1,Bungee jumping from the New River Gorge Bridge is no longer allowed. But when it was, the jump was about 250 m from the take-off point to the bottom of the gorge. Suppose the length of the bungee cord is 100 m and the mass of the jumper is 75 kg. What is the minimum spring constant needed to make the jump safely?
2, We propose to use the energy stored in a flywheel to power our car. The flywheel is a uniform cylinder of mass 600 kg and radius 0.75 m. We use an electric motor to get it spinning at an initial angular speed of 3000 rev/min. Assuming no frictional losses and perfect transfer of energy, how long can our car operate if its average power requirement is 2.5 kW? If you need it the moment of inertia of a cylinder is 1/2 mr^2.
The Correct Answer and Explanation is :
Problem 1: Bungee Jumping and Spring Constant Calculation
To calculate the minimum spring constant for the bungee jump, we need to use the principles of energy conservation. The jumper’s potential energy at the top of the jump is converted into the potential energy stored in the bungee cord (spring).
Given:
- Mass of the jumper, ( m = 75 \, \text{kg} )
- Length of the bungee cord, ( L = 100 \, \text{m} )
- Height of the jump, ( h = 250 \, \text{m} )
- Gravitational acceleration, ( g = 9.8 \, \text{m/s}^2 )
The jumper’s gravitational potential energy at the start is:
[
E_{\text{grav}} = mgh = 75 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 250 \, \text{m} = 183750 \, \text{J}
]
This energy is converted into elastic potential energy in the bungee cord when it stretches. The elastic potential energy stored in the cord is given by:
[
E_{\text{elastic}} = \frac{1}{2} k x^2
]
Where:
- ( k ) is the spring constant
- ( x ) is the extension of the cord (which we assume to be ( x = h – L = 250 \, \text{m} – 100 \, \text{m} = 150 \, \text{m} ))
Equating the gravitational potential energy to the elastic potential energy, we have:
[
183750 \, \text{J} = \frac{1}{2} k (150)^2
]
Solving for ( k ):
[
k = \frac{2 \times 183750}{150^2} = \frac{367500}{22500} = 16.33 \, \text{N/m}
]
Thus, the minimum spring constant required is ( k = 16.33 \, \text{N/m} ).
Problem 2: Flywheel Power Duration
For the flywheel, we need to calculate the time the car can operate using the energy stored in the flywheel. The energy stored in the flywheel is given by its rotational kinetic energy:
[
E_{\text{rot}} = \frac{1}{2} I \omega^2
]
Where:
- ( I ) is the moment of inertia of the flywheel (given as ( I = \frac{1}{2} m r^2 ))
- ( \omega ) is the angular speed of the flywheel in radians per second
- ( m = 600 \, \text{kg} ) (mass of the flywheel)
- ( r = 0.75 \, \text{m} ) (radius of the flywheel)
- Initial angular speed ( \omega = 3000 \, \text{rev/min} = 3000 \times \frac{2\pi}{60} = 314.16 \, \text{rad/s} )
The moment of inertia is:
[
I = \frac{1}{2} \times 600 \times (0.75)^2 = 168.75 \, \text{kg} \cdot \text{m}^2
]
Now, calculating the energy stored in the flywheel:
[
E_{\text{rot}} = \frac{1}{2} \times 168.75 \times (314.16)^2 = 8339295.85 \, \text{J}
]
The car requires 2.5 kW of power, which is 2500 J/s. The duration the flywheel can power the car is:
[
\text{Time} = \frac{E_{\text{rot}}}{P} = \frac{8339295.85}{2500} = 3335.72 \, \text{seconds} \approx 55.6 \, \text{minutes}
]
Thus, the car can operate for approximately 55.6 minutes.
These problems illustrate energy conservation principles in both mechanical and rotational systems. The bungee jump calculation shows the relationship between gravitational potential energy and elastic potential energy, while the flywheel problem focuses on how rotational kinetic energy can be transferred to do work over time.