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A Texas cockroach of mass \(0.17 \mathrm{~kg}\) runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius \(15 \mathrm{~cm}\), rotational inertia \(5.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and frictionless bearings. The cockroach's speed (relative to the ground) is \(2.0 \mathrm{~m} / \mathrm{s}\), and the lazy Susan turns clockwise with angular speed \(\omega_{0}=\) \(2.8 \mathrm{rad} / \mathrm{s}\). The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?

Short Answer

Expert verified
(a) The new angular speed of the lazy Susan is 7.83 rad/s. (b) Mechanical energy is not conserved; energy is lost as the cockroach stops.

Step by step solution

01

Understand and Set Up the Problem

Initially, we need to determine the conservation of angular momentum before and after the cockroach stops. Before stopping, the system includes both the cockroach and the lazy Susan rotating in opposite directions. The key point here is that the angular momentum of the system is conserved because no external torques are acting on the system.
02

Calculate Cockroach's Initial Angular Momentum

The cockroach moves at a linear speed of 2.0 m/s along the rim of the lazy Susan, which has a radius of 0.15 m. The angular momentum of the cockroach can be calculated as: \[ L_{c} = m imes r imes v \] where \( m = 0.17 \) kg, \( r = 0.15 \) m, and \( v = 2.0 \) m/s. Substitute these values to find \( L_{c} \).
03

Calculate Lazy Susan's Initial Angular Momentum

The angular momentum of the lazy Susan is given by: \[ L_{s} = I imes ext{ω}_{0} \] where \( I = 5.0 \times 10^{-3} \) kg⋅m² and \( ext{ω}_{0} = 2.8 \) rad/s. Calculate \( L_{s} \) using these values.
04

Total Initial Angular Momentum

The total initial angular momentum of the system is the sum of the magnitudes of the angular momentums of the cockroach and lazy Susan, taking into consideration their opposite directions: \[ L_{ ext{total, initial}} = L_{c} - L_{s} \]
05

Conservation of Angular Momentum

After the cockroach stops, its angular momentum becomes zero. Thus, the total angular momentum of the system is only due to the lazy Susan, which must be equal to the initial total angular momentum due to the conservation law: \[ I imes ext{ω}_{ ext{final}} = L_{ ext{total, initial}} \] Solve for \( ext{ω}_{ ext{final}} \) to find the new angular speed of the lazy Susan.
06

Mechanical Energy Conservation Check

Before stopping, the cockroach has kinetic energy, both translational and arising from its movement around the axis, along with the kinetic energy of the lazy Susan. After stopping, only the lazy Susan has kinetic energy. Compare the system's total initial and final mechanical energies to determine if energy is conserved. If they are not equal, mechanical energy is not conserved due to work done or dissipated (e.g., through heat or sound).

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Key Concepts

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

Rotational Inertia
Rotational inertia, often called the moment of inertia, is a measure of how easy or difficult it is to change the rotational speed of an object. It's like how mass affects linear motion, but in a circular context. When we talk about an object's rotational inertia, we usually refer to how the mass is distributed relative to the axis of rotation. The greater the spread of mass from the axis, the larger the rotational inertia.

The lazy Susan in our problem has a rotational inertia of \(5.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\). This means it doesn't require much torque to start or stop its rotation compared to objects with larger rotational inertia. When an object has large rotational inertia, it helps stabilize its rotation, needing more torque to alter its angular speed.
  • Influences how the lazy Susan reacts when the cockroach runs on it.
  • Essential for calculating energy changes.
Mechanical Energy Conservation
Mechanical energy conservation refers to the principle that in an isolated system, where only conservative forces act, the total mechanical energy remains constant. In the exercise, mechanical energy before and after the cockroach stops is compared. Before stopping, both the cockroach and lazy Susan possess kinetic energy.

The cockroach contributes two types of kinetic energy:
  • Translational kinetic energy: Due to the cockroach's movement along the lazy Susan.
  • Rotational kinetic energy: As it runs along the rim, rotating around the axle.
After the cockroach stops, only the lazy Susan continues to have kinetic energy. By comparing the system's initial and final kinetic energies, we determine if mechanical energy is conserved. If initial and final energies don't match, energy was likely converted to other forms like heat or sound, indicating that some energy was not conserved in mechanical form.
Linear and Angular Momentum
Momentum, both linear and angular, are foundational concepts in physics describing the motion's persistence under no external changes. Linear momentum relates to straight-line motion, calculated as the product of mass and velocity. Angular momentum pertains to rotational motion and is similar to linear momentum but in circular motion.

In our exercise:
  • Linear Momentum: The cockroach running at 2.0 m/s contributes to its momentum, a factor moving along the circular edge.
  • Angular Momentum: This type of momentum is critical here, defined as \(L = m \times r \times v\) for the cockroach and \(L_{s} = I \times \omega \) for the lazy Susan.
The conservation of angular momentum suggests that without external torques, the initial momentum of the entire system (the laz] Susan and cockroach) is equal to its final momentum. When the cockroach stops, its contribution to the angular momentum disappears; thus, the lazy Susan's new momentum equals the system's initial total angular momentum. This principle was key in solving the problem to find the lazy Susan's final angular speed.

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Most popular questions from this chapter

Two particles, each of mass \(2.90 \times 10^{-4} \mathrm{~kg}\) and speed \(5.46 \mathrm{~m} / \mathrm{s}\), travel in opposite directions along parallel lines separated by \(4.20 \mathrm{~cm} .\) (a) What is the magnitude \(L\) of the angular momentum of the two-particle system around a point midway between the two lines? (b) Is the value different for a different location of the point? If the direction of either particle is reversed, what are the answers for (c) part (a) and (d) part (b)?

An automobile traveling at \(80.0 \mathrm{~km} / \mathrm{h}\) has tires of \(75.0 \mathrm{~cm}\) diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in \(30.0\) complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?

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