Question

Barnes Wallis is famous for creating a 'bouncing bomb' able to destroy the great dams of the Ruhr in Germany. The bomb was cylindrical and bounced along on the water. It was spinning such that as it bounced it slowed down and came to a stop at the dam. The bomb then sank along the dam wall where an internal pneumatic pistol ignited the charge causing the bomb to explode hard against the dam surface. Wallis discovered that he needed the mass of the bomb to be just 4.3 tonnes consisting of 2.7 tonnes of explosives. The density of steel casing was about 7900 kg m^3 and the density of the explosive is about 800 kgm^3 Calculate the average radius of the bomb if the length was fixed at 1.6 meters. The force of impact (average force) of such a bomb on water after falling from a height of 75 m is enormous. First, find the speed of impact, assuming that air resistance is negligible. Then calculate the average force of impact of the bomb if it initially sinks to a depth of 3/4 of its diameter.

i. Use the volume of a cylinder to calculate the size required to encase the explosive. Assume the thickness of the casing is the same at the ends as it is around the outside of the cylinder. You should get two equations one for the size of the inside explosive and another for the outside casing. Some of the variables, however, will be shared between the two equations.

ii. Find the velocity of the bomb when it hits the water.

iii. If you know it goes from this velocity to zero velocity, in a distance equal to 3/4 of its diameter, then
what is the deceleration of the bomb and what is the average force of impact?

Answer 1

@ganeshie8 @atlas please help me :s

Answer 2

@theEric @gurrilabill

Answer 3

@Vincent-Lyon.Fr

Answer 4

thank you guys, hope i will get some help with this :)
<3

Answer 5

@BeautifulMelodies please help .. thanks

Answer 6

I have no idea. Well I do. But I have a semester test to study for.

Answer 7

@BeautifulMelodies no problem but can you please tag someone you know who can help me
Thanks for response :)

Answer 8

@Nannarzz can you please help me :) pleaase i am really stuck with this problem :s

Answer 9

@BillBell sir please help me!

Answer 10

@Chineseboy15

Answer 11

please some one :s

Answer 12

That link may be helpful
http://home.cc.umanitoba.ca/~stinner/stinner/pdfs/1989-dambusters.pdf

Answer 13

That link may be helpful
http://home.cc.umanitoba.ca/~stinner/stinner/pdfs/1989-dambusters.pdf

Answer 14

@Gebooors
thank you sir

Answer 15

@Chineseboy15 please anyone else with more help ,,

Answer 16

Hi! I have some helpful hints.

You know that the total cylinder has the volume of both the case and explosives. I will say this in math now:
\(V=V_c+V_e\)

You know the general density formula is
density = mass / volume
\(\implies\rho=\dfrac mV\implies V=\dfrac m\rho\)
You can find the volumes for each component, now.
\(V_c=\dfrac{m_c}{\rho_c}\)
\(V_e=\dfrac{m_e}{\rho_e}\)

You know the general formula for the volume of a cylinder:
\(V=\pi r^2h\)
Like I said before, \(V=V_c+V_e\). And now you know \(V_c\) and \(V_e\).
Equting the two expressions of \(V\),
\(V=\pi r^2h=V_c+V_e\)
Why do this? Now you can solve for \(r\). Everything but \(r\) is known.

At this point, you've found the dimensions of the entire cylinder.

I still haven't found the inside cylinder dimensions.
However, we know the radius of the entire cylinder, so we can find the diameter for part (iii).

Part (ii) is kinematics for a falling object under gravity.
Part (iii) is kinematics for finding average acceleration. When average acceleration is known, and the total mass is known, the average force can be known.

Answer 17

Please make sure you follow every step! :) And try to see what you can learn!

A lot of times the first step is breaking down the problem so that you can understand it. I drew a picture. And the rest is learning how everything is related. You put those relation into math so that you can see them easily and solve for unknown values.

Answer 18

this is what i have done, i am not so sure if its correct but seems so :)
volume size required to encase the explosive = ?


Density of the explosive = 800 kg/m^3
weight of explosives = 2.7 tonnes = 2700 kg


volume = mass/density
also
Volume = 4/3*pie*(r^3)

=> mass/density = 4/3*pie*(r^3)
=> 2700 /800 = 4/3*pie*(r^3)
=> 3.375 = 4/3*pie*(r^3)
=> 2.53 = pie*(r^3)
=> 0.80 = (r^3)
=> r = 0.93 m^3
so
=> V = 4/3*p*(r^3)
=> V = 3.369 m^3 is the volume size required to encase the explosive
**********



PART (ii)

Velocity = ?
v^2 = 2gh
v = (2gh)^1/2
v = (2 x 9.81 x 75)^1/2
v = 735.75 m/sec
*********



Part(iii)

If you know it goes from this velocity to zero velocity, in a distance equal to 3/4 of its diameter, then
what is the deceleration of the bomb and what is the average force of impact?

vi=735.75 m/s
vf=0
s= 3/4(d)

first we need to measure diameter of the bomb ' d = ?'


Total weight of bomb = 4300 kg
Total density of bomb = 8700 kg/m^3


volume = mass/density
also
Volume = 4/3*pie*(r^3)

=> mass/density = 4/3*p*(r^3)
=> 4300 /8700 = 4/3*p*(r^3)
=> (4300/8700)*3/(4*pie) = (r^3)
=> 0.117 = (r^3)
=> r = 0.5 m
so d = 2r
=> d = 1m

Now

distance after which velocity becomes zero (after impact with water) = s = 3/4 (d)
s=0.75

vf^2=vi^2+2as
0=(735.75)^2+2(a)(0.75)
0.75*a=-(735.75)^2
0.75*a=-541328
a = -721770 m/s^2 = deceleration

Force of impact = f = ma
m = Total weight of bomb = 4300 kg
f = (4300)(-721770)
f = -3103611000 N
(-) sign indicates oppisite direction

Please correct me if you can thanks for responding @theEric

Answer 19

I think you accidentally used the formula for volume of a sphere!

Answer 20

It looks like you didn't write part (ii), but got \(v_f=735.75\) as the solution. I have a different answer, so I don't know who's wrong. I used the same equation that you used in part (iii):\(\quad v_f^2=v_i^2+2ad\) where \(v_i=0\), \(a=-9.8\), and \(d=75\) and final units are \(\rm m/s^2 \).

Answer 21

Oh, I found part (ii)! :P

Answer 22

It looks like part (ii) has a calculation error.

Assuming that your velocity as it hits the water is correct, then the kinematics section is correct except that you lost a \(2\) along the way.
\(vf^2=vi^2+\color{limegreen}2as\\
0=(735.75)^2+\color{limegreen}2(a)(0.75)\\
\color{red}{0.75*a}=-(735.75)^2\\
\color{red}{0.75*a}=-541328\\
a = \color{red}{-721,770}\ {\rm m/s^2} = {\rm deceleration}\)

Answer 23

You also want to make sure you don't drop the diameter:
\(s=\frac34d\cancel\impliess=0.75\) unless \(d=1\).
I don't know what \(d\) is, though. But it can be found!

Other than that, it's good!

Answer 24

Your work was very organized; thank you! ;)

Answer 25

Correction for my \(\LaTeX\) issue:
\(s=\frac34d\cancel\implies s=0.75\)

Answer 26

Take care!