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Concept Questions As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. (a) What applies the force to the person to create the torque causing this acceleration? What is the direction of this force? (b) According to Newton's actionreaction law, what can you say about the direction of the force applied to the carousel by the person and about the nature (clockwise or counterclockwise) of the torque that it creates? (c) Does the torque identified in part (b) increase or decrease the angular speed of the carousel? Problem The carousel has a radius of \(1.50 \mathrm{~m}\), an initial angular speed of \(3.14 \mathrm{rad} / \mathrm{s}\), and a moment of inertia of \(125 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The mass of the person is \(40.0 \mathrm{~kg}\). Find the final angular speed of the carousel after the person climbs aboard. Verify that your answer is consistent with your answers to the Concept Questions.

Step by step solution

01

Identify the Initial Conditions

The carousel has an initial angular speed, \( \omega_i = 3.14 \, \text{rad/s}\), a radius, \( r = 1.50 \, \text{m}\), and a moment of inertia, \( I_i = 125 \, \text{kg} \cdot \text{m}^2 \). The person has a mass, \( m = 40.0 \, \text{kg} \). We assume the person begins with zero angular speed relative to the carousel.

02

Calculate Moment of Inertia After the Person Boards

Once the person climbs aboard at the edge, they contribute to the system's moment of inertia. The moment of inertia for a point mass at a distance \( r \) is \( I = mr^2 \). So, the person's contribution is \( I_p = 40 \, \text{kg} \times (1.5 \, \text{m})^2 = 90 \, \text{kg} \cdot \text{m}^2 \). The total moment of inertia is \( I_f = I_i + I_p = 125 + 90 = 215 \, \text{kg} \cdot \text{m}^2 \).

03

Use Conservation of Angular Momentum

Angular momentum is conserved because the carousel is frictionless. So, the initial angular momentum, \( L_i = I_i \cdot \omega_i \), is equal to the final angular momentum, \( L_f = I_f \cdot \omega_f \). We set up the equation: \[ 125 \, \text{kg} \cdot \text{m}^2 \times 3.14 \, \text{rad/s} = 215 \, \text{kg} \cdot \text{m}^2 \times \omega_f \].

04

Solve for the Final Angular Speed

Rearrange the conservation of angular momentum equation to solve for \( \omega_f \): \[ \omega_f = \frac{125 \times 3.14}{215} = 1.82 \, \text{rad/s} \]. This is the final angular speed of the carousel after the person boards.

05

Verify with Conceptual Questions

(a) The force that applies torque is the force exerted by the carousel on the person, directed tangentially along the direction of rotation. (b) By Newton's third law, the person exerts an equal and opposite force on the carousel, causing a torque in the clockwise direction. (c) This torque decreases the angular speed, which aligns with the calculation showing a reduced \( \omega_f \) compared to \( \omega_i \).

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Key Concepts

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

Angular Acceleration

When the person climbs onto the carousel, they initially have zero angular speed about the carousel’s axis but end up rotating with it. This change in the person's rotation speed is known as **angular acceleration**. Angular acceleration tells us how quickly the angular velocity of an object changes over time.

• It occurs when there is a change in rotational speed.
• In this case, the person speeds up to match the carousel’s rotation.
• A torque is needed to cause this acceleration.

Torque is produced when the carousel exerts a force on the person, pulling them into its circular path. The direction of the force applied by the carousel on the person is tangential and counterclockwise, the same as the carousel's rotation.
This interaction between force and rotational motion defines the concept of angular acceleration.

Moment of Inertia

The **moment of inertia** is a physical quantity that determines how easily an object can be rotated about an axis. It's akin to mass in linear motion, providing a measure of an object's resistance to changes in rotational speed.

• Depends on the object's mass and distribution of mass relative to the axis.
• Increases when more mass is located further from the axis.
• Contributes significantly to conserving angular momentum.

For the problem, when the person joins the carousel, their mass adds to the system’s total moment of inertia because they position themselves at the outer edge:
\[ I_{ ext{person}} = m imes r^2 = 40 imes (1.5)^2 = 90 ext{ kg} ext{ m}^2 \]Therefore, the new moment of inertia for the system comprises the carousel’s initial moment plus the person's contribution. This increased moment results in a reduced angular speed, as shown by the calculation involving the conservation of angular momentum (step 3 in the solution). The system’s rotational motion changes to maintain the overall angular momentum.

Newton’s Third Law

**Newton's third law** is often stated as "For every action, there is an equal and opposite reaction." This simple yet fundamental principle plays a crucial role in understanding interactions involving forces and torques in rotating systems.

• Torque is generated when forces act over a distance from a pivot point.
• When the carousel exerts a force on the person to accelerate them, the person exerts an equal and opposite force back on the carousel.
• This interaction results in a clockwise torque opposing the carousel’s initial rotation.

This clockwise torque exerted by the person slightly reduces the carousel's angular speed, accounting for the decrease observed when the person boards. Hence, the direction and nature of the forces and torques explain why the final angular speed is lower than the initial value. This explanation aligns with the concept questions and verifies the conservation of angular momentum in this scenario.