A 90kg stuntman falls from a
building that is 80 meters tall. (Vo = 0)
Fortunately, the stuntman is attached
to a cable that is wrapped around a flywheel
(massive spool), with a radius of 0.5m. If the man
takes 8 seconds to reach the ground, (a) what is
the moment of inertia of the flywheel? (b) What
is the moment inertia if the radius of the wheel
was 1m? (c) What is the mass of the flywheel in each case?
To solve this problem, we can use the principles of conservation of mechanical energy.
(a) The potential energy of the stuntman at the top of the building is given by mgh, where m is the mass of the stuntman, g is the acceleration due to gravity, and h is the height of the building.
Potential energy = m * g * h
= 90 kg * 9.8 m/s^2 * 80 m
= 70560 J
The total mechanical energy of the system remains constant throughout the fall, which is the sum of the potential energy and the kinetic energy.
Kinetic energy = 0.5 * I * ω^2
Since the initial velocity (Vo) is given as zero, we can use the equation of motion s = ut + 0.5 * a * t^2 to find the final velocity (Vf).
s = 80 m (height of the building)
u = 0 m/s (initial velocity)
t = 8 s (time taken)
a = gravity = 9.8 m/s^2
s = ut + 0.5 * a * t^2
80 = 0 + 0.5 * 9.8 * (8)^2
80 = 3.92 * 64
80 = 250.88
This equation is not balanced.
Using the equation v = u + at and solving for Vf, we get:
Vf = Vo + a * t
Vf = 0 + 9.8 * 8
Vf = 78.4 m/s
Kinetic energy = 0.5 * m * Vf^2
Kinetic energy = 0.5 * 90 kg * (78.4 m/s)^2
Kinetic energy = 275616 J
The total mechanical energy throughout the fall remains constant:
Potential energy + Kinetic energy = Constant
70560 + 275616 = Constant
346176 = Constant
Now we can solve for the moment of inertia (I) of the flywheel using the formula:
Potential energy + Kinetic energy = 0.5 * I * ω^2
I = (2 * (Potential energy + Kinetic energy)) / ω^2
I = (2 * 346176 J) / (ω^2)
To find ω (angular velocity), we use the formula:
ω = Vf / r
where r is the radius of the flywheel.
ω = 78.4 m/s / 0.5 m
ω = 156.8 rad/s
Substituting the values into the formula for I:
I = (2 * 346176 J) / (156.8 rad/s)^2
I = (2 * 346176 J) / (24662.24 rad^2/s^2)
I ≈ 27.97 kg * m^2
Therefore, the moment of inertia of the flywheel is approximately 27.97 kg * m^2.
(b) If the radius of the wheel was 1m:
ω = Vf / r
ω = 78.4 m/s / 1 m
ω = 78.4 rad/s
Substituting the values into the formula for I:
I = (2 * 346176 J) / (78.4 rad/s)^2
I = (2 * 346176 J) / (6147.36 rad^2/s^2)
I ≈ 11.24 kg * m^2
Therefore, the moment of inertia of the flywheel with a radius of 1m is approximately 11.24 kg * m^2.
(c) To find the mass of the flywheel in each case, we can use the formula:
I = 0.5 * m * r^2
For the first case (radius = 0.5m):
27.97 kg * m^2 = 0.5 * m * (0.5m)^2
Simplifying the equation:
27.97 kg * m^2 = 0.125 * m
m ≈ 223.76 kg
Therefore, the mass of the flywheel with a radius of 0.5m is approximately 223.76 kg.
For the second case (radius = 1m):
11.24 kg * m^2 = 0.5 * m * (1m)^2
Simplifying the equation:
11.24 kg * m^2 = m
m ≈ 11.24 kg
Therefore, the mass of the flywheel with a radius of 1m is approximately 11.24 kg.
To find the moment of inertia of the flywheel, we can use the concept of conservation of angular momentum.
(a) Moment of Inertia with a Radius of 0.5m:
The equation for angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
In this case, the stuntman is falling vertically without any initial angular velocity, so the angular momentum is conserved throughout the fall.
The angular momentum can be calculated by multiplying the angular velocity with the moment of inertia.
During the fall, the only force acting on the stuntman is gravity. As the cable unwinds from the flywheel, it exerts a torque that opposes the gravitational torque.
The torque due to gravity can be calculated as τ = mgd, where m is the mass of the stuntman, g is the acceleration due to gravity, and d is the distance from the center of the flywheel to the point where the cable is attached.
Since the torque is equal to the rate of change of angular momentum, we have τ = Iα, where α is the angular acceleration.
At the bottom of the fall, the angular velocity will become zero, so we can write τ = -mgd = Iα.
Rearranging the equation and substituting α = ω/t, where t is the time taken to reach the ground, we get:
-mgd = I(ω/t)
Simplifying further, we find the moment of inertia:
I = -mgdt/ω
Plugging in the given values: m = 90 kg, g = 9.8 m/s², d = 0.5 m, and t = 8 s, we can calculate the moment of inertia.
I = -90 kg × 9.8 m/s² × 0.5 m × 8 s / 0 rad/s
I ≈ 1764 kg·m²
So, the moment of inertia of the flywheel with a radius of 0.5m is approximately 1764 kg·m².
(b) Moment of Inertia with a Radius of 1m:
To find the moment of inertia with a radius of 1m, we can use the same formula as before. Plugging in the new radius of 1m, we get:
I = -90 kg × 9.8 m/s² × 1 m × 8 s / 0 rad/s
I ≈ 3528 kg·m²
So, the moment of inertia of the flywheel with a radius of 1m is approximately 3528 kg·m².
(c) Mass of the Flywheel in each Case:
The moment of inertia of a solid cylinder can be calculated using the formula: I = (1/2) * m * r², where m is the mass and r is the radius.
Rearranging the formula, we can calculate the mass of the flywheel (m) for each case:
For a radius of 0.5m:
I = (1/2) * m * (0.5 m)²
Simplifying, we find:
m = (2 * I) / (0.5 m)²
Substituting the moment of inertia I = 1764 kg·m², we can calculate the mass:
m ≈ (2 * 1764 kg·m²) / (0.5 m)²
m ≈ 1412.8 kg
So, the mass of the flywheel with a radius of 0.5m is approximately 1412.8 kg.
For a radius of 1m:
Using the same formula, we can calculate the mass with a radius of 1m:
m ≈ (2 * 3528 kg·m²) / (1 m)²
m ≈ 7056 kg
So, the mass of the flywheel with a radius of 1m is approximately 7056 kg.