Solutions to Homework #2 for Physics
104, Summer 2003
Chapter 3, Section 3.3
Exercises 36, 41,44,48 Problems 5,7 Case 1,7
36. The normal or support force provided by the rim of a spinning
roulette wheel provides the cental force that keeps the ball in
uniform circular motion. The ball presses against the rim, in
reaction to this force. This sensation of an outward force
throwing the ball out of the orbit is often termed 'centrifugal
force'.As the wheel slows down, the ball rolls down,
because the roulette is inclined.
41. You feel heaviest at the bottom of the swing since the
normal force from the seat (which is your apparent weight) has
to exceed your weight, to produce a net central acceleration.
Alternatively,one can say the normal force has to balance the
sum of your weight and the fictitious centrifugal force you feel.
44. If you undergo free-fall, gravity is still acting on you.
However, since you are not pressing on anything, you feel no
normal force from the floor, and therefore, sense an apparent
weight of zero.
48. At the top of the Ferris wheel, as long as you are still
sitting on your chair, your net acceleration must be towards the
5. To generate artificial gravity, the astronauts need to be
'thrown out against the rim' of the wheel-shaped space station.
To generate the same weight sensation (apparent weight N) as
the earth, N = mg. But N = mv^2/R. Thus:
mg = mv^2/R or simplifying: v = (gR)^(1/2) = [(10m/s^2)(100m)]^(1/2)
or v = 31.6 m/s (Rim must turn with this speed). The circumference
of the wheel is 2 x pi x R = 628 m. Dividing this by the speed,
we get time for 1 revolution = 19.9 seconds. Thus, the station
must turn once every 19.9 seconds.
7. Central Acceleration = v^2/R = (1m/>s)^2/(.05m) = 20 m/s^2
towards the blender's central axis.
(a) During this approach to equilibrium, the scale reading
oscillates about your correct weight since the scaling is bouncing
up and down somewhat in reaction to your sudden application of weight.
(b) Standing on the scale with one foot makes no difference compared
to standing with both feet. Although the pressure on the area under
your foot has been doubled, the weight read is the same.
(c). Should read > real weight since you're applying a force on
scale to propel yourself up.
(d). The weights reported by each scale is half your actual weight
because it has been distributed equally between them.
(e). The top scale reads your real weight. The bottom scale reads
your real weight plus 10 N since the top scale exerts a force equal
to its weight on the bottom one.
Exercises 2, 3, 15, 20, 21 Problems 3, 6, 7, 15 Case 1
2. The density of the log, identical to that of the stick, is less
than that of water. Thus, it floats.
3. Although the automobile is predominantly metal, a sealed
car contains air, making the average density actually less than
that of water. Therefore, if leak-tight, it will float.
15. When the hot air inside cools, it contracts so that the internal
pressure is less than the external atmospheric pressure. As a
result, the net inward pressure bows the lid inward.
20. As the level of tea goes down, the gravitational potential energy
decreases so the kinetic energy of the fluid becomes less at the faucet.
This makes reduces the speed of the outgoing tea.
21. Pressure linearly increases with depth. Thus, the base of the dam
must be kept stronger against these larger pressures.
3. For fixed volume, the pressure of an ideal gas is proportional to
its temperature T. Since T triples, the new pressure must be 3x
atmospheric pressure or 303975 N/m^2.
6. The net force must be the difference in the weights of the log
and displaced water. This is simply (10 kg - 8 kg)x 10m/s^2 = 20 N.
7. The weight of the displaced fluid must be the weight of the boat or
1200N, since the boat was floating. This corresponds to displaced water of
mass = 120 kg.
15. At a depth h = 300 m, the water pressure is atmospheric pressure + (water density)x gh
= 101325 Pa + [(1 g/cm^3)(1kg/1000g)(100 cm/m)^3](10 m/s^2)(300m) = 3.1 x 10^6 Pa
a. Net force = 0 since it's not accelerating anywhere.
b. it would be displacing more air than water, so the average density of
displaced fluid must decrease.
c. The buoyant force will decrease.
d. Since the weight of the ship doesn't change but the average density of
the displaced fluid has decreased, a net force downwards will appear (which
is why it bobs down consequently).
e. If it is displaced below the equilibrium level, the average
density of the displaced fluid will increase and the net force will be upwards.
Combined with the observation in (d), displacing the ship about the equilibrium
level makes it move back to it, making the equilibrium stable (Ship bobs up and
down about it).
f. The new passenger's weight forces the ship to seek a new equilibrium level,
where the buoyant force is larger than before. Thus, it will sink deeper.
g. Knowing the change in depth lets the captain know how much weight it carries,
compared to having an empty ship.
h. The air exerts a buoyant force on the ship. However, since the ship + air
system is denser than air, the ship cannot float in air.
Exercises 3, 8, 11, 16, 20, 32 Problems 2, 6, 8
3. According to Poiseuille's Law, the volume rate flow is proportional
to D^4, for fixed pressure difference, viscosity and length.Thus, A tiny
narrowing of the tube leads to significantly increased pressure in the
8. Near the ground, there is greater friction between land
and the air.
11. While the can was falling, the pressure throughout the liquid
is essentially constant, since the upper layers of applesauce do not
apply a force equal to their weight upon the lower layers (they are
weightless during free fall). Upon hitting the ground, the upper layers
now have weight and increase the pressure on the lower layers. In addition,
they have an impulse force due to the impact that adds to this pressure.
20. The front of the truck encounters high velocity air flow, which creates a
low pressure region, to which the bicycle is being sucked into.
32. There will be no lift due to the difference in air pressures. The
frisbee will fly like a projectile.
2. A 5% narrowing of blood vessel diameter implies that D'/D = .95
where D' = new diameter, D = old diameter. Using Poiseuille's Law,
we know that for the flow rate rate of blood is proportional to D^4
so that if all else remained constant, the flow rate must change by a
factor of (D'/D)^4 or (.95)^4 = 0.8145. But of course,
we don't want blood to flow any faster or slower to our brains. Thus.
for constant viscosity and length of blood vessels, the pressure drop
delta p must increase by the factor of 1/0.8145 = 1.23 or 23%.
6. From Table 5.1.1, the viscosity of water and honey are 0.001 and
1000 Pa-s, respectively. This means the ratio of viscosity of honey to
water is about 1,000,000 or 1x10^6. Poiseuille's Law states that the
flow rate is propotional to 1/viscosity of the fluid. Thus, we expect
that with honey, the flow rate, should decrease by a factor of 1x10^6.
This implies that the flow will take 1x10^6 times as as long compared to water,
or 5x10^6 seconds.
8. The boundary between laminar and turbulent flow occurs at a
Reynolds number of about 2000. Since the Reynolds number is defined by:
Reynolds # = (density x obstacle length x flow speed)/viscosity
using density of air = 1.25 kg/m^3, obstacle length = width of blimp = 15 m,
air viscosity = .0000183 Pa-s (from Table 5.1.1) and Reynolds # = 2000,
then the maximum flow speed for laminar flow must be 0.001952 m/s or
1.952 mm/s (very slow).