You are the technical consultant for an action-adventure film in which a stunt calls for the hero to drop off a 25.1-m tall building and land on the ground safely at a final vertical speed of 3.81 m/s. At the edge of the building's roof, there is a 103-kg drum that is wound with a sufficiently long rope (of negligible mass), has a radius of 0.387 m, and is free to rotate about its cylindrical axis with a moment of inertia I0. The script calls for the 56.9-kg stuntman to tie the rope around his waist and walk off the roof.
a) Determine the required value of the stuntman's acceleration if he is to land safely at a speed of 3.81 m/s, and use this value to calculate the moment of inertia of the drum about its axis.
b) What is the angular acceleration of the drum?
c) How many revolutions does the drum make during the fall?
a)
1. vf^2 = vi^2 + 2ad for linear acceleration
2. a = r*α
3. -F= mg-m(αr)
4. (-rF)/α = I
b) verified in 2. of a)
c)
1. circumference 2pi*r
2. distance/circumfrence = revolutions
a) To determine the required value of the stuntman's acceleration, we can use the principles of conservation of energy. The potential energy of the stuntman at the top of the building (mgh) is converted into both kinetic energy (1/2mv^2) and rotational kinetic energy (1/2Iω^2) at the bottom of the fall, where m is the mass of the stuntman, g is the acceleration due to gravity, h is the height of the building, v is the final vertical speed, I is the moment of inertia of the drum, and ω is the angular speed of the drum.
Applying the conservation of energy equation:
mgh = 1/2mv^2 + 1/2Iω^2
Since the stuntman ties the rope around his waist, the acceleration of the stuntman is the same as the angular acceleration of the drum.
So we have a = α
Next, we need to express the angular speed, ω, in terms of the velocity, v.
The drum is wound with a rope, so the radius, r, of the drum is related to the distance fallen by the stuntman, h, as:
r = h + (L/2)
where L is the length of rope wound on the drum.
The angular speed, ω, can be related to the velocity, v, using the relationship:
v = rω
Substituting for r:
v = (h + L/2)ω
Solving for ω:
ω = v / (h + L/2)
Substituting this into the conservation of energy equation:
mgh = 1/2mv^2 + 1/2I((v/(h + L/2))/(r))^2
Simplifying further, we get:
gh = 1/2v^2 + 1/2I(v/(h + L/2))^2
Rearranging the equation to solve for the moment of inertia, I:
I = (2mgh) / ((v/(h + L/2))^2 - 1)
Now, we can calculate the moment of inertia of the drum.
b) To determine the angular acceleration of the drum, we can use the equation:
α = a / r
where a is the acceleration of the stuntman and r is the radius of the drum.
Using the required value of the stuntman's acceleration calculated in part a, we can now calculate the angular acceleration.
c) To determine the number of revolutions the drum makes during the fall, we need to calculate the angular displacement, θ, of the drum.
The angular displacement, θ, can be calculated using the equation:
θ = ω × t
where t is the time of the fall.
The number of revolutions, n, can be calculated as:
n = θ / (2π)
Substituting the value calculated for θ, we can determine the number of revolutions the drum makes during the fall.