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A 3.5-m-diameter merry-go-round with rotational inertia 120 kg · m² is spinning freely at 0.60 rev/s . Four 25-kg children sit suddenly on the edge of the merry-go-round. Express your answer using two significant figures. V = 0.17 rev/s Submit Previous Answers All attempts used; correct answer displayed Part B Determine the total energy lost to friction between the children and the merry-go-round. Express your answer using two significant figures.

A 3.5-m-diameter merry-go-round with rotational inertia 120 kg · m² is spinning freely at 0.60 rev/s . Four 25-kg children sit suddenly on the edge of the merry-go-round. Express your answer using two significant figures. V = 0.17 rev/s Submit Previous Answers All attempts used; correct answer displayed Part B Determine the total energy lost to friction between the children and the merry-go-round. Express your answer using two significant figures.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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### Physics Problem - Rotational Motion and Energy #### Problem Statement A 3.5-m-diameter merry-go-round with rotational inertia \(120 \, \text{kg} \cdot \text{m}^2\) is spinning freely at \(0.60 \, \text{rev/s}\). Four 25-kg children sit suddenly on the edge of the merry-go-round. #### Part A After the children sit on the edge of the merry-go-round, the system's new rotational speed needs to be found. **Given:** - Diameter of the merry-go-round: \(3.5 \, \text{m}\) - Rotational inertia of the merry-go-round: \(120 \, \text{kg} \cdot \text{m}^2\) - Initial angular speed: \(0.60 \, \text{rev/s}\) - Number of children: 4 - Mass of each child: \(25 \, \text{kg}\) **Calculation Submissions:** - The correct answer for the new angular speed is \( \nu = 0.17 \, \text{rev/s} \). #### Part B Determine the total energy lost to friction between the children and the merry-go-round. **Express your answer using two significant figures:** - \( \Delta K = 331.923 \, \text{J} \) This indicates that a change in kinetic energy \(\Delta K\) is calculated, highlighting the energy lost due to friction when the children sit on the merry-go-round. ### Explanation - The problem illustrates rotational inertia and angular momentum conservation principles. - The rotational inertia of the merry-go-round changes when the children sit on it, affecting the rotational speed. - The calculation of energy loss involves understanding the changes in kinetic energy due to the added mass at a radius from the axis of rotation. This example problem and its solved parts are crucial for understanding rotational dynamics and energy principles in physics.
Transcribed Image Text:### Physics Problem - Rotational Motion and Energy #### Problem Statement A 3.5-m-diameter merry-go-round with rotational inertia \(120 \, \text{kg} \cdot \text{m}^2\) is spinning freely at \(0.60 \, \text{rev/s}\). Four 25-kg children sit suddenly on the edge of the merry-go-round. #### Part A After the children sit on the edge of the merry-go-round, the system's new rotational speed needs to be found. **Given:** - Diameter of the merry-go-round: \(3.5 \, \text{m}\) - Rotational inertia of the merry-go-round: \(120 \, \text{kg} \cdot \text{m}^2\) - Initial angular speed: \(0.60 \, \text{rev/s}\) - Number of children: 4 - Mass of each child: \(25 \, \text{kg}\) **Calculation Submissions:** - The correct answer for the new angular speed is \( \nu = 0.17 \, \text{rev/s} \). #### Part B Determine the total energy lost to friction between the children and the merry-go-round. **Express your answer using two significant figures:** - \( \Delta K = 331.923 \, \text{J} \) This indicates that a change in kinetic energy \(\Delta K\) is calculated, highlighting the energy lost due to friction when the children sit on the merry-go-round. ### Explanation - The problem illustrates rotational inertia and angular momentum conservation principles. - The rotational inertia of the merry-go-round changes when the children sit on it, affecting the rotational speed. - The calculation of energy loss involves understanding the changes in kinetic energy due to the added mass at a radius from the axis of rotation. This example problem and its solved parts are crucial for understanding rotational dynamics and energy principles in physics.
### Center of Mass Calculation for a Nonuniform Horizontal Bar #### Problem Statement A nonuniform, horizontal bar of mass \( m \) is supported by two massless wires against gravity. The left wire makes an angle \( \phi_1 \) with the horizontal, and the right wire makes an angle \( \phi_2 \). The bar has length \( L \). The objective is to find the position of the center of mass of the bar, \( x \), measured from the bar's left end. Express the center of mass in terms of \( L \), \( \phi_1 \), and \( \phi_2 \). Note that any trigonometric function entered in your answer must be followed by an argument in parentheses. #### Diagram Description The diagram provided (Figure 1) shows: - A horizontal bar indicated by a shaded orange rectangle with a length \( L \). - Two wires, one on the left making an angle \( \phi_1 \) with the horizontal and one on the right making an angle \( \phi_2 \). - The center of mass is at a distance \( x \) from the bar's left end. - Arrows indicating the direction of tension forces in the wires and the distance \( L \) from left to right. #### Provided Mathematical Expression To express the center of mass \( x \) in terms of \( L \), \( \phi_1 \), and \( \phi_2 \): \[ x = \frac{L}{(\tan(\phi_1) \cdot \cot(\phi_2)) + 1} \] #### Instructions for Solution - Ensure that the variables \( \phi_1 \) and \( \phi_2 \) are correctly replaced with their respective value inputs. - Make sure that the parentheses are used appropriately around the trigonometric arguments. This calculation is crucial for understanding the physics of nonuniform bodies and the application of torque balance in mechanical systems.
Transcribed Image Text:### Center of Mass Calculation for a Nonuniform Horizontal Bar #### Problem Statement A nonuniform, horizontal bar of mass \( m \) is supported by two massless wires against gravity. The left wire makes an angle \( \phi_1 \) with the horizontal, and the right wire makes an angle \( \phi_2 \). The bar has length \( L \). The objective is to find the position of the center of mass of the bar, \( x \), measured from the bar's left end. Express the center of mass in terms of \( L \), \( \phi_1 \), and \( \phi_2 \). Note that any trigonometric function entered in your answer must be followed by an argument in parentheses. #### Diagram Description The diagram provided (Figure 1) shows: - A horizontal bar indicated by a shaded orange rectangle with a length \( L \). - Two wires, one on the left making an angle \( \phi_1 \) with the horizontal and one on the right making an angle \( \phi_2 \). - The center of mass is at a distance \( x \) from the bar's left end. - Arrows indicating the direction of tension forces in the wires and the distance \( L \) from left to right. #### Provided Mathematical Expression To express the center of mass \( x \) in terms of \( L \), \( \phi_1 \), and \( \phi_2 \): \[ x = \frac{L}{(\tan(\phi_1) \cdot \cot(\phi_2)) + 1} \] #### Instructions for Solution - Ensure that the variables \( \phi_1 \) and \( \phi_2 \) are correctly replaced with their respective value inputs. - Make sure that the parentheses are used appropriately around the trigonometric arguments. This calculation is crucial for understanding the physics of nonuniform bodies and the application of torque balance in mechanical systems.
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