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Unformatted text preview: Massachusetts Institute of Technology  Physics Department Physics8.01 Fall 1999 Solutions for Assignment # 9
by Dru Renner December 1, 1999 11:25 pm Problem 9.1
(a) There is only one thing that interacts with the car: the road. (Ignoring air drag.) The road exerts a force on the car through friction. This propels the car forward. (Answer (i)) (b) The force of the motor is an internal force, but it does exert a force on the car and the car does travel a distance hence the motor does do work. The friction acts on the edge of the tire which is momentarily at rest if the car rolls without slipping hence friction does no work. (Answer (ii)) (c) 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 understand why John becomes exhausted if he does no work, please see the last paragraph on page 162.) (d) Viewed from the rest frame of the building, Mary does do work on the backpack. Joan's argument is correct in that frame, but viewed from Mary's rest frame, Mary does no work on the backpack since the backpack does not move in Mary's rest frame. Work can be di erent in di erent frames. (To understand why Mary does not feel any di erence between standing in the elevator or on the sidewalk, please see the second paragraph on page 163.) (e) Jean is correct: the rope did do work on Joe. The only force Joe feels is that due to the rope, so the only work done on Joe is done by the rope. (f ) When the y ies in the jar with constant speed the net force on the y must be zero. Thus the air pushes upwards on the y, and the upward force is the weight of the y. According to Newton's third law, the y must therefore push on the air with the same force. Since the air is enclosed in the jar, this downward force is felt by the scale. Thus the scale reading remains the same. There is a transient period while the y takes o (it pushes against the bottom of the jar) as it is being accelerated. During that period the scale will read a higher value than before. (g) When the jar is open, the situation is very di erent. Imagine that the y is sitting on a tray, and the tray is on the scale, and then it starts to y. Clearly the scale will read a lower value as the y is "gone". 1 In an open jar, with the y ying in the jar, the scale will read a lower value. The y is still pushing down on the air with a force equal to its own weight, but now this force is not 100% transferred to the bottom of the jar. There is \leakage" to the outside world (outside the jar). (h) We can re the rocket with such an impulse as to bring our speed to zero. Then earth's gravity will bring us back without any further e ort. If our speed in the orbit is 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 re our rockets when we are farthest from earth. (i) Let MR be the mass of the rock, MB be the mass of the boat and anything else in it, W be the density of water, and R be the density of the rock. With the rock in the boat, the amount of water displaced is V1 = MB + MR = MB + MR
W W W With the rock at the bottom of the pool, the amount of water displaced is V2 = MB + MR
W R Since the rock sinks, we know that R < W hence V1 > V2 and the water level will go down. (j) Let M be the mass of ice and W be the density of water. With the ice initially oating, the amount of water displaced is V1 = M W The ice melts, and the corresponding volume of water is V2 = M W Problem 9.2 Thus V1 = V2 and the water level will remain the same. (Ohanian, page 372, problem 16) Let m = 80 kg, l = 0:6 m, and h = 1:2 m. We only need to concern ourselves with a crosssection 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 oor with the center of mass. The angle between this line and the bottom of the box is given by 2 tan = = h l 2 h 2 l =) = tan1 h l ! 63:4 The angle of this line relative to the oor is v !2 u u l L=t 2 + + . The length of this line is given by ! h 2 0:671 m 2 Therefore the height of the center of mass is given by H = L sin( + ) 0:671 sin( + 63:4)
Therefore the potential energy is given by U = mgH 5:26 102 sin( + 63:4)
where the units of U are J and must be measured in degrees. A plot of U as a function of is shown below.
Gravitational Potenetial Energy
550 500 450 400 U()
350 300 250 200 0 10 20 30 40 50 60 70 80 90 The height of the center of mass is measured relative to the ground. When = 0 the center of mass is at a height h and U 6= 0. 2 (b) The critical angle C occurs when
C + = 90 =) C = 26:6 (U is an increasing function of for < C and a decreasing function of for > C .) (c) For a conservative force the work is the change in energy. W = U ( C ) U (0) = 5:26 102 sin(90) 5:26 102 sin(63:4) 55:7 J
3 Problem 9.3
w (Ohanian, page 372, problem 18) Let be the width of each book. We will use the scheme illustrated by Figure 14.36 on page 372. You rst place the top book. Then you place the next book so that its right most edge is directly under the center of mass of the rst book. Then you continue adding books at the bottom in this fashion: place each book so that its right most edge is directly under the center of mass for all the books above. We can explicitly work out the rst ve books in the sequence the pattern will be clear. Let denote the right most edge of the th book, and choose 1 = 0. Then the shift between each book is = +1 and the center of the th book is given by + 2 . Then
xi i x
i xi xi i xi w x1 =0 =)
2 x2 =1 1
x1 x1 +2 =1 2
w x2 w w = x2 1=1 2
x w x3 1 =2
x1 +2+
w x2 +2 =3 4 +
x3 w =)
w 2 = x3 2=1 4
x w x4 =1 3
x1 + + 2
w x + w 2 + w = 11 2 12
w =)
w 3 = x4 3=1 6
x w + 2 + 2+ 2 + The total protrusion is
x5 w w x5 x x x x 1 =4 x3 +2+
w x x4 25 + 2 = 24
x x =) + + 4 = x5 1 4=8
x : w w 1= 5 4+ 4 3+ 3 2+ 2 1=
x x w w 4 3 2 + 1 1 04 From the above it seems clear that the general rule is = 1 =1 2 2 For an in nite number of books the total protrusion is =1 =1 =1 X X X = 1 2 = 2 1 =1 =1 =1
i i i i i w w i i i i i i i The second piece above is the sum of the harmonic series, 1 , =1 X1
i i i=1 i !1 which is known to diverge therefore the maximum protrusion for an in nite number of books is an in nite distance. 4 Problem 9.4 (Ohanian, page 377, problem 50) From Table 14.1 on page 366 we know that the Young's modulus for bone is Y = 3:2 1010 N=m2. From Equation (27) on page 366 we know that L= 1F L YA
We know that L = 38 cm = 38 102 m, A = 10 cm2 = 10 104 m2 , and F = 1 68 g. 2 1 (Each leg supports 2 of the total weight.) This gives 1 L = Y F L 4:0 106 m A Problem 9.5 (Ohanian, page 408, problem 31) First we will calculate the initial moment of inertia I1 and the initial angular frequency of the torsional pendulum !1. 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 108 kg m2
where M = 0:6 g = 0:6 103 kg and R = 1 1:0 cm = 5:0 103 m. The period of 2 the torsional pendulum is supposed to coincide with one second, but the watch makes 1:2 60 = 72 more periods per day than it should. The frequency is then given by
1 60 60 + = 24 24 60 60 72 1 + 8:3333 104 Hz
1 and !1 = 2 = 2 + 5:2359 103 radian=s We can use I1 and !1 to calculate from Equation (75) on page 397.
