Physics-8.01Fall1999
Solutions for Assignment # 9
December1,199911:25pm
Problem 9.1
(a) John does no work on the computer monitor. John does exert a force on the
computer monitor, but the monitor does not move; hence there is no work done by John on the monitor. (To understandwhy John becomes exhausted if he does no work, please see the last paragraph on page 162.)
(b) Whenthe fly fliesinthejarwith constant speedthenetforceonthe fly mustbe
zero. Thus the air pushes upwards on the fly, and the upward force is the weight of the fly. According toNewton’sthirdlaw,thefly mustthereforepushontheairwith the same force. Sincetheairisenclosedinthejar,thisdownwardforceisfeltby the scale. Thus the scale reading remains the same.
Thereisatransientperiod whilethe fly takesoff(itpushesagainstthebottomof thejar) asitisbeing accelerated. During thatperiodthe scalewill read ahighervalue thanbefore.
(c) Whenthejarisopen,the situationisvery different. Imaginethatthe fly is sitting
onatray,and thetrayisonthe scale,and thenitstartstofly. Clearly the scalewill read a lower value as the fly is ”gone”.
Inanopenjar,withtheflyflyinginthejar,the scalewill read alowervalue. The flyis stillpushingdownontheairwith aforceequal toitsownweight,but nowthisforce isnot100% transferred tothebottomof thejar. Thereis “leakage” totheoutsideworld (outsidethejar).
(d) We can fire the rocket with such an impulse as to bring our speed to zero. Then
earth’sgravity willbring usback without any further effort. If our speedintheorbitis v then the impulse required to come to a temporary stop is I = mv; thus a smaller orbital speed requires a smaller impulse which requires less fuel. Our speed is largest when we are closest to earth, and our speed is smallest when we are farthest from earth; therefore we should fire our rockets when we are farthest from earth.
(e) Let MR be the mass of the rock, MB be the mass of the boat and anything else in
it, ρW be the density of water, and ρR be the density of the rock. With the rock in the boat, the amount of water displaced is
V1 =
MB + MR MB MR
=+
ρW ρWρW
1
With the rock at the bottom of the pool, the amount of water displaced is
V2 =
MB MR
+ ρWρR
Since the rock sinks, we know that ρR >ρW ;hence V1 >V2 and the water level will go
down.
(f) Let M be the mass of ice and ρW be the density of water. With the ice initially
floating, the amount of water displaced is
V1 =
M ρW
The ice melts, and the corresponding volume of water is
V2 =
M ρW
Thus V1 = V2 and the water level will remain the same.
Problem 9.2
(Ohanian,page372,problem16)
Let m =80kg, l =0.6 m, and h =1.2 m. We only need to concern ourselves with a cross-section of the box as shown in Figure 14.34 on page 372.
(a) Now imagine a line connecting the edge of the box which remains in contact with
the floor with the center of mass. The angle between this line and the bottom of the box isgivenby
tanφ =
h 2 l 2 =
hl
=⇒
φ =tan−1
��
h l
≈ 63.4◦
The angle of this line relative to the floor is θ+ φ. The length of this line is given by
� �� �2 � l�
��2
L =
2
+
h 2
≈ 0.671 m
Therefore the height of the center of mass is given by
H = Lsin(θ+ φ)≈ 0.671· sin(θ+63.4◦)
Therefore the potential energy is given by
U = mgH ≈ 5.26× 102 · sin(θ+63.4◦)
2
where the units of U are J and θ must be measured in degrees. The height of the center of massis measured relativetotheground. When θ =0 the center of mass is at a height h
and U = 0. 2 (b) The critical angle θC occurs when
θC + φ =90◦
=⇒
θC =26.6◦
(U is an increasing function of θ for θ<θC and a decreasing function of θ for θ>θC.)
(c) For a conservative force the work is the change in energy.
