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Chapter 5: Problem 57

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 while in circular motion?

Short Answer

Expert verified
The magnitude of the centripetal acceleration is `162.76 m/s^2`.

Step by step solution

01

- Analyzing the Motion after the String Breaks

First, determine the time taken for the stone to hit the ground after the string breaks. Use the free fall equation for the vertical motion: \[ y = v_{iy}t + \frac{1}{2}gt^2 \]Here, \( y = 2 \, \text{m} \) (the height), \( v_{iy} = 0 \, \text{m/s} \) (since the stone flies off horizontally), and \( g = 9.8 \, \text{m/s}^2 \). Solving for \( t \)\[ 2 = \frac{1}{2} \times 9.8 \times t^2 \]\[ 2 = 4.9t^2 \]\[ t^2 = \frac{2}{4.9} \]\[ t = \sqrt{\frac{2}{4.9}} \]\[ t = 0.64 \, \text{s} \]
02

- Determine the Horizontal Velocity

Now, use the horizontal distance traveled and the time to find the horizontal velocity of the stone:\[ d = v_xt \]\[ 10 \, \text{m} = v_x \times 0.64 \, \text{s} \]\[ v_x = \frac{10 \, \text{m}}{0.64 \, \text{s}} \]\[ v_x = 15.625 \, \text{m/s} \]This is the horizontal speed at which the stone flies off, which was also the tangential speed during the circular motion.
03

- Calculating the Centripetal Acceleration

Use the tangential speed and radius to find the centripetal acceleration:\[ a_c = \frac{v^2}{r} \]Here, \( v = 15.625 \, \text{m/s} \) and \( r = 1.5 \, \text{m} \):\[ a_c = \frac{(15.625)^2}{1.5} \]\[ a_c = \frac{244.14}{1.5} \]\[ a_c = 162.76 \, \text{m/s}^2 \]

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Key Concepts

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

circular motion
Understanding circular motion is key for analyzing this exercise. When an object moves in a circle, it experiences a force that keeps it moving along the circular path. This force is called the centripetal force.
Many examples of circular motion are seen in daily life, such as a car turning around a bend, a satellite orbiting a planet, or in this case, a stone being whirled in a circle.

The stone is continuously changing direction, even if it's moving at a constant speed. This change in direction means the stone is accelerating toward the center of the circle, called centripetal acceleration. This doesn't change the stone's speed but changes its direction.
The formula for centripetal acceleration is:
\[ a_c = \frac{v^2}{r} \]
where \( a_c \) is the centripetal acceleration, \( v \) is the tangential speed, and \( r \) is the radius of the circle.
free fall equation
Free fall describes the motion of objects under the influence of gravity alone. In this exercise, once the string breaks, the stone enters free fall.
The free fall equation is used to determine how long it takes for an object to hit the ground from a certain height:
\[ y = v_{iy}t + \frac{1}{2}gt^2 \]
Here, \( y \) is the height, \( v_{iy} \) is the initial vertical velocity (0 m/s in this case), and \( g \) is the acceleration due to gravity (9.8 m/s\(^2\)).
By knowing the height of the stone from the ground (2 meters) and solving for \(t \) (time), we find the time taken for the stone to hit the ground.
horizontal motion
In horizontal motion, the object moves along a straight path parallel to the ground. The rate of horizontal motion (velocity) remains constant if air resistance is negligible.
For the stone, once the string breaks and it flies off horizontally, its horizontal velocity can be determined using the distance it travels and the time it takes to hit the ground.
Using the formula: \[ d = v_xt \]
Here, \( d \) is the horizontal distance, \( v_x \) is the horizontal velocity, and \( t \) is the time. We can rearrange this to solve for \( v_x \) using the distances given.
This horizontal velocity is the same as the stone's tangential speed during its circular motion before the string broke.

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Most popular questions from this chapter

(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? (b) What would the period of rotation of Earth have to be for objects on the equator to have a centripetal acceleration with a magnitude of \(9.8 \mathrm{~m} / \mathrm{s}^{2} ?\)

Particle Leaves Origin A particle leaves the origin with an initial velocity \(\vec{v}_{1}=(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and a constant acceleration \(\vec{a}=\) \(\left(-1.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-0.500 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). When the particle reaches its maximum \(x\) coordinate, what are (a) its velocity and (b) its position vector?

A particle moves so that its position as a function of time is \(\vec{r}(t)=(1 \mathrm{~m}) \hat{\mathrm{i}}+\left[\left(4 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\mathrm{j}}\). Write expressions for (a) its velocity and (b) its acceleration as functions of time.

. Stairway A ball rolls horizontally off the top of a stairway with a speed of \(1.52 \mathrm{~m} / \mathrm{s}\). The steps are \(20.3 \mathrm{~cm}\) high and \(20.3 \mathrm{~cm}\) wide. Which step does the ball hit first?

Two Seconds Later Two seconds after being projected from ground level, a projectile is displaced \(40 \mathrm{~m}\) horizontally and \(53 \mathrm{~m}\) vertically above its point of projection. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from its point of projection?
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