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A space station shaped like a giant wheel has a radius of \(100 \mathrm{~m}\) and a moment of inertia of \(5.00 \times 10^{8} \mathrm{~kg} \cdot \mathrm{m}^{2}\) A crew of 150 lives on the rim, and the station is rotating so that the crew experiences an apparent acceleration of \(1 g\) (Fig. P8.64).When 100 people move to the center of the station for a union meeting, the angular speed changes. What apparent acceleration is experienced by the managers remaining at the rim? Assume the average mass of a crew member is \(65.0 \mathrm{~kg}\).

Short Answer

Expert verified
To find the final apparent acceleration, it needs to be first determined how the space station's moment of inertia changes as the people move, and then using conservation of angular momentum, find the new angular velocity. From it, the new apparent acceleration can be calculated.

Step by step solution

01

Compute Initial Angular Momentum

The initial angular momentum \(L_{i}\) can be calculated using the relation \(L_{i} = I_{i} \cdot \omega_{i}\), where \(I_{i}\) is the initial moment of inertia and \(\omega_{i}\) is the initial angular velocity. The apparent acceleration \(a_{i}\) experienced by the crew due to the initial rotation can be related with the radius of the space station \(R\) and the initial velocity \(\omega_{i}\) via the relation \(a_{i} = R \cdot \omega_{i}^{2}\). Substituting \(\omega_{i}\) from this equation into the formula for \(L_{i}\) gives \(L_{i} = I_{i} \cdot (a_{i} / R)\)
02

Compute Final Angular Momentum

When 100 people move to the center, the final moment of inertia \(I_{f}\) changes, but the angular momentum is conserved. The final moment of inertia \(I_{f}\) equals the initial moment of inertia \(I_{i}\) minus the moment of inertia of 100 people moving to the centre, which can be computed as \(100 \cdot m \cdot R^{2}\), where \(m\) is the mass of one person. Hence, \(I_{f} = I_{i} - 100 \cdot m \cdot R^{2}\). The final angular momentum \(L_{f}\) hence equals \(L_{i} = I_{f} \cdot \omega_{f}\), where \(\omega_{f}\) is the final angular velocity to solve for.
03

Compute Final Acceleration

As the conservation of angular momentum indicates that the initial and final angular momenta are equal, the previous two steps give us \(I_{i} \cdot (a_{i}/R) = I_{f} \cdot \omega_{f}\). By rearranging, the final angular velocity \(\omega_{f}\) results in \(\omega_{f} = (I_{i} \cdot (a_{i}/R) ) / I_{f}\). Subsequently, the final acceleration \(a_{f}\) can be given by \(a_{f} = R \cdot \omega_{f}^{2}\)

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Key Concepts

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

Moment of Inertia
Moment of inertia, often represented as 'I', is a measure of an object's resistance to changes in its rotational motion. It's similar to mass in linear dynamics; the greater the moment of inertia, the harder it is to start or stop the object spinning. In our space station example, the moment of inertia is initially quite high due to the distribution of mass - crew members - at a greater radius.

When considering the change in moment of inertia after crew members move to the center, it is crucial to understand that the moment of inertia is not only dependent on the total mass but also on the distribution of that mass relative to the axis of rotation. The formula for the moment of inertia of point masses is given by the sum of the product of mass and the square of its distance from the rotation axis, mathematically expressed as \( I = \text{Σ} m_r \times r^2 \), where \( m_r \) is each point mass and \( r \) its distance from the axis.

In the space station scenario, when people move towards the center, the distance \( r \) decreases, hence decreasing the moment of inertia. This concept is pivotal in understanding how the space station's rotational behavior will change as the crew moves.
Angular Velocity
Angular velocity, symbolized by \( \omega \), is a measure of how fast an object rotates. It's the rate at which the object's angle changes with respect to time, analogous to linear velocity in straight-line motion. In the context of the space station, angular velocity determines how quickly the station is spinning and, consequently, the artificial gravity experienced by the inhabitants.

