Physics 2022 SQ

SQ-1

The diagram shows a simple diagram of the hot-air balloon. A hot air balloon consists of a bag, called an envelope, which contains heated air. Suspended beneath is the load which comprises the basket carrying equipment, passengers and a burner (an open flame produced by burning liquid propane). The variation of density of air with temperature inside the envelope is given in the table below. You can use $g=9.8N/kg$.

1. Using the information given in the table, calculate the upthrust per ${\text{m}}^3$ ($U$) and the weight of air per ${\text{m}}^3$ ($W_A$), when the surrounding air temperature is $20\text{°C}$ with the balloon air temperature at $100\text{°C}$. Show that lifting force per unit volume (${\text{m}}^3$) $L$ ($=U-W_A$) is about $2.5{\text{ N/m}}^3$.
2. If the volume of the envelope is fixed $1800{\text{ m}}^3$ and the surrounding air temperature is ${20}^{\circ }C$ with the balloon air temperature at ${100}^{\circ }C$, determine the acceleration of the hot-air balloon system if total mass of the load and the skin of the balloon is $320\text{ kg}$.
3. In reality, the actual acceleration when the balloon is ascending is lesser than the answer you calculated in part (b). Explain why this is so.

ANSWER

Part 1: Calculating the Lifting Force per Unit Volume

This part of the problem asks us to think about the forces acting on just one cubic metre ($1{\text{ m}}^3$) of air. The key principle at play here is Archimedes' Principle, which states that the upward buoyant force (or upthrust) on an object submerged in a fluid is equal to the weight of the fluid that the object displaces.

In our case, the "object" is the hot air inside the balloon, and the "fluid" is the cooler air surrounding it.

Step 1: Calculate the Upthrust per m³ ($U$)

The upthrust is the weight of the displaced surrounding air. We are told the surrounding air is at $20\text{°C}$.

• From the table, the density of air at $20\text{°C}$ is ${\rho }_{surrounding}=1.204{\text{ kg/m}}^3$.
• For a volume of $V=1{\text{ m}}^3$, the mass of this displaced air is:
  $m_{surrounding}={\rho }_{surrounding}\times V=1.204{\text{ kg/m}}^3\times 1{\text{ m}}^3=1.204\text{ kg}$.
• The weight of this displaced air gives us the upthrust per m³:
  $U=m_{surrounding}\times g=1.204\text{ kg}\times 9.8\text{ N/kg}=11.7992\text{ N}$.

So, the upthrust per cubic metre is approximately $11.8{\text{ N/m}}^3$.

Step 2: Calculate the Weight of Air per m³ ($W_A$)

This is the weight of the hot air inside the balloon. The balloon's air temperature is $100\text{°C}$.

• From the table, the density of air at $100\text{°C}$ is ${\rho }_{hot}=0.9461{\text{ kg/m}}^3$.
• For a volume of $V=1{\text{ m}}^3$, the mass of the hot air is:
  $m_{hot}={\rho }_{hot}\times V=0.9461{\text{ kg/m}}^3\times 1{\text{ m}}^3=0.9461\text{ kg}$.
• The weight of this hot air is:
  $W_A=m_{hot}\times g=0.9461\text{ kg}\times 9.8\text{ N/kg}=9.27178\text{ N}$.

So, the weight of the hot air per cubic metre is approximately $9.27{\text{ N/m}}^3$.

Step 3: Show the Lifting Force per m³ ($L$) is about 2.5 N/m³

The lifting force is the net effect of the upthrust pushing up and the weight of the hot air pulling down.

• $L=U-W_A$
• $L=11.7992{\text{ N/m}}^3-9.27178{\text{ N/m}}^3=2.52742{\text{ N/m}}^3$.

This value, $2.52742{\text{ N/m}}^3$, is indeed about $2.5{\text{ N/m}}^3$. We have successfully shown this.

Think of it this way: every cubic metre of hot air inside the balloon is about $2.5\text{ N}$ lighter than the cubic metre of cooler air it has replaced. It's this difference that generates the lift!

Part 2: Calculating the Acceleration

Now, we'll apply Newton's Second Law of Motion ($F_{net}=ma$) to the entire balloon system. We need to find all the forces acting on the balloon and the total mass that is being accelerated.

