/*! 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] | 91Ó°ÊÓ

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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 162.98 m/s².

Step by step solution

01

Determine Time of Flight

Since the stone flies off horizontally, we first need to find the time it takes for it to reach the ground. Using the equation for vertical motion, use\[ h = \frac{1}{2} g t^2 \]where \( h = 2 \text{ m} \) is the height and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Solve for \( t \):\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 2}{9.8}} \approx 0.64 \text{ s} \]
02

Calculate Horizontal Speed

The horizontal distance the stone travels is \( 10 \text{ m} \). Use the formula:\[ d = v_{\text{horizontal}} \cdot t \]Solve for \( v_{\text{horizontal}} \):\[ v_{\text{horizontal}} = \frac{d}{t} = \frac{10}{0.64} \approx 15.63 \text{ m/s} \]
03

Find Centripetal Acceleration

The centripetal acceleration can be calculated using the formula:\[ a_{c} = \frac{v^2}{r} \]where \( v = 15.63 \) m/s is the horizontal speed and \( r = 1.5 \text{ m} \) is the radius of the circle:\[ a_{c} = \frac{(15.63)^2}{1.5} \approx 162.98 \text{ m/s}^2 \]

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Key Concepts

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

Circular Motion
In circular motion, an object moves along a circular path. This motion is prevalent in various scenarios, from planets orbiting stars to a stone tied to a string being twirled around. Understanding circular motion involves knowing certain forces and equations. For instance, while the stone is in circular motion, it experiences a force directed towards the center of the circle. This force helps maintain the stone's path. This force is centripetal force, and it provides the centripetal acceleration that changes the stone's direction without changing its speed. The formula for centripetal acceleration is given by:
  • \( a_{c} = \frac{v^2}{r} \)
where \( v \) is the velocity of the object and \( r \) is the radius of the circle. The key takeaway here is that centripetal acceleration keeps the stone moving in its circular path before the string breaks.
Horizontal Projectile Motion
Horizontal projectile motion occurs when an object is propelled horizontally and then affected only by gravity vertically. In the case of the stone, once the string breaks, it flies off in a horizontal path before gravity pulls it down to the ground. Initially, the horizontal speed of the stone remains constant چون러운 air resistance is negligible. The distance traveled horizontally over time helps determine this speed.
  • The time of flight and the initial horizontal speed determine the horizontal distance traveled.
  • Horizontal motion can be analyzed independently of vertical motion.
This separation comes in handy because it simplifies calculations by allowing us to focus on horizontal movements and vertical movements separately.
Kinematics
Kinematics is the branch of mechanics that describes the motion of objects without necessarily discussing the forces that lead to the motion. In this scenario, we applied kinematic equations to decipher the motion of the stone. The kinematic equations aid in understanding the relationship between time, velocity, acceleration, and displacement. By using the vertical motion equation:
  • \( h = \frac{1}{2} g t^2 \).
We solved the time it took for the stone to reach the ground. Subsequently, this data helped to find the horizontal velocity. Kinematics breaks down motion into manageable parts, easing the understanding of how objects travel from one point to another under gravity's influence.
Gravity
Gravity is the force that pulls objects toward the center of the Earth. It plays a critical role in projectile motion, affecting the vertical component of an object's motion. For the stone in question, gravity pulls it vertically downward as it travels horizontally after the string breaks.
  • The acceleration due to gravity is usually denoted by \( g \), with a standard value of \( 9.8 \text{ m/s}^2 \).
  • Gravity acts continuously, affecting objects, bringing them back down to the earth's surface.
In the problem at hand, gravity determines how long the stone continues descending after being released from circular motion. It emphasizes how vital gravity is in projectile motion scenarios, helping calculate how far and how fast objects drop when in motion.

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Most popular questions from this chapter

An astronaut is rotated in a horizontal centrifuge at a radius of \(5.0 \mathrm{~m} .\) (a) What is the astronaut's speed if the centripetal acceleration has a magnitude of \(7.0 g ?\) (b) How many revolutions per minute are required to produce this acceleration? (c) What is the period of the motion?

The position \(\vec{r}\) of a particle moving in an \(x y\) plane is given by \(\vec{r}=\left(2.00 t^{3}-5.00 t\right) \hat{\mathrm{i}}+\left(6.00-7.00 t^{4}\right) \hat{\mathrm{j}},\) with \(\vec{r}\) in meters and \(t\) in seconds. In unit-vector notation, calculate (a) \(\vec{r},(\mathrm{~b}) \vec{v},\) and (c) \(\vec{a}\) for \(t=2.00 \mathrm{~s}\). (d) What is the angle between the positive direction of the \(x\) axis and a line tangent to the particle's path at \(t=2.00 \mathrm{~s} ?\)

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking \(2.50 \mathrm{~s}\). Then security agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking \(10.0 \mathrm{~s}\). What is the ratio of the man's running speed to the sidewalk's speed?

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 particle \(P\) travels with constant speed on a circle of radius \(r=3.00 \mathrm{~m}\) (Fig. \(4-56\) ) and completes one revolution in \(20.0 \mathrm{~s}\). The particle passes through \(O\) at time \(t=0 .\) State the following vectors in magnitudeangle notation (angle relative to the positive direction of \(x\) ). With respect to \(O,\) find the particle's position vector at the times \(t\) of (a) \(5.00 \mathrm{~s}\), (b) \(7.50 \mathrm{~s},\) and (c) \(10.0 \mathrm{~s}\). (d) For the \(5.00 \mathrm{~s}\) interval from the end of the fifth second to the end of the tenth second, find the particle's displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.

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