/*! 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 19 An adapted golf device attaches ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An adapted golf device attaches to a wheelchair to help people with mobility impairments play putt-putt. The stationary frame OD is attached to the wheelchair, and a club holder OB is attached to the pin at O. Holder OB is 6 in. long and weighs 8 oz, and the distance between \(O\) and \(D\) is \(x=1\) ft. The putter shaft has a length of \(f\) \(L=36\) in. and weighs 10 oz, while the putter head at \(A\) weighs 12 oz. Knowing that the 1 -lb/in. spring between \(D\) and \(B\) is unstretched when \(\theta=90^{\circ}\) and that the putter is released from rest at \(\theta=0\), determine the putter head speed when it hits the golf ball.

Short Answer

Expert verified
The putter head speed is determined by solving potential energy conversion to kinetic form; using inertia and energy conservation principles.

Step by step solution

01

Convert Units and Assign Values

First, we will make sure all measurements are in consistent units. We are given holder OB which is 6 in long and weighs 8 oz, OD which is 12 in, shaft length L as 36 in, shaft weight as 10 oz, and putter head weight as 12 oz. Note that 1 lb = 16 oz. Set original lengths and weights for calculations: - Weight of holder OB: 0.5 lb - Weight of shaft: 0.625 lb - Weight of putter head A: 0.75 lb - Spring constant: 1 lb/in - OD length: 12 in.
02

Calculate Total Moment of Inertia

Calculate the moment of inertia about point O for all components. This includes the holder OB, the shaft, and the putter head. - Moment of inertia for holder OB: Let length = 6 in, total moment of inertia is \ \( I_{OB} = \frac{1}{3} \times \text{Weight of OB} \times (6)^2 \)- Moment of inertia for shaft: Let shaft length = 36 in, assumed at midpoint \ \( I_{shaft} = \frac{1}{3} \times \text{Weight of shaft} \times (36)^2 \)- Moment of inertia for putter head: Let head be a point mass \ \( I_{head} = \text{Weight of head} \times (36)^2 \)- Sum all: \( I_{total} = I_{OB} + I_{shaft} + I_{head} \)
03

Establish Energy Conservation

Since the system is released from rest and the spring is unstressed at \ \(\theta = 90^{\circ} \) we can apply energy conservation: \eneInitial Spring Energy + Initial Gravitational Potential Energy = Final Kinetic Energy at putter head. - Initial Spring Energy is 0, and Initial Gravitational Potential Energy is related to its height change from \ \theta=0 \- Calculate the change in potential energy and equate it with the kinetic energy when \ \theta=90^{\circ} \- Simplify to find linear velocity terms.
04

Solve for Putter Head Speed

Solve the energy equation to find the velocity \( v \) of the putter head.- The computation for energy includes converting the stored energy from potential energies to kinetic resulting from the motion by holder and putter head.- Converting potential energy change into kinetic energy, solve for \( v \) through simple algebraic method.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Energy Conservation in Mechanics
Energy conservation is a pivotal concept in mechanics as it helps predict how systems behave over time. When dealing with devices like the adapted golf device attached to a wheelchair, understanding energy conservation allows us to calculate the speed at which the putter head reaches the golf ball.
The core idea here is that energy within an isolated system remains constant.
  • Initially, when the device is brought to rest, the spring is unstretched and positioned at an angle of heta = 90^{\circ} \text{,} with a net thermal energy and potential energy attributed to the gravitational height at that instant.
  • As the putter is released, the potential energy housed in the spring begins to convert into kinetic energy.
  • The gravitational potential energy also decreases only to reformulate this potential energy into kinetic energy when the angle shifts to 0° as the putter swings down.

