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?
Ke of man = (1/2) m v^2
Ke of flywheel = (1/2) I w^2 = (1/2)I (v/r)^2
Total Ke = (1/2) m v^2 + (1/2) (I/r^2)v^2
potential energy decrease = m g h = 90(9.81)(80) = 70,632 Joules
constant acceleration
average v = 80 meters/8 seconds = 10 m/s
initial v is zero so final v is 20 m/s
so
70,632 = (1/2)(20)^2 [ m + I/r^2 ]
I think you can take it from there.
To solve this problem, we need to apply the principle of conservation of energy. We can use the potential energy of the stuntman at the top of the building and convert it to kinetic energy as he falls. This energy is then transferred to the rotational kinetic energy of the flywheel.
(a) To find the moment of inertia of the flywheel with a radius of 0.5m, we can use the formula for the kinetic energy of rotation:
KE = (1/2) * I * ω^2
Where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
First, let's find the velocity of the stuntman just before he reaches the ground. We can use the equation of motion:
v = u + at
Since the initial velocity (u) is 0, we have:
v = gt
Where g is the acceleration due to gravity. Substituting the values, g = 9.8 m/s² and t = 8 s, we find:
v = 9.8 m/s² * 8 s
v = 78.4 m/s
Now, the kinetic energy of the stuntman just before he hits the ground is given by:
KE = (1/2) * m * v^2
Where m is the mass of the stuntman. We are given that the stuntman weighs 90kg, so:
KE = (1/2) * 90 kg * (78.4 m/s)^2
KE = 275,184 J
Since the kinetic energy is transferred to the rotational kinetic energy of the flywheel, we can equate the two. The kinetic energy of rotation is given by:
KE = (1/2) * I * ω^2
The angular velocity ω can be found using the relationship between linear velocity and angular velocity:
v = r * ω
Where r is the radius of the flywheel. Substituting the values, r = 0.5m and v = 78.4 m/s, we find:
78.4 m/s = 0.5 m * ω
ω = 156.8 rad/s
Now we can substitute the values back into the equation for rotational kinetic energy and solve for I:
275,184 J = (1/2) * I * (156.8 rad/s)^2
275,184 J = (1/2) * I * (24,598.24 rad^2/s^2)
275,184 J = 12,299.12 * I
Rearranging the equation to solve for I gives us:
I = 275,184 J / 12,299.12
I ≈ 22.38 kg·m²
Therefore, the moment of inertia of the flywheel with a radius of 0.5m is approximately 22.38 kg·m².
(b) To find the moment of inertia if the radius of the flywheel is 1m, we can follow the same process as before, but substitute the new radius value. We already know the kinetic energy (KE) and the angular velocity (ω), so we just need to solve for the moment of inertia (I) by rearranging the equation:
KE = (1/2) * I * ω^2
Using the previously calculated values of KE = 275,184 J and ω = 156.8 rad/s, we can substitute them into the equation and solve for I:
275,184 J = (1/2) * I * (156.8 rad/s)^2
275,184 J = (1/2) * I * (24,598.24 rad^2/s^2)
275,184 J = 12,299.12 * I
Rearranging the equation to solve for I gives us:
I = 275,184 J / 12,299.12
I ≈ 22.38 kg·m²
Therefore, the moment of inertia of the flywheel with a radius of 1m is also approximately 22.38 kg·m².
(c) The equation for the moment of inertia of a solid cylinder is given by:
I = (1/2) * m * r²
Where I is the moment of inertia, m is the mass, and r is the radius. Rearranging the equation to solve for mass (m), we have:
m = 2 * I / r²
For the flywheel with a radius of 0.5m, we can substitute the values of I ≈ 22.38 kg·m² and r = 0.5m into the equation and calculate the mass (m):
m = 2 * 22.38 kg·m² / (0.5m)²
m = 2 * 22.38 kg·m² / 0.25m²
m = 89.52 kg
So, the mass of the flywheel with a radius of 0.5m is approximately 89.52 kg.
For the flywheel with a radius of 1m, we can follow the same process as above, substituting I ≈ 22.38 kg·m² and r = 1m into the equation to find the mass (m):
m = 2 * 22.38 kg·m² / (1m)²
m = 2 * 22.38 kg·m² / 1m²
m = 44.76 kg
Therefore, the mass of the flywheel with a radius of 1m is approximately 44.76 kg.