Question
Before a small party balloon is inflated, it is placed on an electronic weighing scale which records its mass as 1.3g. The balloon is now inflated with air so that it is spherical in shape with a diameter of 22cm. If the density of air in the room is 1.21kg/m^3.
Before a small party balloon is inflated, it is placed on an electronic weighing scale which records its mass as 1.3g. The balloon is now inflated with air so that it is spherical in shape with a diameter of 22cm. If the density of air in the room is 1.21kg/m^3.
i) Calculate the mass of air displaced by the balloon,
ii) The inflated balloon in i) is placed on the scale which now gives a reading of 1.55g. Calculate the mass of air in the balloon and suggest a reason as to why this value is more than the value you calculated in i),
iii) A 2.1g nut is now tied to the bottom of the inflated balloon using a light cotton thread. The balloon is dropped from a height of 3m above the ground.
a) Show that the magnitude of the acceleration of the balloon and nut at the start of their descent is 3.45m/s^2
b) Explain why the value of this acceleration will approach zero as the balloon decends,
c) Calculate the magnitude of the drag force acting on the balloon and nut when they are moving at their terminal velocity of 2m/s,
d) The value of the drag force Fd is given by : Fd = (CdrouAv^2)/2 where Cd is the drag coefficient , rou is the density of air and A is the cross-sectional area of the balloon. Calculate the value of the drag coefficient Cd for the balloon,
iv)The same balloon used in i) and ii) is now inflated with helium (density of helium in the balloon = 0.166 kg/m^3) instead of air so that it has the same volume as the air balloon,
a) Calculate the downward force that would be required to hold this balloon alone (no nut attached) at a constant height of 3m above the ground,
b)The downward force is removed and the helium balloon is allowed to rise.
Explain how the conservation of energy is applicable to the motion.
Answer
i)
Volume of air displaced
= (4/3)πr^3
= (4/3)π * (0.11)^3 m^3
= 0.0055753 m^3
Mass of air diaplaced
= 1210 * 0.0055753 g
= 6.74 g
ii)
Let m = mass of air in the balloon
=> final weight of balloon
= weight of deflated balloon + weight of air filled in - weight of air displaced
=> 1.30g + mg - 6.74g = 1.55g
=> m = 6.74 + 1.55 - 1.30 g = 6.99 g
The air filled in and air displaced have the same volume, but the air filled in has higher pressure than the pressure of air outside which is at atmospheric pressure. Hence, inside air has higher density and higher mass.
iii) a)
Apparent weight of the balloon + nut
= (1.55 + 2.1) * 10^-3 * 9.81 N = 0.0358065 N
Mass of balloon + air + nut
= 1.3 + 6.99 + 2.1 = 10.39 * 10^-3 kg
Acceleration at the start
= (0.0358065) /(10.39 x 10^-3) m/s^2
= 3.45 m/s^2
iii) b)
As the balloon descends, three forces are acting on it.
1) its weight downwards,
2) force of buoyancy upwards and
3) force of air resistance upwards.
Of these, the first two forces are constant.
Air resistance is proportioanl to the downward velocity of the balloon. As the initial velocity at the start, downward velocity is zero, there is no air resistance and the balloon has the maximum downward acceleration. As it gains velocity due to this acceleration, the upward air resistance increases with velocity till it balances the first two constant forces. At that instant as all the three forces balance each other (taken with appropriate signs), there is no net force on the balloon and it continues to move down with whatever velocity it has gained by that time called the terminal velocity and has zero acceleration due to zero net force on it.
From the instant velocity reaches terminal velocity,
Buoyant force - Weight of balloon = Downward resistive force
[Note: Please note that this discussion is valid near the surface of the earth, where we can consider acceleration due to gravity "g" and air density to remain almost constant. At great heights, both "g" and air density vary and hence terminal velocity will change and not remain constant. It will reduce and ultimately become zero.]
iii) c)
Drag force
= apparent weight
= (1.55 + 2.1) x 10^-3 * 9.81 N
= 3.58 x 10^-2 N.
iii) d)
Fd=(CdrouAv^2)/2
=> 3.58 x 10^-2 = Cd x 1.21 x π(0.11)^2 * 2^2/2
=> Cd
= (0.0358) / (2.42 π * 0.0121)
= 0.389.
iv) a)
Mass of He filled
= volume x density
= 0.0055753 m^3 x 0.166 kg/m^2
= 0.0009255 kg
If F = downward force to be applied, then
F = weight of air displaced - weight of balloon
= (0.00674 - 0.0013 - 0.0009255) * 9.81 N
= 0.0443 N.
iv) b)
The He balloon on the surface of the earth has positive potential energy stored in it due to the force of buoyancy. As it rises, this energy is getting converted into gravitational potential energy and kinetic energy and a part is lost to overcome the resisitive force. Thus,
ΔPE(buoyancy) = ΔPE(gravitational) + ΔKE + work done against the resistive force.
