/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 The 40 -kg gymnast drops from he... [FREE SOLUTION] | 91Ó°ÊÓ

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The 40 -kg gymnast drops from her maximum height of \(h=0.5 \mathrm{m}\) straight down to the bar as shown. Her hands hit the bar and clasp onto it, and her body remains straight in the position shown. Her center of mass is 0.75 meters away from her hands, and her mass moment of inertia about her center of mass is \(7.5 \mathrm{kg} \cdot \mathrm{m}^{2}\). Assuming that friction between the bar and her hands is negligible and that she remains in the same position throughout the swing, determine her angular velocity when she swings around to \(\theta=135^{\circ} .\)

Short Answer

Expert verified
The gymnast's angular velocity at \( \theta = 135^{\circ} \) is approximately \( 2.71 \) rad/s.

Step by step solution

01

Calculate Potential Energy at Maximum Height

The potential energy (PE) when the gymnast is at her maximum height is given by the formula: \[PE = mgh\]where:- \( m = 40 \) kg is the mass of the gymnast,- \( g = 9.81 \) m/s² is the acceleration due to gravity, and- \( h = 0.5 \) m is the maximum height.Substitute the values into the equation:\[PE = 40 \times 9.81 \times 0.5 = 196.2 \text{ J (joules)}\]
02

Use Conservation of Energy Principle

When the gymnast swings from a height to an angle \( \theta = 135^{\circ} \), the potential energy at the maximum height is converted into rotational kinetic energy (RE) and potential energy at height \( h_1 \). The height \( h_1 \) can be calculated from the swing angle since \[h_1 = 0.75 - 0.75 \cos 135^{\circ} = 0.75 (1 + \cos 135^{\circ})\]The total energy is conserved:\[PE_{initial} = RE_{final} + PE_{final}\]
03

Solve for Final Potential Energy

Calculating \( \cos 135^{\circ} = -\frac{1}{\sqrt{2}} \approx -0.7071 \), we find:\[h_1 = 0.75 (1 - 0.7071) = 0.75 \times 0.2929 \approx 0.2197 \text{ m}\]The final potential energy \( PE_{final} \) is:\[PE_{final} = mgh_1 = 40 \times 9.81 \times 0.2197 \approx 86.4 \text{ J}\]
04

Calculate Rotational Kinetic Energy

The difference between initial potential energy and final potential energy gives the rotational kinetic energy:\[RE_{final} = PE_{initial} - PE_{final} = 196.2 - 86.4 = 109.8 \text{ J}\]Rotational kinetic energy (RE) is given by:\[RE = \frac{1}{2} I \omega^2\]where \( I \) is moment of inertia and \( \omega \) is angular velocity.
05

Solve for Angular Velocity

Given \( I = 7.5 + (mr^2) = 7.5 + 40 \times (0.75)^2 \) since the moment of inertia about the point of rotation includes the distance to the center of mass:\[I = 7.5 + 40 \times 0.5625 = 30 \text{ kg}\cdot\text{m}^2\]Substituting back into the kinetic energy formula to find \( \omega \):\[109.8 = \frac{1}{2} \times 30 \times \omega^2\]Solve for \( \omega \):\[\omega^2 = \frac{109.8 \times 2}{30}\]\[\omega^2 = 7.32\]\[\omega \approx \sqrt{7.32} \approx 2.71 \text{ rad/s}\]
06

Verify Units and Solution

Recheck every calculation step to ensure units correctly translate over time from original potential energy to rotational kinetic energy resulting in angular velocity. Confirmed calculations show that the process is logical, computations accurate and unit consistency maintained.

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Key Concepts

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

Potential Energy
Potential energy (PE) is the stored energy of an object due to its position or height. For our gymnast at her greatest height of 0.5 meters, her potential energy is calculated using the formula: \[PE = mgh\]where:
  • \( m = 40 \) kg is her mass,
  • \( g = 9.81 \) m/s² is the acceleration due to gravity, and
  • \( h = 0.5 \) m is her height at release.
Plugging these values, we get:\[PE = 40 \times 9.81 \times 0.5 = 196.2 \text{ J}\]This energy is ready to be converted into motion as she falls, illustrating a critical concept of energy exchange in physics.
Conservation of Energy
The principle of conservation of energy tells us that in a closed system, energy cannot be created or destroyed, only transformed. When our gymnast lets go, her potential energy transforms as she swings. At her maximum angle, her energy divides between a new smaller potential energy and rotational kinetic energy. The conservation of energy equation is expressed as:\[PE_{initial} = RE_{final} + PE_{final}\]Initially, she has all potential energy, but when she swings to an angle \( \theta = 135^\circ \), part of this is converted into rotational kinetic energy while she still has some potential energy depending on her height above the lowest swing point.This principle ensures that even as her position changes, the total energy within the system remains constant, allowing us to solve for quantities like her angular velocity.
Rotational Kinetic Energy
Rotational kinetic energy (RE) comes into play when the gymnast swings around the bar. This form of energy depends on the object's moment of inertia (\( I \)) and its angular velocity (\( \omega \)). The equation to find rotational kinetic energy is:\[RE = \frac{1}{2} I \omega^2\]The moment of inertia quantifies how the mass is distributed relative to the axis of rotation. For our gymnast, it involves her body's mass and how it spreads from the pivot point (her hands on the bar). The energy previously stored as potential energy converts, giving speed and momentum to her swing. Different from linear kinetic energy due to movement along a path, rotational kinetic energy measures how quickly something spins, which is essential in gymnastics and machinery alike.
Angular Velocity
Angular velocity (\( \omega \)) measures how fast an object rotates around a fixed point, such as a gymnast swinging on a bar. It's akin to speed but in rotational terms—instead of kilometers per hour, we deal with radians per second.To find the gymnast's angular velocity, use the rotational kinetic energy and her moment of inertia:The equation \( RE = \frac{1}{2} I \omega^2 \) rearranges to solve for angular velocity:\[\omega = \sqrt{\frac{2 \times RE}{I}}\]Plugging in our values:- \( RE = 109.8 \) J- \( I = 30 \text{ kg} \cdot \text{m}^2 \)\[\omega \approx \sqrt{\frac{2 \times 109.8}{30}} \approx 2.71 \text{ rad/s}\]This figure represents her speed of rotation and highlights how energy changes within a swinging motion to produce physical dynamics encountered in everyday rotations.

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