W = U(θC)−U(0)=5.26× 102 · sin(90◦)−5.26× 102 · sin(63.4◦)≈ 55.7J
Problem 9.3
(Ohanian,page372,problem18)
Let w bethewidth of eachbook. Wewill usethe schemeillustratedbyFigure14.36 onpage372. You firstplacethetopbook. Thenyouplacethenextbook sothatitsright mostedgeisdirectly underthecenterof massof the firstbook. Thenyoucontinueadding booksatthebottominthisfashion:placeeachbook sothatitsright mostedgeisdirectly under the center of mass for all the books above. We can explicitly work out the first five booksinthe sequence; thepatternwillbeclear. Let xi denote the right most edge of the ith book, and choose x1 =0. Then the shift between each book is Δi = xi+1 −xi and the center of the ith bookisgivenby xi + w. Then 2
x1 =0
1 w
x2 = x1 +
12
�
�
�
1
= w 2
�
=⇒
1
Δ2 = x2 −x1 = w
2
1
Δ2 = x3 −x2 = w
4
1
Δ3 = x4 −x3 = w
6
1
Δ4 = x5 −x4 = w
8
1 ww x3 = x1 ++ x2 +
222
�
3
= w 4
�
=⇒
1 www x4 = x1 ++ x2 ++ x3 +
3222
�
=
11
w 12
�
=⇒
1 wwww x5 = x1 ++ x2 ++ x3 ++ x4 +
42222Thetotalprotrusionis
=
25
w 24
=⇒
x5 −x1 = x5 −x4 + x4 −x3 + x3 −x2 + x2 −x1 =Δ4 +Δ3 +Δ2 +Δ1 ≈ 1.04· w
3
¿From the above it seems clear that the general rule is
Δi =
11 w w = · 2ii 2
For an infinite number of books the total protrusion is
i�=∞
=∞ 1 1 ww i�
Δi = · = ·
22 i=1 ii=1 i=1 i
i�=∞
, The second piece above is the sum of the harmonic series, 1ii�=∞
1
→∞ ii=1
which is known to diverge; therefore the maximum protrusion for an infinite number of
books is an infinite distance.
Problem 9.4
(Ohanian,page377,problem50)
From Table 14.1 on page 366 we know that the Young’s modulus for bone is Y = 3.2× 1010 N/m2 .FromEquation(27) onpage366 weknowthat
ΔL 1 F
= L YA
Weknowthat L =38 cm =38× 10−2 m, A =10 cm2 =10×10−4 m2, and F =
1
(Each legsupports 2of the total weight.) This gives
ΔL =
1 F
· L ≈ 4.0× 10−6 m YA
12
68· g.
Problem 9.5
(Ohanian,page408,problem31)
First we will calculate the initial moment of inertia I1 and the initial angular frequency of the torsional pendulum ω1. We will make the approximation that the balance wheel is a thin hoop. The moment of inertia for a thin hoop is given in Table 12.1 on page 309 as
I1 = MR2 =1.5000× 10−8 kg· m 2
where M =0.6g =0.6 × 10−3 kg and R = 1· 1.0cm = 5.0 × 10−3 m. The period of 2the torsional pendulum is supposed to coincide with one second, but the watch makes 1.2· 60 =72 moreperiodsperday thanitshould. Thefrequency isthengivenby
ν1 =
24· 60· 60+72
≈ 1+8.3333× 10−4 Hz
24· 60· 60
4
and
ω1 =2πν1 =2π +5.2359× 10−3 radian/s
We can use I1 and ω1 to calculate κ fromEquation(75) onpage397.
2
κ = I1ω1≈ 5.9316× 10−7 m · N/radian
We want to adjust a screw so that ν2 = 1 Hz and hence ω2 =2π radian/s. The required
moment of inertia is
I2 =
κ
≈ 1.5025× 10−8 kg· m 2 2 ω2
We must therefore increase the moment of inertia by
ΔI = I2 −I1 ≈ 2.5× 10−11 kg· m 2
Nowlet Ihoop be the moment of inertia of the wheel without the screw. Then
2
I1 = Ihoop + mr 1
and
2
I2 = Ihoop + mr 2
and
2 2
ΔI = m(r2−r1)
where m =0.020g =2.0× 10−5 kg is the mass of one screw and r1 and r2 are the two
radii of the screw. We will make the approximation that r1 ≈ 1· 1.0cm =5.0× 10−3 m 2 and then calculate r2.
r2 =
� 2 r1+
ΔI
≈ 5.12× 10−3 m m
Therefore we need to move the screw out a distance
Δr = r2 −r1 ≈ 1.2× 10−4 m =1.2× 10−2 cm
Problem 9.6
(Ohanian,page488,problem17)
(a)We can use Bernoulli’s equation to solve this problem.
12
ρv+ ρgz + p=constant 2
5
where ρ =1055kg/m3 . The flow of blood is slow enough that we can consider v =0 throughout the body. The above equation becomes
ρgz + p=constant
Now define the origin of z to coincide with the heart, and let p0 = 110 mmHg = 1.46× 104 Pa. Then the constant is given by
p0 =constant
and
ρgz + p= p0
=⇒
p= p0 −ρgz
for other points in the body. For the feet, z = −140 cm = −1.4 m, and
pfeet =1.46× 104 −1055· 9.8·−1.4 ≈ 2.91× 104 Pa ≈ 221 mmHg
where g =9.8m/s2 . For the brain, z =40 cm =0.4 m, and
pbrain = p0 −1055· 0.4· g =1.46× 104 −1055· 0.4· 9.8 ≈ 1.05× 104 Pa ≈ 79.8 mmHg where g =9.8m/s2 .
