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

Exercises

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 wheel's center.

Problems

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. Case:

1. (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.

## Chapter 4.

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.

Problems:

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

Case 1:

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.

## Chapter 5.

Exercises 3, 8, 11, 16, 20, 32 Problems 2, 6, 8

Exercises

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 blood vessels.

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.

Problems

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).