By summing up these conversions, one can employ the energy conservation equation:\[ \text{Initial Spring Energy} + \text{Initial Gravitational Potential Energy} = \text{Final Kinetic Energy} \] This equates the stored potential energies to kinetic energies as motion occurs.
These transformations allow for systematic energy balance checks which simplify the solution to finding velocities when the mechanical action takes place within constraints, ensuring no energy is lost, only transformed.
Moment of Inertia Calculations
Moment of inertia is a concept in mechanics that deals with the distribution of mass in relation to an axis, often referred to as the rotational "mass".
It is crucial to understand how different parts of the adapted golf device contribute to the total moment of inertia, which ultimately affects the putter head's swing.
  • For the holder OB, which behaves as a rod rotating around an end, its moment of inertia is calculated as:\[ I_{OB} = \frac{1}{3} \times \text{Weight of OB} \times (\text{Length of OB})^2 \]
  • The shaft is another major contributor, modeled as a rod with its mass "centered" or distributed equally. This is calculated as:\[ I_{shaft} = \frac{1}{3} \times \text{Weight of shaft} \times (\text{Length of shaft})^2 \]
  • The putter head, being a point mass at the end of the shaft, adds inertia calculated by:\[ I_{head} = \text{Weight of head} \times (\text{Distance from center O to head})^2 \]
Understanding these factors and how they interact helps shape our analysis allowing prediction of how smoothly or rapidly the putter head can translate or rotate under mechanical operations.
Unit Conversion in Mechanics
Unit conversion is the backbone of solving almost any mechanics problem correctly.
In this case, consistent unit usage ensures precise calculation of the forces and moments present in the adapted golf device system.
  • Initially, the measurements were provided in inconsistent units like inches for length and ounces for weight. These were converted into more convenient mechanics-friendly units, such as pounds and feet, before starting calculations.
  • For weight conversions, remember that 1 pound equals 16 ounces. So, for example, the holder OB's weight of 8 oz becomes 0.5 lbs, aligning with the requirement.
  • For length conversion, 1 foot equals 12 inches. Hence, OD length 1 ft directly converts to 12 inches for consistent calculations with shaft lengths etc.

These conversions are crucial to handle computations involving a mix of linear measurements, angles, and forces, thus simplifying complex physics into manageable steps.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 9-in.-radius cylinder of weight 18 lb rests on a 6-lb carriage. The system is at rest when a force P of magnitude 2.5 lb is applied as shown for 1.2 s. Knowing that the cylinder rolls without sliding on the carriage and neglecting the mass of the wheels of the carriage, determine the resulting velocity of (a) the carriage, (b) the center of the cylinder.

Member \(A B C\) has a mass of \(2.4 \mathrm{kg}\) and is attached to a pin support at \(B\). An \(800-\mathrm{g}\) sphere \(D\) strikes the end of member \(A B C\) with a vertical velocity \(\mathrm{v}_{1}\) of \(3 \mathrm{m} / \mathrm{s}\). Knowing that \(L=750 \mathrm{mm}\) and that the coefficient of restitution between the sphere and member \(A B C\) is \(0.5,\) determine immediately after the impact \((a)\) the angular velocity of member \(A B C,(b)\) the velocity of the sphere.

A uniform slender bar of length \(L=200 \mathrm{mm}\) and mass \(m=0.5 \mathrm{kg}\) is supported by a frictionless horizontal table. Initially the bar is spinning about its mass center \(G\) with a constant angular speed \(\omega_{1}=6 \mathrm{rad} / \mathrm{s}\). Suddenly latch \(D\) is moved to the right and is struck by end \(A\) of the bar. Knowing that the coefficient of restitution between \(A\) and \(D\) is \(e=0.6\), determine the angular velocity of the bar and the velocity of its mass center immediately after the impact.

A small grinding wheel is attached to the shaft of an electric motor that has a rated speed of 3600 rpm. When the power is turned off, the unit coasts to rest in 70 s. The grinding wheel and rotor have a combined weight of 6 lb and a combined radius of gyration of 2 in. Determine the average magnitude of the couple due to kinetic friction in the bearings of the motor.

A \(5-m-\) -long ladder has a mass of \(15 \mathrm{kg}\) and is placed against a house at an angle \(\theta=20^{\circ} .\) Knowing that the ladder is released from rest, determine the angular velocity of the ladder and the velocity of end \(A\) when \(\theta=45^{\circ}\) Assume the ladder can slide freely on the horizontal ground and on the vertical wall.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.