(b) The pressure in the brain is still given by the expression above
pbrain = p0 −1055· 0.4· g
Now g =61 m/s2,therefore
pbrain = p0 −1055· 0.4· 61 = p0 −2.57× 104 Pa = p0 −196 mmHg
The heart could at most create a pressure p0 = 190 mmHg, so the pressure in the brain will be negative.
Problem 9.7
(Ohanian,page489,problem18)
The pressure at the surface of the liquid is p =1atm =1.013 × 105 Pa. This pressure must be able to support the mass of the column of water raised in the tube. Let A be the cross-section of the tube and ρ =1.0× 103 kg be the density of water. Then
pA= F = Mg = ρAhg
6
=⇒
h =
p
≈ 10.3m ρg
Problem 9.8
(Ohanian,page489,problem24)
Let ρI =920kg/m3 is the density of ice and ρW =1025kg/m3 is the density of sea water. Also let M be the mass of the ice. The volume of the ice is
VI =
The volume of water displaced by the ice is
VW =
The volume above the water is
ΔV = VI −VW = M
� M ρI
M ρW
1 1 − ρIρW
�
=30· 400· 400 m 3 =4.8× 106 m 3
Therefore the total mass of the ice is
M = �ΔV
1
ρI
−
1ρW
� = �4.8× 106 m3
1 1
− 1025920 � ≈ 4.31× 1010 kg
and the total volume of the ice is
4.31× 1010 M
=≈ 4.68× 107 m 3 VI = ρI 920
(b) The total mass of the ice is
M ≈ 4.31× 1010 kg
Problem 9.9
(Ohanian,page491,problem35)
Theblock will sink untilitdisplacesanequal massof oil(ρO =8.5×102 kg/m3)and
water(ρW =1.0×103 kg/m3). (Ifthe block is too heavy then it will sink to the bottom.) Let H =10 cm =10−1 m be the height of the box; A =30 cm×20 cm =6×10−2 m2 be the area of the bottom of the box; and M =5.5 kg be the mass of the box. Also let h be the distance between the bottom of the box and the oil/water interface. The condition for equilibrium is
M = hAρW +(H −h)AρO
=⇒
h =
M −HAρO
≈ 4.4 cm
A(ρW −ρO )
7
Problem 9.10 (Ohanian,page493,problem50)
Example 9 on page 481 illustrates a portion of this problem.
We can use Bernoulli’s equation to relate the water at the top of the tank to the water emerging from the hole.
1 2
ρv+ ρgz + patom = ρgh+ patom 2
=⇒
v =
� 2g(h−z)
The water at both the top of the tank and the hole is in contact with the atmosphere; hence the pressure is the same, patom. The velocity at the top of the tank is zero. Note that the velocity of the water emerging from the hole is given as if it fell the distance h−z. Now the water flows horizontally from the hole with speed v above. The time it takes to hit the floor is given by
1 2
gT= z 2
The horizontal distance it moves is
d = vT =
� =⇒ T =
� 2z g
2g(h−z)·
� � 2z = 2 (h−z)z g
The maximum distance occurs when
dd 1
� = 2· (h−2zmax)= 0
dz (h−z)z
The maximum distance is
dmax = 2 (h−zmax)zmax = 2
� � � =⇒
1
zmax = h
2
1 1
h− h h = h
22
� Problem 9.11 (a) Theballdisplacesavolume V of theliquid ofdensity ρ;thereforethemassofliquid
which overflows is
M = ρV
(b) At first, when the rod is approaching the water but when the ball is still above the
water, the rod is carrying the full weight mg of the ball. However, as the ball is pushed
into the water, the rod will carry less weight and when the ball is fully immersed, the weight even becomes negative; the rod is pushing down with a force ρVg − mg tokeep theballimmersed. The water exerts aforce(thebuoyantforce) of ρVg upwards on the
8
ball. Since action equals minus reaction, the ball will exert a force downwards onto the water of ρVg. The weight of the container with the remaining water is W −ρVg. Hence the scale will read W −ρVg+ ρVg = W. Notice that this is independent of the mass m of the ball. This is not surprising as the rod is carrying the weight of the ball.
(c) The buoyant force upwards on the ball is ρVg and the weight of the ball is mg.
Thusthetension(pulling upwardsatthe scale) T = ρVg −mg. The buoyant force ρVg is upwards on the ball, thus the ball will exert a force downwards onto the water of ρVg. The weight of the container with the remaining water is W −ρVg. Hence the scale will read W − ρVg − T + ρVg = W − ρVg + mg. Notice that in the case that ρVg = mg (neutralbuoyancy)theanswerunderpart(b) and(c) shouldbethe same. Thetension inthe string(c) isthenzero,and theforceneededfortherod tokeep theballimmersed (b)is also zero. The scale reads W in both cases.
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