Hanging Hot Air Balloon Model
March 4th, 2010 | 
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Blue Royal Aero Hot Air Balloon Authentic Model Hanging Aviation Decor Large 13" | |
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AUTHENTIC MODELS Blue Hanging Royal Aero Hot Air Balloon 32 cm Aviation Flight | |
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Hanging Hot Air Balloon Model
Help with Isolated System?
A daredevil plans to bungee-jump from a hot-air balloon 65.0 meters above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 meters above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test, hanging at rest from a 5.00 meter length of the cord, the daredevil finds his body weight stretches the cord by 1.50 meters. He intends to drop from rest at the point where the top end of a longer section of the cord is attached to the stationary hot-air balloon.
A) What length of cord should he use?
B) What maximum acceleration will he experience?
(Please show work so I can understand this problem)
When in equilibrium the cord stretches , d = 1.5 m , This allows one to express the spring constant ,K, in terms of man's weight;
Kd = mg
K = mg/d
Using conservation of energy. Measuring down as positive from the rim of the balloon. He begins with zero, velocity, stretch & position so his initial energy is zero. As he falls he loses gravitational PE "mgY" , gains elastic PE (once the cord extends beyond its no stretch length "L") "(1/2)K(Y-L)^2 " and aquires KE = (!/2)mV^2 . THe sum must remain zero;
0 = (1/2)mV^2 - mgY + (1/2)(mg/d)(Y-L)^2
At Y = 65 - 10 = 55m we require V = 0
(Y-L) = SqRt[2dY]
L = 55 - SqRt[(2)(1.5)(55)]
= 42.2 m , This is the unstretched length of cord
b) Untill the cord begins stretching, at Y = L , the person falls with accel. "g" , after the cord begins to stretch the net force is;
mg - K(Y-L) = ma
Since Y > L or = L , the 2nd term is always positive or zero and it is seen that "a" will be a max at Y = L and a(max) = g
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Blue Royal Aero Hot Air Balloon Authentic Model Hanging Aviation Decor Large 13" | |
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Royal Aero Yellow 13" Hot Air Balloon Model Decorative Hanging Aviation Decor | |
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Blue Royal Aero Hot Air Balloon Authentic Model Hanging Aviation Decor Large 13" | |
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US $74.99 | 16d 29m |
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AUTHENTIC MODELS Blue Hanging Royal Aero Hot Air Balloon 32 cm Aviation Flight | |
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Red Travels Light 7" Hot Air Balloon Model Decorative Hanging Aviation Decor NIB | |
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Royal Aero Yellow 13" Hot Air Balloon Model Decorative Hanging Aviation Decor | |
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Green Royal Aero 13" Hot Air Balloon Model Decorative Hanging Aviation Decor New | |
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Hot-air balloon ride over Valley of the Kings, Luxor 10/30/09
Bungee-jumping (kinetic energy) physics problem?
A daredevil plans to bungee-jump from a hot-air balloon 65m above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10m above the ground. Model his body as a particle/cord having negligible mass & obeying Hooke's law. In a preliminary test, hanging at rest from a 5m length of the cord, he finds his body weight stretches the cord by 1.5m.
a. What length of cord should he use?
b. What maximum acceleration will he experience?
a.
For the cord, k = mg/1.5, where m = mass of the jumper.
Let L = length of the cord required.
When he falls through the height L, he loses PE = mgL and the cord becomes taut. Thereafter, it starts stretching and stretches till the PE lost by the jumper is stored in the cord in the form of PE of the cord.
Let the cord stretch by x meter
=> mgL = (1/2)kx^2 and L+x = 65 - 10 = 55
=> mgL = (1/2)*(mg/1.5)*(55-L)^2
=> 3L = L^2 - 110L + 3025
=> L^2 - 113L + 3025 = 0
=> L = (1/2)[113 ± √(113^2 - 4*3025)]
=> L = (1/2)[113 ± 25.865] = 69.4 or 43.6 m
=> length of cord should be 43.6 m.
b.
At the lowest point, restoring force in the cord in the upward direction
= k*elongation of the cord
= (mg/1.5)*(55-43.6)
= mg * (11.4/1.5)
Downward force = mg
Net force upwards = mg*(11.4/1.5) - mg = mg(9.9/1.5)
Maxm. acceleration upwards
= Net Force/mass
= mg(9.9/1.5)/m
= 6.6g m/s^2
= 6.6 * 9.8 m/s^2
= 64.68 m/s^2.
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