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?
(d) How many revolutions does the drum make during the fall?
(a) To determine the expression for the stuntman's linear acceleration, we need to apply Newton's second law. The force applied to the stuntman is the tension in the rope, and the opposing force is the weight of the stuntman given by mg, where m is his mass and g is the acceleration due to gravity.
The net force acting on the stuntman is given by:
F_net = T - mg
Using Newton's second law, F_net = ma, we can solve for the acceleration:
T - mg = ma
a = (T - mg) / m
Now, let's consider the torque on the drum. The tension in the rope causes a torque on the drum, given by T * r, where r is the radius of the drum. The torque causes an angular acceleration on the drum, given by the equation:
τ = I0 * α
Since the drum is free to rotate about its cylindrical axis, the torque on the drum is equal to the moment of inertia times the angular acceleration. Therefore:
T * r = I0 * α
We can rearrange the equation for α as:
α = (T * r) / I0
Now, let's substitute the expression for T from the linear acceleration equation into the angular acceleration equation:
α = ((mg + ma) * r) / I0
(b) To determine the required value of the stuntman's acceleration for him to land safely at a speed of 5 m/s, we set the final vertical velocity to 5 m/s. The initial vertical velocity is assumed to be zero.
The equation for final velocity in terms of initial velocity, acceleration, and displacement is given by:
v^2 = u^2 + 2as
Here, v is the final velocity (5 m/s), u is the initial velocity (0 m/s), and s is the displacement (18 m). Solving for acceleration, we get:
a = (v^2 - u^2) / (2s)
Substituting the given values, we find:
a = (5^2 - 0^2) / (2 * 18) = 25 / 36 m/s^2
Now, let's use this value of acceleration to calculate the moment of inertia of the drum.
Using the equation for α from part (a), we have:
α = ((mg + ma) * r) / I0
Substituting the given values, we find:
25/36 = ((69 * 9.8 + 69 * 25 / 36) * 0.4) / I0
Simplifying the equation and solving for I0, we get:
I0 = ((69 * 9.8 + 69 * 25 / 36) * 0.4) / (25/36)
(c) To find the angular acceleration of the drum, we can substitute the known values into the equation for α:
α = ((mg + ma) * r) / I0
Using the given values, we find:
α = ((69 * 9.8 + 69 * 25 / 36) * 0.4) / I0
(d) To determine the number of revolutions the drum makes during the fall, we can use the kinematic equation for angular displacement:
θ = ω_0 * t + 0.5 * α * t^2
Since the initial angular velocity is zero (ω_0 = 0) and final angular velocity is given by ω = α * t, we can simplify the equation to:
θ = 0.5 * α * t^2
The time it takes for the stuntman to fall from the building can be found using the equation:
y = (1/2) * g * t^2
Rearranging for t:
t = sqrt(2 * y / g)
Substituting the given values for y (18 m) and g (9.8 m/s^2), we can find t. Then we can substitute the values of α and t into the equation for θ to calculate the angular displacement in radians. Finally, we can convert the angular displacement to revolutions by dividing by 2π (since there are 2π radians in one revolution).