You are the technical consultant for an action-adventure film in which a stunt calls for the hero to drop off a 20.0 -m-tall building and land on the ground safely at a final vertical speed of \(4.00 \mathrm{~m} / \mathrm{s}\). At the edge of the building's roof, there is a \(100 .-\mathrm{kg}\) drum that is wound with a sufficiently long rope (of negligible mass), has a radius of \(0.500 \mathrm{~m}\), and is free to rotate about its cylindrical axis with a moment of inertia \(I_{0}\). The script calls for the 50.0 -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 \(I_{0}\). b) Determine the required value of the stuntman's acceleration if he is to land safely at a speed of \(4.00 \mathrm{~m} / \mathrm{s},\) and 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?

Short Answer

Expert verified
Answer: The expression for the stuntman's linear acceleration is \(a = \frac{I_0}{m r^2} g\). The required moment of inertia of the drum is \(I_0 = 40.8\, \mathrm{kg}\,\mathrm{m}^2\). The angular acceleration of the drum is \(8.00\, \mathrm{s}^{-2}\), and the drum makes approximately 6.37 revolutions during the fall.

Step by step solution

01

Expressing forces and torques

We will first set up Newton's 2nd law equation for the forces acting on the stuntman and the torque acting on the drum. Let T be the tension in the rope, m be the mass of the stuntman, a be the linear acceleration of the stuntman, r be the radius of the drum, and I be the moment of inertia of the drum. We have: - For the stuntman: T - mg = ma (1), - For the drum(detaltag): Tr = I(alpha) (2), where (alpha) is the angular acceleration of the drum. From these equations, we can express T and substitute it into equation (2) to get an expression that relates the linear acceleration of the stuntman, the radius of the drum, and the moment of inertia of the drum.
02

Determine the linear acceleration of the stuntman

From equation (1), we get: \(T = m(a+g)\) (3) Now substituting equation (3) into (2), we get: \(m(a+g)r = I(\alpha)\) (4) Since the linear acceleration of the stuntman is related to the angular acceleration of the drum via: \(a = r(\alpha)\) (5) Now substituting equation (5) into (4) and solving for a, we get: \( a = \frac{I_0}{m r^2} g \) (6)
03

Determine the required moment of inertia

Using h as the height of the fall, if the stuntman is to land safely, we know that the final vertical speed is \(4.00 \mathrm{~m} / \mathrm{s}\). We can use height h, initial speed \(v_0\) as 0, final speed \(v_f\) as \(4.00 \mathrm{~m} / \mathrm{s}\) and acceleration a to determine the moment of inertia. We have: \(h = \frac{1}{2}a(t_f)^2 = \frac{v_f^2 - v_0^2}{2a}\), Solving (6) for a, we substitute \(I_0\) from the values of h, \(v_0\), \(v_f\), m, upon rearranging we get: \(I_0 = \frac{2a m r^2}{g}\) (7) Substituting the known values into equation (7): \(I_{0} = \frac{2 (4.00\mathrm{~m/s}) (50.0\mathrm{~kg}) (0.5\mathrm{~m})^2}{9.81 \mathrm{~m/s^2}} = 40.8 \mathrm{~kg}\,\mathrm{m}^2\)
04

Determine the angular acceleration of the drum

Using equation (5), we can determine the angular acceleration of the drum: \((\alpha)= \frac{a}{r} = \frac{4.00 \mathrm{~m/s}}{0.5\mathrm{~m}} = 8.00\, \mathrm{s}^{-2}\)
05

Determine the number of revolutions during the fall

To determine the number of revolutions the drum makes during the fall, we can use the angular displacement equation, knowing the angular acceleration and the total time \((t_f)\) of the fall: \(\theta = \omega_0 (t_f) + \frac{1}{2}(\alpha)(t_f)^2\), where \(\omega_0\) is the initial angular velocity which is zero. Solving for \(t_f\) from \(h = \frac{1}{2}a(t_f)^2\): \(t_f = \sqrt{\frac{2h}{a}}\) In this case, the height is \(20.0\mathrm{~m}\), so: \(t_f = \sqrt{\frac{2 (20.0\,\mathrm{m})}{4.00 \,\mathrm{m/s^2}}} = 3.16\,\mathrm{s}\) Now, we can determine the angular displacement: \(\theta = \frac{1}{2}(\alpha)(t_f)^2 = \frac{1}{2}(8.00\,\mathrm{s^{-2}})(3.16\,\mathrm{s})^2 = 40.0\,\mathrm{rad}\) Since the number of revolutions in radian is given by \(\theta = n \times 2\pi\), we can determine the number of revolutions, n: \(n = \frac{\theta}{2\pi} = \frac{40.0\,\mathrm{rad}}{2\pi\,\mathrm{rad}} = 6.37\) The drum makes approximately 6.37 revolutions during the fall.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Acceleration
Angular acceleration is a fundamental concept in rotational kinematics, describing how the rate of rotation of an object changes over time. A real-world example of this might be a figure skater spinning on the ice – as they pull their arms in, they spin faster due to an increase in angular acceleration. In our exercise, we calculate the angular acceleration of a drum as a stuntman falls, directly linking it to the linear acceleration of the fall via the relationship:

\(\alpha = \frac{a}{r}\)