2 = I1 !1 5:9316 107 m N=radian We want to adjust a screw so that moment of inertia is
2 2 = 1 Hz and hence !2 = 2 radian=s. The required I2 = !2 1:5025 108 kg m2
We must therefore increase the moment of inertia by 5 2 2 I1 = Ihoop + mr1 and I2 = Ihoop + mr2 and
2 2 I = m(r2 r1 ) where m = 0:020 g = 2:0 105 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 1:0 cm = 5:0 103 m 2 and then calculate r2 .
2 r2 = r1 + mI 5:12 103 m s Therefore we need to move the screw out a distance Problem 9.6 r = r2 r1 1:2 104 m = 1:2 102 cm 1 v2 + gz + p = constant 2 where = 1055 kg= m3. The ow of blood is slow enough that we can consider v = 0 throughout the body. The above equation becomes gz + p = constant
Now de ne 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 gz + p = p0
6 =) p = p0 gz (a) (Ohanian, page 488, problem 17) We can use Bernoulli's equation to solve this problem. Now let Ihoop be the moment of inertia of the wheel without the screw. Then I = I2 I1 2:5 1011 kg m2 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 = 9:8 m=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 = 9:8 m=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 Problem 9.7 The heart could at most create a pressure p0 = 190 mmHg, so the pressure in the brain will be negative. (Ohanian, page 489, problem 18) The pressure at the surface of the liquid is p = 1 atm = 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 crosssection of the tube and = 1:0 103 kg be the density of water. Then pA = F = Mg = Ahg =) h = pg 10:3 m
W Problem 9.8
I (Ohanian, page 489, problem 24) Let = 920 kg=m3 is the density of ice and = 1025 kg=m3 is the density of sea water. Also let M be the mass of the ice. The volume of the ice is VI = M
I The volume of water displaced by the ice is VW = M
The volume above the water is 7 W V = VI VW = M 1 1
I ! W = 30 400 400 m3 = 4:8 106 m3 Therefore the total mass of the ice is M= 1
I V 1
W = 4:81 10 1 m 4:31 1010 kg
6 3 920 1025
10 and the total volume of the ice is VI = M = 4:3192010 4:68 107 m3
I (b) The total mass of the ice is Problem 9.9
W M 4:31 1010 kg
O (Ohanian, page 491, problem 35) The block will sink until it displaces an equal mass of oil ( = 8:5 102 kg=m3) and water ( = 1:0 103 kg=m3). (If the block is too heavy then it will sink to the bottom.) Let H = 10 cm = 101 m be the height of the box A = 30 cm 20 cm = 6 102 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 = hA + (H h)A
W O =) M h = A ( H A ) 4:4 cm O W O Problem 9.10 (Ohanian, page 493, problem 50) 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. q 1 v2 + gz + p = gh + p =) v = 2g(h z) atom atom 2 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 ows horizontally from the hole with speed v above. The time it takes to hit the oor is given by 8 1 gT 2 = z 2 The horizontal distance it moves is
q =) T = 2gz s q d = vT = 2g(h z) 2gz = 2 (h z)z s The maximum distance occurs when dd = 2 q 1 (h 2zmax) = 0 dz (h z)z
The maximum distance is
q =) 1 zmax = 2 h dmax = 2 (h zmax)zmax = 2 s h 1h 1h = h 2 2 Problem 9.11
(a) The ball displaces a volume V of the liquid of density therefore the mass of liquid which over ows is M= V
The remaining liquid exerts a net force of Mg upward on the ball, but the force of gravity exerts a force of mg downward on the ball. Since the ball originally oated, we know that M > m therefore the rigid rod is required to hold the ball submerged. This rod must apply an appropriate force F = (M m)g downward on the ball. Thus the mass that is missing because the ball is less dense, is e ectively replaced by the rod. Hence, the scale will read the same, W . (c) The tension must now provide the force that holds the ball in place thus T = (M m)g. The scale is blind to the internal arrangement of the mass, hence the scale will read W = W Mg + mg = W (M m)g. This result is identical to W = W T .
(b)
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