/*! 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 67 A boy whirls a stone in a horizo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 67

A boy whirls a stone in a horizontal circle of radius \(1.5 \mathrm{~m}\) and at height \(2.0 \mathrm{~m}\) above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of \(10 \mathrm{~m}\). What is the magnitude of the centripetal acceleration of the stone during the circular motion?

Short Answer

Expert verified
The centripetal acceleration is approximately 162.76 m/s虏.

Step by step solution

01

Identify Given Information

We know the initial height of the stone is 2.0 meters. Upon striking the ground, it covers a horizontal distance of 10 meters. We also know the radius of the circle is 1.5 meters.
02

Determine Time of Flight

First, calculate the time it takes for the stone to fall to the ground using the vertical motion equation. Start with the formula for free fall \[ h = \frac{1}{2} g t^2 \]where \( h = 2 \) m and \( g = 9.81 \) m/s虏. Solve for \( t \): \[ 2 = \frac{1}{2} \times 9.81 \times t^2 \]leading to\[ t^2 = \frac{4}{9.81} \]and \[ t = \sqrt{\frac{4}{9.81}} \approx 0.64 \text{ seconds}. \]
03

Calculate Initial Horizontal Velocity

Use the horizontal displacement to find the horizontal velocity. The stone travels 10 meters horizontally while taking 0.64 seconds to land:\[ x = v_x t \]\[ 10 = v_x \times 0.64 \]so,\[ v_x = \frac{10}{0.64} \approx 15.625 \text{ m/s}. \]
04

Relate Horizontal Velocity to Circular Motion

The initial horizontal velocity \( v_x \) is also the tangential speed of the stone in circular motion. Thus, the tangential speed of the stone at the moment the string breaks is \( 15.625 \text{ m/s} \).
05

Calculate Centripetal Acceleration

Use the centripetal acceleration formula,\[ a_c = \frac{v^2}{r} \]where \( v = 15.625 \text{ m/s} \) and \( r = 1.5 \text{ m} \):\[ a_c = \frac{(15.625)^2}{1.5} \approx 162.76 \text{ m/s}^2. \]

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!

Key Concepts

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

Circular Motion
In our scenario, circular motion refers to the movement of the stone as it whirls around in a horizontal circle. This motion is uniform, meaning the stone travels at a constant speed along a circular path. The key element in circular motion is the centripetal force, which acts towards the center of the circle and keeps the stone moving in its circular path.
Without this force, the stone would fly off tangentially, which is what happens when the string breaks in this problem, causing the stone to strike the ground. To understand circular motion better, consider these points:
  • **Centripetal Force:** This force helps the stone maintain its path; when the string breaks, the centripetal force is removed.
  • **Radius of the circle (r):** Here, it's given as 1.5 meters, determining the size of the circle.
  • **Tangential Speed (v):** This speed is perpendicular to the radius at all points along the circle and was calculated to be 15.625 m/s for this situation.
Utilizing these aspects, we calculated the centripetal acceleration, which measures how quickly the direction of the stone's velocity changes, even though its speed may remain constant.
Free Fall
Free fall describes the motion of the stone after the string breaks, wherein it begins to drop solely under the influence of gravity. This is a crucial part of our problem as it determines how long it takes the stone to reach the ground.
The stone begins its free fall from a height of 2 meters, and its descent can be analyzed using the kinematic equation for vertical motion:\[ h = \frac{1}{2} g t^2 \]This equation tells us that the distance fallen under gravity is proportional to the square of the time of fall, with gravity's acceleration \( g = 9.81 \text{ m/s}^2 \) as a key factor. From solving the equation, we determine the time \( t \) it takes for the stone to complete its descent, which is approximately 0.64 seconds. This time also dictates how long the stone travels horizontally before hitting the ground.
Horizontal Velocity
Horizontal velocity pertains to the speed of the stone as it travels along the horizontal axis. In this exercise, the horizontal velocity is a critical component since it is a result of the circular motion's tangential speed at the moment the string breaks.
From the problem, we know:
  • **Horizontal Distance (x):** The stone covers this distance across the ground, which measures 10 meters here.
  • **Time of Flight (t):** The time was calculated to be 0.64 seconds using free fall principles.
  • **Equation for Horizontal Velocity:** Given by \( x = v_x t \), where \(x\) is horizontal distance and \(v_x\) is horizontal velocity.
Utilizing these values, we solve for horizontal velocity, revealing it to be approximately 15.625 m/s. This velocity is the same as the tangential speed the stone had just before the string broke. Knowing the stone's horizontal velocity allows us to examine its behavior after being freed from circular motion.

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

\(A\) is \(90 \mathrm{~km}\) due west of oasis \(B\). A desert camel leaves \(A\) and takes \(50 \mathrm{~h}\) to walk \(75 \mathrm{~km}\) at \(37^{\circ}\) north of due east. Next it takes \(35 \mathrm{~h}\) to walk \(65 \mathrm{~km}\) due south. Then it rests for \(5.0 \mathrm{~h}\). What are the (a) magnitude and (b) direction of the camel's displacement relative to \(A\) at the resting point? From the time the camel leaves \(A\) until the end of the rest period, what are the (c) magnitude and (d) direction of its average velocity and (e) its average speed? The camel's last drink was at \(A\); it must be at \(B\) no more than \(120 \mathrm{~h}\) later for its next drink. If it is to reach \(B\) just in time, what must be the (f) magnitude and (g) direction of its average velocity after the rest period?

A graphing surprise. At time \(t=0\), a burrito is launched from level ground, with an initial speed of \(16.0 \mathrm{~m} / \mathrm{s}\) and launch angle \(\theta_{0}\). Imagine a position vector \(\vec{r}\) continuously directed from the launching point to the burrito during the flight. Graph the magnitude \(r\) of the position vector for (a) \(\theta_{0}=40.0^{\circ}\) and (b) \(\theta_{0}=80.0^{\circ}\). For \(\theta_{0}=40.0^{\circ}\), (c) when does \(r\) reach its maximum value, (d) what is that value, and how far (e) horizontally and (f) vertically is the burrito from the launch point? For \(\theta_{0}=80.0^{\circ},(\mathrm{g})\) when does \(r\) reach its maximum value, (h) what is that value, and how far (i) horizontally and (j) vertically is the burrito from the launch point?

A lowly high diver pushes off horizontally with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) from the platform edge \(10.0 \mathrm{~m}\) above the surface of the water. (a) At what horizontal distance from the edge is the diver \(0.800 \mathrm{~s}\) after pushing off? (b) At what vertical distance above the surface of the water is the diver just then? (c) At what horizontal distance from the edge does the diver strike the water?

A projectile is fired horizontally from a gun that is \(45.0 \mathrm{~m}\) above flat ground, emerging from the gun with a speed of \(250 \mathrm{~m} / \mathrm{s}\). (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

A small ball rolls horizontally off the edge of a tabletop that is \(1.20 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally from the table edge. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Magnetism

Read Explanation

Translational Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Force

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.