Step 1: Identify and Calculate the Forces

• Total Upthrust ($F_{upthrust}$): This is the lifting force per unit volume ($L$) we just found, multiplied by the total volume of the balloon's envelope.
  • Wait, let's be careful. The net lift on the air is $L\times V$. But it's clearer to calculate the total upthrust and total weight separately.
  • $F_{upthrust}$ = Weight of $1800{\text{ m}}^3$ of surrounding air at $20\text{°C}$.
  • $F_{upthrust}=({\rho }_{surrounding}\times V)\times g=(1.204{\text{ kg/m}}^3\times 1800{\text{ m}}^3)\times 9.8\text{ N/kg}$
  • $F_{upthrust}=2167.2\text{ kg}\times 9.8\text{ N/kg}=21238.56\text{ N}$.
• Total Weight ($W_{total}$): This is the sum of two downward forces: the weight of the hot air inside the envelope and the weight of the load (skin, basket, passengers, etc.).
  • Weight of hot air ($W_{hot\mathrm{\_}air}$):
    $W_{hot\mathrm{\_}air}=({\rho }_{hot}\times V)\times g=(0.9461{\text{ kg/m}}^3\times 1800{\text{ m}}^3)\times 9.8\text{ N/kg}$
    $W_{hot\mathrm{\_}air}=1702.98\text{ kg}\times 9.8\text{ N/kg}=16689.204\text{ N}$.
  • Weight of the load ($W_{load}$):
    $W_{load}=m_{load}\times g=320\text{ kg}\times 9.8\text{ N/kg}=3136\text{ N}$.
  • $W_{total}=W_{hot\mathrm{\_}air}+W_{load}=16689.204\text{ N}+3136\text{ N}=19825.204\text{ N}$.

Step 2: Calculate the Net Force ($F_{net}$)

• $F_{net}=F_{upthrust}-W_{total}$
• $F_{net}=21238.56\text{ N}-19825.204\text{ N}=1413.356\text{ N}$.

Step 3: Calculate the Total Mass ($m_{total}$)

The total mass being accelerated includes the hot air inside and the load.

• Mass of hot air ($m_{hot\mathrm{\_}air}$) = ${\rho }_{hot}\times V=0.9461{\text{ kg/m}}^3\times 1800{\text{ m}}^3=1702.98\text{ kg}$.
• Mass of the load ($m_{load}$) = $320\text{ kg}$.
• $m_{total}=m_{hot\mathrm{\_}air}+m_{load}=1702.98\text{ kg}+320\text{ kg}=2022.98\text{ kg}$.

Step 4: Calculate the Acceleration ($a$)

• $F_{net}=m_{total}\times a$
• $a=\frac{F_{net}}{m_{total}}=\frac{1413.356\text{ N}}{2022.98\text{ kg}}\approx 0.6986{\text{ m/s}}^2$.

So, the initial acceleration of the hot-air balloon system is approximately $0.70{\text{ m/s}}^2$.

Part 3: Real-World Considerations

This is a great critical thinking question. Our calculation in part (b) is based on an ideal model. In reality, the actual acceleration is less. Why?

The main reason is Air Resistance (or Drag).

Our calculation only considered two forces: gravity (weight) and buoyancy (upthrust). We completely ignored any forces that might oppose the motion. As the balloon begins to ascend, it has to push through the surrounding air. This creates a resistive force, known as air drag, that acts in the opposite direction to the motion.

So, the true net force is:

Since the drag force ($F_{drag}$) is a positive value that subtracts from the upward force, the real net force is smaller than what we calculated. According to Newton's Second Law ($a=F_{net}/m$), a smaller net force results in a smaller acceleration.

Furthermore, this drag force isn't constant; it increases as the balloon's speed increases. This means that as the balloon picks up speed, the drag force grows until it balances the net lift. At that point, the net force becomes zero, the acceleration becomes zero, and the balloon continues to rise at a constant terminal velocity.

SQ-2

There are a number of factors to consider when calculating the sag of overhead electrical power transmission cables. These include weight, weather and thermal aspects of the conductor cables.

In a certain set-up, the span of the two towers supporting the cables is $80.0\text{ m}$ and the conducting cable is made of aluminum with $1.2\text{ cm}$ diameter.

The temperature in this region range between $-{50}^{\circ }C$ and $+{40}^{\circ }C$.

The density of aluminum is $ \rho = 2.7 \text{ g/cm}^3 $.

