/*! 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 104 A \(4-16\) ball \(B\) is traveli... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(4-16\) ball \(B\) is traveling around in a circle of radius \(r_{1}=3 \mathrm{ft}\) with a speed \(\left(v_{B}\right)_{1}=6 \mathrm{ft} / \mathrm{s}\). If the attached cord is pulled down through the hole with a constant speed \(v_{r}=2 \mathrm{ft} / \mathrm{s},\) determine how much time is required for the ball to reach a speed of \(12 \mathrm{ft} / \mathrm{s}\). How far \(r_{2}\) is the ball from the hole when this occurs? Neglect friction and the size of the ball.

Short Answer

Expert verified
Thus, it requires 0.75 seconds to reach a speed of 12 ft/s, and when this occurs, the ball is 1.5 feet from the hole.

Step by step solution

01

Identify given quantities and find angular speed

The initial radius \(r_1 = 3\) ft, initial velocity \(v_{B_1} = 6\) ft/s, and final velocity \(v_{B_2} = 12\) ft/s are given. We can start by determining the initial angular speed,\(\omega_1\), of the ball, which is \(v_{B_1} / r_1 = 6/3 = 2\) rad/s.
02

Use conservation of angular momentum

Angular momentum is conserved because the net external torque on the ball is zero. So, \(m * r_1 * \omega_1 = m * r_2 * \omega_2\). Use this equation to find the relationship between initial and final angular speeds, which gives \(\omega_2 = r_1 * \omega_1 / r_2\).
03

Express the final speed in terms of angular speed

Express the final speed of the ball \(v_{B_2}\) in terms of final angular speed \(v_{B_2} = r_2 * \omega_2\). Substituting in the expression for \(\omega_2\) from step 2, we get \(v_{B_2} = r_1 * \omega_1\). This gives us \(r_2 = r_1 * v_{B_1} / v_{B_2} = 3 * 6 / 12 = 1.5\) ft.
04

Determine the time required

We know that the cord is pulled down at a constant speed, \(v_r = 2\) ft/s. So, the total time taken to pull the cord from \(r_1\) to \(r_2\) can be found as \((r_1 - r_2) / v_r = (3 - 1.5) / 2 = 0.75\) s.

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Key Concepts

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

Circular Motion
Circular motion occurs when an object moves along a circular path at a constant or changing speed.
In this case, the ball travels around in a circular path of radius 3 feet.
As it moves, the direction of the velocity changes continuously, while the speed can vary depending on the dynamics of the system.In circular motion, two main forces come into play:
  • **Centripetal Force**: Keeps the object moving in a circular path by acting towards the center of the circle.
  • **Centrifugal Force**: An apparent force that seems to push the object away from the center when observed in a rotating frame.
Remember, centripetal acceleration, the acceleration directed toward the center of the circle, is given by the formula: \[a_c = \frac{v^2}{r}\]where \(v\) is the linear speed and \(r\) is the radius.
Understanding these forces and motions helps to predict how speed and radius changes affect the object's motion.
Conservation of Angular Momentum
The principle of conservation of angular momentum is a powerful tool in rotational dynamics.
It states that if no external torque acts on a system, its angular momentum remains constant.
Angular momentum \(L\) for a particle is the product of its mass \(m\), velocity \(v\), and radius \(r\):
\[L = m \, v \, r\]In the context of the exercise,
  • The initial angular momentum \((L_1)\) is given by \(m \, v_{B_1} \, r_1\).
  • The final angular momentum \((L_2)\) is \(m \, v_{B_2} \, r_2\).
Since no external forces act on the ball, \(L_1 = L_2\). Conservation of angular momentum allows us to relate the initial and final conditions:
\[r_1 \, v_{B_1} = r_2 \, v_{B_2}\]This equation shows the dependency of speed and radius, and how they adjust according to changes to conserve angular momentum.
This connection helps solve for unknowns in problems involving rotational systems.
Angular Speed
Angular speed, or angular velocity, refers to how fast an object rotates around a specific point.
It is expressed in radians per second (rad/s).
In the given exercise, it provides insight into the ball's rotational motion as it is pulled closer to the center. Angular speed \(\omega\) is computed as \[\omega = \frac{v}{r}\]where \(v\) is linear speed and \(r\) is the radius of the path.
Initially, the angular speed of the ball can be determined using its velocity and initial radius \[\omega_1 = \frac{6 \, \text{ft/s}}{3 \, \text{ft}} = 2 \, \text{rad/s}\]Ensuring conservation of angular momentum, the angular speed increases as the radius decreases.
This is reflective of the inverse relationship between radius and angular speed when momentum is conserved.
Angular speed directly influences the rotational dynamics of the system, showing how speed adapts as the constraints change.
Rotational Dynamics
Rotational dynamics examines the motion of objects when subjected to forces causing rotation.
It's the counterpart of linear dynamics and deals with the effects of torques and angular motion.In the system described in the exercise:
  • The ball moves in a rotating path, which shrinks as the cord is pulled, altering its radius.
  • The forces involved cause changes in the ball's velocity and angular speed.
Key equations and principles such as torque and moment of inertia come into play:-The torque \(\tau\) acting on a rotating system is the product of force \(F\) and radius \(r\):\[\tau = F \, r\]-The moment of inertia \(I\) is a measure of an object's resistance to changes in rotational motion.Understanding the interaction of these factors is vital for predicting motion in systems like the one in this problem.
The exercise demonstrates how pulling the cord changes these dynamics, affecting the speed and radius of the circle.
Analyzing this interchange helps explain rotational motion in varied contexts.

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Most popular questions from this chapter

A power lawn mower hovers very close over the ground. This is done by drawing air in at a speed of \(6 \mathrm{m} / \mathrm{s}\) through an intake unit \(A,\) which has a cross-sectional area of \(A_{A}=0.25 \mathrm{m}^{2},\) and then discharging it at the ground, \(B\) where the cross-sectional area is \(A_{B}=0.35 \mathrm{m}^{2} .\) If air at \(A\) is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has a mass of \(15 \mathrm{kg}\) with center of mass at \(G .\) Assume that air has a constant density of \(\rho_{a}=1.22 \mathrm{kg} / \mathrm{m}^{3}\)

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