In the given scenario, the angular acceleration \(\alpha\) of the drum is determined after calculating the stuntman's linear acceleration \(a\). It is essential to comprehend that angular acceleration is not just a different 'type' of acceleration but rather the rotational analogue to linear acceleration. Angular acceleration is typically measured in radians per second squared (\(\text{rad/s}^2\)) and in our case, the calculation resulted in an angular acceleration of \(8.00\, \text{s}^{-2}\). It's particularly interesting to note that when we talk about circular motion, linear acceleration and angular acceleration are intimately connected; one can often be converted into the other as demonstrated in the solution.
Moment of Inertia
The moment of inertia, often represented by \(I\), is a measure of an object's resistance to changes in its rotational motion. In simpler terms, it's akin to 'rotational mass.' Imagine trying to spin a wheel by pushing on its rim; a heavier wheel requires more effort to spin, which signifies a greater moment of inertia.

In this stuntman scenario, \(I_0\) is the moment of inertia of the drum, which factors into how easily the drum will begin rotating when the stuntman commences his descent. Applying Newton's second law for rotation, \(T r = I(\alpha)\), where \(T\) is the tension in the rope, we establish a relationship between the moment of inertia and the linear acceleration of the stuntman. With the derived formula, we calculate the drum's moment of inertia as \(40.8\, \mathrm{kg\cdot m^2}\) using the required acceleration for a safe landing.

Understanding the moment of inertia is critical in calculating the overall dynamics of rotational systems and, as shown in the exercise, is vital for engineering applications where rotational motion plays a key role.
Linear Acceleration
Linear acceleration is the rate at which an object's velocity changes concerning time in a straight line. It's one of the most commonly understood forms of acceleration and can be experienced in everyday life, such as when a car speeds up or slows down.

The exercise tackled the concept of linear acceleration as it pertains to a stuntman descending from a building. The calculation involved the forces on the stuntman and the drum. Understanding linear acceleration in this context is essential, as it determines whether the stuntman lands safely. Mathematically, we represented the stuntman's linear acceleration with the equation:

\(a = \frac{I_0}{m r^2} g\)

Substituting known quantities into the equation revealed the magnitude of linear acceleration necessary for a safe landing. The result also led to discovering the moment of inertia for the drum, showcasing a practical application of how linear acceleration is not only a concept of motion but also a bridge to connecting various aspects of physical dynamics.

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Most popular questions from this chapter

In a tire-throwing competition, a man holding a \(23.5-\mathrm{kg}\) car tire quickly swings the tire through three full turns and releases it, much like a discus thrower. The tire starts from rest and is then accelerated in a circular path. The orbital radius \(r\) for the tire's center of mass is \(1.10 \mathrm{~m},\) and the path is horizontal to the ground. The figure shows a top view of the tire's circular path, and the dot at the center marks the rotation axis. The man applies a constant torque of \(20.0 \mathrm{~N} \mathrm{~m}\) to accelerate the tire at a constant angular acceleration. Assume that all of the tire's mass is at a radius \(R=0.35 \mathrm{~m}\) from its center. a) What is the time, \(t_{\text {throw }}\) required for the tire to complete three full revolutions? b) What is the final linear speed of the tire's center of mass (after three full revolutions)? c) If, instead of assuming that all of the mass of the tire is at a distance \(0.35 \mathrm{~m}\) from its center, you treat the tire as a hollow disk of inner radius \(0.30 \mathrm{~m}\) and outer radius \(0.40 \mathrm{~m}\), how does this change your answers to parts (a) and (b)?

A sheet of plywood \(1.3 \mathrm{~cm}\) thick is used to make a cabinet door \(55 \mathrm{~cm}\) wide by \(79 \mathrm{~cm}\) tall, with hinges mounted on the vertical edge. A small 150 - \(\mathrm{g}\) handle is mounted \(45 \mathrm{~cm}\) from the lower hinge at the same height as that hinge. If the density of the plywood is \(550 \mathrm{~kg} / \mathrm{m}^{3},\) what is the moment of inertia of the door about the hinges? Neglect the contribution of hinge components to the moment of inertia.

A uniform solid sphere of mass \(m\) and radius \(r\) is placed on a ramp inclined at an angle \(\theta\) to the horizontal. The coefficient of static friction between sphere and ramp is \(\mu_{s} .\) Find the maximum value of \(\theta\) for which the sphere will roll without slipping, starting from rest, in terms of the other quantities.

A uniform solid sphere of mass \(M\) and radius \(R\) is rolling without sliding along a level plane with a speed \(v=3.00 \mathrm{~m} / \mathrm{s}\) when it encounters a ramp that is at an angle \(\theta=23.0^{\circ}\) above the horizontal. Find the maximum distance that the sphere travels up the ramp in each case: a) The ramp is frictionless, so the sphere continues to rotate with its initial angular speed until it reaches its maximum height. b) The ramp provides enough friction to prevent the sphere from sliding, so both the linear and rotational motion stop (instantaneously).

Determine the moment of inertia for three children weighing \(60.0 \mathrm{lb}, 45.0 \mathrm{lb}\) and \(80.0 \mathrm{lb}\) sitting at different points on the edge of a rotating merry-go-round, which has a radius of \(12.0 \mathrm{ft}\).

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