The coefficient of linear expansion of aluminum is $ \alpha = 25 \times 10^{-6} \text{ }^{\circ}\text{C}^{-1} $ and its specific heat capacity is $ 900 \text{ J/kg}^\circ\text{C} $.

a. If a cable was strung without sag, it would snap during cold weather. Determine the minimum extra length of the aluminum cable (at 20°C) needed so as to avoid the cable from snapping. You can assume that the span is constant at 80.0 m for all temperatures.
b. If 80.4 m of 1.2 cm diameter aluminium cable is used, determine the mass $M$ of the cable.
c. If the power lost as heat in the 80.4 m of cable is 470 J / s, determine the increase in temperature per minute. Assume no heat is lost to the surrounding.
d. The tension ($T$) in the cable can be estimated in a very simplified way by assuming that the weight ($Mg$) of the aluminium cable just acts in the center of cable and that the rest of the cable is massless. With the above assumptions, determine the tension in the cable.
e. State explicitly two weather conditions that will increase the tension in the overhead transmission cables.

ANSWER

a. Minimum Extra Length of the Cable

Concept:

The core principle here is thermal contraction. Materials shrink when they get cold. If a cable is strung perfectly taut during a moderate day (say, ${20}^{\circ }\text{C}$), it will try to contract when the temperature drops. Since the towers are fixed, the cable can't get shorter, which creates immense tension and can cause it to snap.

To prevent this, we need to install a cable that is slightly longer than the span. The crucial condition is that at the absolute minimum temperature, the cable's length should be exactly equal to the span, so it is perfectly straight but under no tension.

Calculation:

We'll use the formula for linear thermal expansion: $\mathrm{\Delta }L=\alpha L_0\mathrm{\Delta }T$, where:

• $\mathrm{\Delta }L$ is the change in length.
• $\alpha$ is the coefficient of linear expansion ($25\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$).
• $L_0$ is the original length (the length we need to find at ${20}^{\circ }\text{C}$).
• $\mathrm{\Delta }T$ is the change in temperature.

The final length, $L_f$, is $L_0+\mathrm{\Delta }L$. We want this final length to be $80.0\text{ m}$ at the coldest temperature, $-{50}^{\circ }\text{C}$.

The temperature change is $\mathrm{\Delta }T=(\text{final temperature})-(\text{initial temperature})=-{50}^{\circ }\text{C}-{20}^{\circ }\text{C}=-{70}^{\circ }\text{C}$.

So, we have the equation:
$80.0\text{ m}=L_0+\alpha L_0\mathrm{\Delta }T=L_0(1+\alpha \mathrm{\Delta }T)$

Let's solve for $L_0$:
$L_0=\frac{80.0\text{ m}}{1+(25\times {10}^{-6}{}^{\circ }{\text{C}}^{-1})(-{70}^{\circ }\text{C})}$
The calculation shows that the length of the cable at ${20}^{\circ }\text{C}$ must be approximately $80.14\text{ m}$.

Answer (a): The minimum extra length needed is $80.14\text{ m}-80.0\text{ m}=\mathbf{0.14}\text{ m}$, or $14.0\text{ cm}$.

b. Mass of the Cable

Concept:

To find the mass, we'll use the fundamental relationship: $\text{Mass}=\text{Density}\times \text{Volume}$. We are given the density of aluminum and the dimensions of the cable (length and diameter), so we can calculate its volume assuming it's a cylinder.

Calculation:

1. Convert Units: It's best practice to work in standard SI units (meters, kilograms).

   • Density: $\rho =2.7{\text{ g/cm}}^3=2700{\text{ kg/m}}^3$
   • Diameter: $1.2\text{ cm}=0.012\text{ m}$, so the radius is $r=0.006\text{ m}$.
   • Length: $L=80.4\text{ m}$.

2. Calculate Volume: The volume $V$ of a cylinder is $V=\pi r^{2}L$.
   $V=\pi (0.006\text{ m}{)}^2(80.4\text{ m})$

3. Calculate Mass: $M=\rho V$
   $M=(2700{\text{ kg/m}}^3)\times V$

Answer (b): The mass $M$ of the $80.4\text{ m}$ cable is approximately $\mathbf{24.55}\text{ kg}$.

c. Increase in Temperature per Minute

Concept:

This part deals with specific heat capacity. The power lost in the cable is dissipated as heat, which raises the cable's temperature. The formula that connects heat energy and temperature change is $Q=mc\mathrm{\Delta }T$, where:

• $Q$ is the heat energy added.
• $m$ is the mass of the object.
• $c$ is the specific heat capacity ($900{\text{ J/kg}}^{\circ }\text{C}$).
• $\mathrm{\Delta }T$ is the change in temperature.

