/*! 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 10 The spool has a mass of \(40 \ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 10

The spool has a mass of \(40 \mathrm{~kg}\) and a radius of gyration of \(k_{O}=0.3 \mathrm{~m}\). If the \(10-\mathrm{kg}\) block is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity \(\omega=15 \mathrm{rad} / \mathrm{s}\). Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

Short Answer

Expert verified
The block must fall approximately 5.73 m for the spool to reach an angular velocity of 15 rad/s. The tension in the cord while the block is in motion is about 57.19 N.

Step by step solution

01

Define variables

Let's define:- Initial Kinetic Energy \(KE_{i}\) as 0, because the block starts from rest.- Final Kinetic Energy \(KE_{f}\) equals the kinetic energy of the falling block plus the kinetic energy of the rotating spool.- Potential Energy \(PE\) equals the weight of the block times the height \(h\) it has fallen.- The total energy at the start and at the end remains the same because no external work is done on the system.
02

Apply the Principle of Conservation of Energy

We equate the total initial energy \(KE_{i} + PE_{i}\) with the final energy \(KE_{f} + PE_{f}\):\[0 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}Ik\omega^2\]The term \(\frac{1}{2}mv^2\) is the kinetic energy of the falling block and the term \(\frac{1}{2}Ik\omega^2\) is the kinetic energy of the rotating spool.
03

Substitute the given values into the equation

The fallen height is \(h = \frac{\frac{1}{2}mv^2 + \frac{1}{2}Ik\omega^2}{mg}\). Here:- \(m = 10kg\) is the mass of the block,- \(v = \omega r\) is the speed of the block (where \(r = 0.3m\) is the radius of the gyration and \(\omega = 15 rad/s\) is the angular velocity),- \(I = mk^2\) is the mass moment of inertia of the spool (where \(m = 40kg\) is the mass of the spool and \(k = 0.3m\) is the radius of gyration),- \(g = 9.81m/s^2\) is the acceleration due to gravity.
04

Calculate the Tension in the cord

The tension \(T\) in the cord can be calculated using Newton鈥檚 second law:\[T = m(g 鈭� a)\]where,- \(a = r伪 = r\omega^2/r = \omega^2\), and- \(伪 = \frac{ \omega^2}{r}\) is the angular acceleration of the block.
05

Substitute the given values into the equation

Substitute \(m = 10kg\), \(\omega = 15 rad/s\) and \(r = 0.3m\) into the equation to find the tension \(T\).

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 91影视!

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

The wheel has a mass of \(100 \mathrm{~kg}\) and a radius of gyration of \(k_{O}=0.2 \mathrm{~m} .\) A motor supplies a torque \(M=(40 \theta+900) \mathrm{N} \cdot \mathrm{m}\), where \(\theta\) is in radians, about the drive shaft at \(O .\) Determine the speed of the loading car, which has a mass of \(300 \mathrm{~kg}\), after it travels \(s=4 \mathrm{~m}\). Initially the car is at rest when \(s=0\) and \(\theta=0^{\circ} .\) Neglect the mass of the attached cable and the mass of the car's wheels.

The spool has a mass of \(50 \mathrm{~kg}\) and a radius of gyration of \(k_{o}=0.280 \mathrm{~m}\). If the \(20-\mathrm{kg}\) block \(A\) is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity \(\omega=5 \mathrm{rad} / \mathrm{s}\). Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

The spool has a mass of \(20 \mathrm{~kg}\) and a radius of gyration of \(k_{O}=160 \mathrm{~mm}\). If the \(15-\mathrm{kg}\) block \(A\) is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity \(\omega=8 \mathrm{rad} / \mathrm{s}\) Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord.

The force of \(T=20 \mathrm{~N}\) is applied to the cord of negligible mass. Determine the angular velocity of the \(20-\mathrm{kg}\) wheel when it has rotated 4 revolutions starting from rest. The wheel has a radius of gyration of \(k_{O}=0.3 \mathrm{~m}\).

If the gear is released from rest, determine its angular velocity after its center of gravity \(O\) has descended a distance of \(4 \mathrm{ft}\). The gear has a weight of \(100 \mathrm{lb}\) and a radius of gyration about its center of gravity of \(k=0.75 \mathrm{ft}\).
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electrostatics

Read Explanation

Famous Physicists

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
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.