The relationship between linear velocity \( v \) at the rim and the angular velocity is given by \( v = \text{R} \times \omega \), where \( R \) is the radius of the circular path. The initial angular velocity can be directly linked to the apparent acceleration experienced by the inhabitants through the equation \( a = \text{R} \times \omega^2 \), which is derived from centripetal acceleration. Understanding angular velocity helps us appreciate how the station's rotation rate changes as people move, which directly affects the artificial gravity or 'apparent acceleration' they feel.
Conservation of Angular Momentum
The conservation of angular momentum principle states that if no external torque acts on a system, the total angular momentum of the system remains constant. This is a rotational analogue to the conservation of linear momentum and is crucial in problems involving rotational dynamics.

In our space station, once it is set spinning, external torques are negligible in the vacuum of space, and thus its angular momentum will be conserved. This concept aids in solving for the new angular velocity after the crew members' repositioning. Starting with the initial angular momentum \( L_i = I_i \times \omega_i \) and understanding that \( L_f = L_i \), we can figure out how the system's angular velocity must adjust to conserve angular momentum when the moment of inertia changes because of the crew's movement. This leads to the insight that the angular velocity will increase once the moment of inertia decreases, as the two quantities are inversely proportional in the conservation of angular momentum.
Apparent Acceleration
Apparent acceleration in the context of rotational motion refers to the 'felt' acceleration, which in the case of our space station is the artificial gravity experienced by the inhabitants. This apparent acceleration is a result of the centripetal force required to keep a mass moving in a circular path, and it's directed towards the center of the rotation axis.

Initially, the crew experiences an apparent acceleration of 1g, creating a sensation equivalent to Earth's gravity. When the distribution of mass in the station changes, the apparent acceleration will also change, as it depends on the angular velocity and the radius from the center of rotation. The final apparent acceleration can be derived from the new angular velocity and is given by \( a_f = \text{R} \times \omega_f^2 \). Understanding this helps in determining how comfortable the living conditions remain for the crew after the repositioning - a fundamental aspect of space station design and crew safety.

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Most popular questions from this chapter

QIC S A solid uniform sphere of mass \(m\) and radius \(R\) rolls without slipping down an incline of height \(h\). (a) What forms of mechanical energy are associated with the sphere at any point along the incline when its angular speed is \(\omega\) ? Answer in words and symbolically in terms of the quantities \(m, g, y, I, \omega\), and \(v\). (b) What force acting on the sphere causes it to roll rather than slip down the incline? (c) Determine the ratio of the sphere's rotational kinetic energy to its total kinetic energy at any instant.

A \(10.0-\mathrm{kg}\) cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of \(10.0 \mathrm{~m} / \mathrm{s}\), determine (a) the translational kinetic energy of its center of gravity, (b) the rotational kinetic energy about its center of gravity, and (c) its total kinetic energy.

S A uniform solid cylinder of mass \(M\) and radius \(R\) rotates on a frictionless horizontal axle (Fig. P8.81). Two objects with equal masses \(m\) hang from light cords wrapped around the cylinder. If the system is released from rest, find (a) the tension in eachcord and (b) the acceleration of each object after the objects have descended a distance \(h\).

QIC A \(0.00500-\mathrm{kg}\) bullet traveling horizontally with a speed of \(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) enters an \(18.0-\mathrm{kg}\) door, embedding itself \(10.0 \mathrm{~cm}\) from the side opposite the hinges as in Figure P8.56. The \(1.00-\mathrm{m}\)-wide door is free to swing on its hinges. (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Explain. (b) Is mechanical energy con-served in this collision? Answer without doing a calculation. (c) At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.) (d) Calculate the energy of the door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.

A solid, uniform disk of radius \(0.250 \mathrm{~m}\) and mass \(55.0 \mathrm{~kg}\) rolls down a ramp of length \(4.50 \mathrm{~m}\) that makes an angle of \(15.0^{\circ}\) with the horizontal. The disk starts from rest from the top of the ramp. Find (a) the speed of the disk's center of mass when it reaches the bottom of the ramp and (b) the angular speed of the disk at the bottom of the ramp.

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