Calculation:

1. Find Heat Energy (Q): We are given the power (energy per second), which is $470\text{ J/s}$. We need the total energy in one minute (60 seconds).
   $Q=\text{Power}\times \text{time}=470\text{ J/s}\times 60\text{ s}=28200\text{ J}$

2. Calculate Temperature Increase: We rearrange the formula to solve for $\mathrm{\Delta }T$:
   $\mathrm{\Delta }T=\frac{Q}{mc}$
   Using the mass we found in part (b), $m=24.55\text{ kg}$.
   $\mathrm{\Delta }T=\frac{28200\text{ J}}{(24.55\text{ kg})(900{\text{ J/kg}}^{\circ }\text{C})}$

Answer (c): The increase in temperature per minute is approximately ${\mathbf{1.28}}^{\circ }\text{C}$.

d. Tension in the Cable

Concept:

This is a problem in statics. We are using a simplified model where the entire weight of the cable ($Mg$) acts at its center. The cable sags downwards, forming a shallow 'V' shape. The tension in the cable pulls upwards and outwards, counteracting the downward pull of gravity.

Calculation:

1. Find the Sag (d): The cable's length ($80.4\text{ m}$) forms the hypotenuse (sisi miring) of two back-to-back right-angled triangles (segitiga siku siku). The span ($80.0\text{ m}$) forms the horizontal base. The sag ($d$) is the vertical height.

   • Half the cable length: $80.4/2=40.2\text{ m}$.
   • Half the span: $80.0/2=40.0\text{ m}$.
     Using Pythagoras' theorem ($a^2+b^2=c^2$):
     $d^2+(40.0{)}^2=(40.2{)}^2$
     $d=\sqrt{(40.2{)}^2-(40.0{)}^2}=\sqrt{16.04}\approx 4.005\text{ m}$

2. Analyze the Forces: At the center point, the downward force is the cable's weight, $W=Mg$. This is balanced by the vertical components of the tension ($T$) from each side of the cable.

   • $W=(24.55\text{ kg})(9.81{\text{ m/s}}^2)\approx 240.8\text{ N}$.
   • Let $\theta$ be the angle the cable makes with the horizontal. The vertical component of tension from one side is $T\sin \theta$. Since there are two sides, the total upward force is $2T\sin \theta$.
   • For equilibrium: $2T\sin \theta =W$.

3. Find $\sin \theta$ and Solve for T: From the geometry of our right-angled triangle:
   $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\text{sag}}{\text{half cable length}}=\frac{4.005\text{ m}}{40.2\text{ m}}$
   Now, we can find the tension $T$:
   $T=\frac{W}{2\sin \theta }=\frac{240.8\text{ N}}{2\times (\frac{4.005}{40.2})}$

Answer (d): The tension in the cable is approximately $\mathbf{1209}\text{ N}$.

e. Weather Conditions

Based on the principles we've discussed, here are two weather conditions that will increase the tension in the cables, along with the physical reasoning.

1. Low Environmental Temperatures

• Reasoning: As we established in part (a), the principle of thermal contraction is a primary factor. When the surrounding air gets colder, the aluminum cable cools down and attempts to contract or shorten. Since the cable is fixed between two towers that are a constant distance apart, it cannot freely shrink. This resistance to contraction creates a pulling force, or tension, within the cable. The colder it gets, the more the cable wants to contract, and the higher the tension becomes.

1. Ice or Snow Accumulation (Icing)

• Reasoning: In part (d), we saw that the tension in the cable is directly related to its total weight ($W=Mg$). Under certain weather conditions, such as freezing rain or wet snow, a thick layer of ice can form around the cable. This accumulated ice can be surprisingly heavy, adding significant mass to the cable. To support this extra weight and prevent the cable from sagging excessively, the tension must increase dramatically. This is one of the most dangerous conditions for power lines and can lead to catastrophic failure of the cables or even the support towers.

A strong wind would be another valid answer, as it exerts a sideways force that adds to the total load the cable must bear, thereby increasing tension. However, cold and icing are typically the most significant weather-related factors.