Archimedes' principle
Two cases object submerged or floating.
m obj
Object: mass mobj , volume Vobj , density ρobj =
Vobj
Fully submerged in fluid of density, ρfl .
Displaced fluid volume equal to Vobj
Vobj
Vobj
Mass of displaced
fluid is m fl = ρfl Vobj
Buoyant force on object (upwards)
is:
FB = m fl g = ρfl Vobjg
Once you have the buoyant force its an equilibrium or kinematics
problem that includes the buoyant force.
Archimedes' principle
If floating, displaced fluid volume Vfl is less than object volume Vobj .
Here the mass of the fluid displaced equals the mass of the object
m obj
m obj = m fl
ρobjVobj =
ρfl Vfl
m fl
Suppose given that 1/10 of
object volume floats above
the water line and want ρobj .
Then,
So,
9
Vfl = Vobj
10
9
ρobjVobj =
ρfl Vobj
10
ρobj =
9
kg
ρfl = 900 3
10
m
Conservation of energy
A small ball of mass m at the end of a massless string of length L
is found to have a tangential speed vo when it is at angle θ.
What is the ball’s speed when the string is
vertical?
L
m
vo
θo
Use conservation of energy:
∆KE + ∆PE =Wnc
KEf − KEi + PEf − PEi =
0
L
Choose the height of the ball when the
string is vertical as the zero of potential
energy. Then,
L
m
vo
1
1
mv f2 − mv 2o + 0 − mgh =
0
2
2
1
1
2
=
mv f
mv o2 + mgh
2
2
2
2
v=
v
f
o + 2gh
=
vf
θ
v o2 + 2gh
=
vf
v 2o + 2gh
But we were not given h we must
solve in the variables given.
L
Draw a horizontal line from the ball’s
original position to the vertical and
note that,
θ
L
m
L −=
h L cos θ
h
vo
And that h is the length of the string L minus L − h
then,
h= L − (L − h)= L − L cos θ= L(1 − cos θ)
So the answer is,
vf =
L−h
v 2o + 2gL(1 − cos θ)
HITT
You are driving along at 20 m/s when you hear a siren that sounds
to you like 450 Hz. Checking your rearview you see a fire truck
coming up on you at 30 m/s.
The general Doppler equation with fo the source frequency
(at rest) is
v ± vD
f = fo
v ± vS
In this case the sign on
A) +
vD is:
B) –
HITT
You are driving along at 20 m/s when you hear a siren that sounds
to you like 450 Hz. Checking your rearview you see a fire truck
coming up on you at 30 m/s.
The general Doppler equation with fo the source frequency
(at rest) is
v ± vD
f = fo
v ± vS
In this case the sign on
A) +
vS is:
B) –
HITT
You are driving along at 20 m/s when you hear a siren that sounds
to you like 450 Hz. Checking your rearview you see a fire truck
coming up on you at 30 m/s.
The general Doppler equation with fo the source frequency
(at rest) is
v ± vD
f = fo
v ± vS
If the speed of sound is 330 m/s the frequency of the siren is:
A) 465 Hz
B) 436 Hz
Ans.
You are driving along at 20 m/s when you hear a siren that sounds
to you like 450 Hz. Checking your rearview you see a fire truck
coming up on you at 30 m/s.
v − vD
f = fo
v − vS
detector heading away from source
source heading toward detector
330 − 20
=
f f=
f o (1.0333)
o
330 − 30
But we were given f = 450 Hz
450 Hz = f o (1.0333)
450 Hz
=
f o = 436 Hz
1.0333
A billiard ball (ball 1) moves along the +x axis with a velocity of 2 m/s.
A second identical billiard ball (ball 2) moves in the same direction
at 4 m/s, catches up to ball 1 and collides elastically with it. What are
the speeds of ball 1 and 2 After the collision?
Conservation of linear momentum says:
pf = pi
m1v1f + m 2 v 2f =m1v1i + m 2 v 2i
But the balls are identical so the masses are all the same and
cancel out giving that,
v1f + v 2f = v1i + v 2i
Now because the collision was elastic, kinetic energy is conserved
so we can also write that,
Or,
1
1
1
1
2
2
2
m1v1f + m 2 v 2f = m1v1i + m 2 v 22i
2
2
2
2
v1f2 + v 22f = v1i2 + v 22i
v1f2 − v1i2 = v 22i − v 22f
(v1f − v1i )(v1f + v1i ) = (v 2i − v 2f )(v 2i + v 2f ) *
The momentum equation result was:
Rearranging
v1f + v 2f = v1i + v 2i
v1f − v1i = v 2i − v 2f
Divide this into the * eqn giving,
v1f + v1i = v 2i + v 2f
(1)
v1f − v1i = v 2i − v 2f
Solve (1) for v2f :
(2) v1f + v1i = v 2i + v 2f
(3) v 2f = v1i + v 2i − v1f
Use (3) in (2): v1f + v1i = v 2i + v1i + v 2i − v1f
=
v1f 2v 2i − v1f
2v1f = 2v 2i
(
(
m
v1f = v 2i
v=
4
( v=
1f
2i
s
Use this result in (3) v 2f = v1i + v 2i − v 2i
m
v=
2
v 2f = v1i
( v=
2f
1i
s
So the billiard balls exchange velocities after this collision (just
like textbook example 6.6 for a head on collision).
HW problem 14.8:
A stone is dropped into a well from rest and the splash is heard
precisely 2 s later. How deep is the well?
Two times here: the time for the stone to reach the water at
the bottom of the well: t1 ; and the time for the sound to reach
your ear from the bottom of the well: t2
2 s is the info given.
The sum of these times t1 + t 2 =
1 2
y o + v oy t1 − gt1
For t1 : y =
2
Take upwards as positive, set the drop point as y = 0 and
the position of the water as y = –d (for depth). Then,
1 2
−d = 0 + 0 − gt1
2
1 2
→ d = gt1
2
y yo + vS t 2
For t2 : =
where t2 = 0 when yo = –d and we need
t2 when y = 0 so,
0 =−d + v S t 2
→ d =v S t 2
So we have d = v S t 2
,
Equate the equations for d:
1
d = gt12
2
&
t1 + t 2 =
2s *
1 2
v S t 2 = gt1
2
Solve the * equation for t2 : t 2= 2 − t1 and substitute
1 2
v S ( 2 − t1 ) =gt1
2
1 2
quadratic in t1
gt1 + v S t1 − 2v S =
0
2
− b ± b 2 − 4ac
Has general solution: x =
2a
1
− v s ± v s2 − 4( g)( −2v s )
2
t1 =
1
2 g
2
− v s ± v s2 + 4gv s
t1 =
g
only the + solution gives a
positive t1
v s2 + 4gv s − v s
t1 =
g
use numerical result from this
above to solve for d.