In the process, total energy of the balloon and environment is conserved, but the mechanical energy of the balloon is reducing. This process of interconversion of the energies comes to a stop when the total
mechanical energy of the balloon (= PE due to buoyancy + PE due to gravity + KE) becomes zero.
ii) The inflated balloon in i) is placed on the scale which now gives a reading of 1.55g. Calculate the mass of air in the balloon and suggest a reason as to why this value is more than the value you calculated in i),
iii) A 2.1g nut is now tied to the bottom of the inflated balloon using a light cotton thread. The balloon is dropped from a height of 3m above the ground.
a) Show that the magnitude of the acceleration of the balloon and nut at the start of their descent is 3.45m/s^2
b) Explain why the value of this acceleration will approach zero as the balloon decends,
c) Calculate the magnitude of the drag force acting on the balloon and nut when they are moving at their terminal velocity of 2m/s,
d) The value of the drag force Fd is given by : Fd = (CdrouAv^2)/2 where Cd is the drag coefficient , rou is the density of air and A is the cross-sectional area of the balloon. Calculate the value of the drag coefficient Cd for the balloon,
iv)The same balloon used in i) and ii) is now inflated with helium (density of helium in the balloon = 0.166 kg/m^3) instead of air so that it has the same volume as the air balloon,
a) Calculate the downward force that would be required to hold this balloon alone (no nut attached) at a constant height of 3m above the ground,
b)The downward force is removed and the helium balloon is allowed to rise.
Explain how the conservation of energy is applicable to the motion.
Answer
i)
Volume of air displaced
= (4/3)πr^3
= (4/3)π * (0.11)^3 m^3
= 0.0055753 m^3
Mass of air diaplaced
= 1210 * 0.0055753 g
= 6.74 g
ii)
Let m = mass of air in the balloon
=> final weight of balloon
= weight of deflated balloon + weight of air filled in - weight of air displaced
=> 1.30g + mg - 6.74g = 1.55g
=> m = 6.74 + 1.55 - 1.30 g = 6.99 g
The air filled in and air displaced have the same volume, but the air filled in has higher pressure than the pressure of air outside which is at atmospheric pressure. Hence, inside air has higher density and higher mass.
iii) a)
Apparent weight of the balloon + nut
= (1.55 + 2.1) * 10^-3 * 9.81 N = 0.0358065 N
Mass of balloon + air + nut
= 1.3 + 6.99 + 2.1 = 10.39 * 10^-3 kg
Acceleration at the start
= (0.0358065) /(10.39 x 10^-3) m/s^2
= 3.45 m/s^2
iii) b)
As the balloon descends, three forces are acting on it.
1) its weight downwards,
2) force of buoyancy upwards and
3) force of air resistance upwards.
Of these, the first two forces are constant.
Air resistance is proportioanl to the downward velocity of the balloon. As the initial velocity at the start, downward velocity is zero, there is no air resistance and the balloon has the maximum downward acceleration. As it gains velocity due to this acceleration, the upward air resistance increases with velocity till it balances the first two constant forces. At that instant as all the three forces balance each other (taken with appropriate signs), there is no net force on the balloon and it continues to move down with whatever velocity it has gained by that time called the terminal velocity and has zero acceleration due to zero net force on it.
From the instant velocity reaches terminal velocity,
Buoyant force - Weight of balloon = Downward resistive force
[Note: Please note that this discussion is valid near the surface of the earth, where we can consider acceleration due to gravity "g" and air density to remain almost constant. At great heights, both "g" and air density vary and hence terminal velocity will change and not remain constant. It will reduce and ultimately become zero.]
iii) c)
Drag force
= apparent weight
= (1.55 + 2.1) x 10^-3 * 9.81 N
= 3.58 x 10^-2 N.
iii) d)
Fd=(CdrouAv^2)/2
=> 3.58 x 10^-2 = Cd x 1.21 x π(0.11)^2 * 2^2/2
=> Cd
= (0.0358) / (2.42 π * 0.0121)
= 0.389.
iv) a)
Mass of He filled
= volume x density
= 0.0055753 m^3 x 0.166 kg/m^2
= 0.0009255 kg
If F = downward force to be applied, then
F = weight of air displaced - weight of balloon
= (0.00674 - 0.0013 - 0.0009255) * 9.81 N
= 0.0443 N.
iv) b)
The He balloon on the surface of the earth has positive potential energy stored in it due to the force of buoyancy. As it rises, this energy is getting converted into gravitational potential energy and kinetic energy and a part is lost to overcome the resisitive force. Thus,
ΔPE(buoyancy) = ΔPE(gravitational) + ΔKE + work done against the resistive force.
In the process, total energy of the balloon and environment is conserved, but the mechanical energy of the balloon is reducing. This process of interconversion of the energies comes to a stop when the total
mechanical energy of the balloon (= PE due to buoyancy + PE due to gravity + KE) becomes zero.
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