, and solve for the velocity vs time first. Also, what happened to the buoyant force?

 

[SteamKing]

The point about drag coefficients is that they are usually based on the cross sectional or frontal area of the body, not the total surface area.

The OP has used (4/3)pi*r^2, which is the total area of the pebble.

 

[haruspex]

The point about drag coefficients is that they are usually based on the cross sectional or frontal area of the body, not the total surface area.

The OP has used (4/3)pi*r^2, which is the total area of the pebble.

That's not how I read it. Seems to have written the mass as (4/3)pi*r^3*density, then canceled one r and π with the expression for cross-sectional area. Not that there was much point in doing that: the mass is given, the density is not.

 

[unscientific]

That's not how I read it. Seems to have written the mass as (4/3)pi*r^3*density, then canceled one r and π with the expression for cross-sectional area. Not that there was much point in doing that: the mass is given, the density is not.

sorry, density of pebble is 2.5g/cm³!

 

[unscientific]

That's not how I read it. Seems to have written the mass as (4/3)pi*r^3*density, then canceled one r and π with the expression for cross-sectional area. Not that there was much point in doing that: the mass is given, the density is not.

I'm more concerned about the integration part; to find the work done against resistance. Is there anything wrong with that?

 

[haruspex]

I'm more concerned about the integration part; to find the work done against resistance. Is there anything wrong with that?

When you lost k from your equations, you might have realized you were off course. K must feature in the answer. There are errors further along, but they're not relevant to finding a correct solution.
i don't think you can hope to integrate v²dx straight off. Work with the forces and get the differential equation for the acceleration.

 

[unscientific]

When you lost k from your equations, you might have realized you were off course. K must feature in the answer. There are errors further along, but they're not relevant to finding a correct solution.
i don't think you can hope to integrate v²dx straight off. Work with the forces and get the differential equation for the acceleration.

There are 2 ways to approach this problem:

Method 1

Substitute v² using the differential equation and integrate; which was what I did but I'm not sure what's wrong with my working...


Method 2

a) Solve the differential equation to obtain v in terms of x. show that when x = 0.1 m, v = v_term

 

[haruspex]

Substitute v² using the differential equation and integrate; which was what I did but I'm not sure what's wrong with my working...

As I said, that substitution lost k, and as soon as that happened you were going nowhere. At best you'd end up with tautology. As it happens, you must have made some other error, but I don't think it's interesting to figure out what.

a) Solve the differential equation to obtain v in terms of x. show that when x = 0.1 m, v = v_term

You have \ddot{x}=\dot{v}=g-kv^2. You could solve that to find v in terms of t, but there is a trick to eliminate t first and solve for v in terms of x. Is that what you mean?

 

[Chestermiller]

Guys,

You've left out the buoyant force on the pebble.

Chet

 

[haruspex]

Guys,

You've left out the buoyant force on the pebble.

Chet

Good catch!

 

[unscientific]

As I said, that substitution lost k, and as soon as that happened you were going nowhere. At best you'd end up with tautology. As it happens, you must have made some other error, but I don't think it's interesting to figure out what.


You have \ddot{x}=\dot{v}=g-kv^2. You could solve that to find v in terms of t, but there is a trick to eliminate t first and solve for v in terms of x. Is that what you mean?

yeah. replace dv/dt by v(dv/dx) then solve the diferential equation, putting initial condition as v = √2gh.

then substitute v = v_term to find that x = 0.1 m...

 

[unscientific]

Guys,

You've left out the buoyant force on the pebble.

Chet

Ignore the buoyant force on the pebble.

 

[unscientific]

I did some working here using method 2, to express v in terms of distance traveled x:

k = b/m = 0.4398 and 1 < α ≤ 10k

Next, i let α = 1.1 ≈ 4% above v_term

I get x = 4m = 40cm...

 

[haruspex]

There's something screwy in your second line, but after that it looks right.
It's a bit hard to guess what they consider to be effectively terminal velocity. How about Plugging in x=0.1 and see what you get for alpha?
And how do you know to ignore buoyancy? Have you checked how much difference it makes?

 

[unscientific]

There's something screwy in your second line, but after that it looks right.
It's a bit hard to guess what they consider to be effectively terminal velocity. How about Plugging in x=0.1 and see what you get for alpha?
And how do you know to ignore buoyancy? Have you checked how much difference it makes?

This is a prelim question in maths paper, and we're told to ignore buoyance force..

I got a crazy a = 4.. which is slightly below the maximum a = 10k = 4.398

 

[voko]

Are you considering that the first five meters are in the air above the lake?

 

[haruspex]

I get slightly different numbers... 2k = Cρ_w A / ρ_p = 3 Cρ_w / (ρ_p 4r) = 3*0.7*1000/(2500*0.02) , k = 5.25. (With buoyancy, k = 8.75.)
(v/v_t)²-1 = ((v₀/v_t)²-1)e^-2kx = (10k-1)e^-2kx
So with x =.1, (v/v_t)²-1 = 51.5e^-1.05 = 18. As you say, much too large. With buoyancy, 15, hardly better (about twice terminal velocity.).

FWIW, imagining it as a real experiment, I don't believe a 10cm wide pebble dropped 5m would slow to around terminal velocity in 10cm.

Just had a thought... I was always bothered by the question asking for when it slowed to 'about terminal velocity' - much too vague. It would make a lot more sense as 'slowed to about twice terminal velocity' (or some other multiple). Is it possible the question is quoted wrongly?

EDIT: Where I wrote "about twice terminal velocity" above I should have written "about four times terminal velocity". Must have square rooted twice.

 

[mfb]

FWIW, imagining it as a real experiment, I don't believe a 10cm wide pebble dropped 5m would slow to around terminal velocity in 10cm.

Same here.
Combining the terminal velocity and the gravitational acceleration, I get an initial deceleration of $g\cdot(10/1.367)^2-g = 510\frac{m}{s^2}$. If acceleration would be constant, the stone would stop after 10m/51 = 20cm. This gives a velocity of ~7m/s at a depth of 10cm. In reality, deceleration is reduced as the speed reduces, therefore the stone is even quicker there. That is hardly "reaching its terminal velocity".

This 20cm happens to be a typical length of the system (it is independent of the height of 5m in my example), the stone approaches its terminal velocity quickly within a few (20cm) - but not in the first 10cm.

 

[unscientific]

No this is exactly how the question was phrased, it's a preliminary examination for first year physics students on "Mathematical Methods"

But nevermind, is my method of approach in this problem correct?

 

[haruspex]

No this is exactly how the question was phrased, it's a preliminary examination for first year physics students on "Mathematical Methods"

But nevermind, is my method of approach in this problem correct?

Yes, it's fine.