You are the technical consultant for an action-adventure film in which a stunt calls for the hero to drop off a 18-m-tall building and land on the ground safely at a final vertical speed of 5 m/s. At the edge of the building's roof, there is a 100-kg drum that is wound with a sufficiently long rope (of negligible mass), has a radius of 0.4 m, and is free to rotate about its cylindrical axis with a moment of inertia I0.The script calls for the 69-kg stuntman to tie the rope around his waist and walk off the roof.
(a) Determine an expression for the stuntman's linear acceleration in terms of his mass m, the drum's radius r, and moment of inertia I0.(Use any variable or symbol stated above along with the following as necessary: g.)
a =
(b) Determine the required value of the stuntman's acceleration if he is to land safely at a speed of 5 m/s.
Use this value to calculate the moment of inertia of the drum about its axis.
(c) What is the angular acceleration of the drum?
(a) To determine the expression for the stuntman's linear acceleration, we can start by analyzing the forces acting on the system.
First, there is the force of gravity acting on both the stuntman and the drum. The weight of the stuntman is given by mg, where m is the mass of the stuntman and g is the acceleration due to gravity (9.8 m/s²). The weight of the drum is given by its mass (100 kg) times g.
Next, there is the tension in the rope. Since the rope is wound around the drum and tied to the stuntman, it creates tension in the rope as the stuntman moves downward. This tension force creates the linear acceleration of the system.
Considering these forces, we can write the equation of motion for the system:
ma = mg - (100 kg)g - T
Where:
a is the linear acceleration of the stuntman (what we are trying to find),
m is the mass of the stuntman (69 kg),
g is the acceleration due to gravity (9.8 m/s²),
(100 kg)g is the weight of the drum, and
T is the tension in the rope.
Rearranging the equation, we can solve for a:
a = (mg - (100 kg)g - T) / m
Substituting the given values, the expression for the linear acceleration becomes:
a = (69 kg * 9.8 m/s² - 100 kg * 9.8 m/s² - T) / 69 kg
(b) To determine the required value of the stuntman's acceleration for a safe landing speed of 5 m/s, we know that the final vertical speed is 5 m/s. Therefore, the final velocity, vf, is 5 m/s, and the initial velocity, vi, is 0 m/s (as the stuntman starts from rest).
We can use the equation of motion relating final and initial velocities, acceleration, and displacement:
vf² = vi² + 2aΔy
Where:
vf is the final velocity (5 m/s),
vi is the initial velocity (0 m/s),
a is the linear acceleration of the stuntman (what we are trying to find), and
Δy is the vertical displacement (the height of the building, which is 18 m).
Rearranging the equation, we can solve for a:
a = (vf² - vi²) / (2Δy)
Since vi is 0 m/s, the equation simplifies to:
a = vf² / (2Δy)
Substituting the given values and solving for a:
a = (5 m/s)² / (2 * 18 m)
(c) To find the moment of inertia of the drum, we need to determine the angular acceleration of the drum.
The relationship between linear acceleration (a) and angular acceleration (α) for objects rotating about an axis with a moment of inertia (I) is given by:
a = rα
Where:
r is the radius of the drum (0.4 m).
Since we have already calculated the linear acceleration (a) in part (b), we can substitute it into the equation to find the angular acceleration:
rα = a
Solving for α, we have:
α = a / r
Substituting the calculated value for a, we can determine the angular acceleration.