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Chapter 8: Problem 33

A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The angular speed of the child has a constant valuc of 0.250 \(\mathrm{rad} / \mathrm{s}\) . At the instant the child spots the horse, onc-quarter of a turn away, the merry-go-round begins to move (in the direction the child is runing) with a constant angular acceleration of 0.0100 \(\mathrm{rad} / \mathrm{s}^{2}\) . What is the shortest time it takes for the child to catch up with the horse?

Short Answer

Expert verified
The shortest time is approximately 42.6 seconds.

Step by step solution

01

Convert the child's distance to radians

A quarter of a turn is equivalent to one-fourth of the full circle. Since a full circle is \(2\pi\) radians, then one-quarter of a turn is equal to \(\frac{2\pi}{4} = \frac{\pi}{2}\) radians.
02

Write the angular position equation for the child

The child moves with a constant angular speed, so the angular position \(\theta_c\) of the child as a function of time \(t\) can be given as: \[\theta_c = \omega_c \cdot t\] Where \(\omega_c = 0.250\, \mathrm{rad}/\mathrm{s}\) is the angular speed of the child.
03

Write the angular position equation for the merry-go-round

The merry-go-round starts from rest and has a constant angular acceleration, so its angular position \(\theta_m\) as a function of time \(t\) is given by: \[\theta_m = \frac{1}{2} \cdot \alpha \cdot t^2\] where \(\alpha = 0.0100 \, \mathrm{rad}/\mathrm{s}^2\) is the angular acceleration.
04

Set up the equation for the child catching the horse

For the child to catch the horse, their angular positions must be equal, considering the initial \(\frac{\pi}{2}\) rad offset. Therefore, \[\omega_c \cdot t = \frac{\pi}{2} + \frac{1}{2} \cdot \alpha \cdot t^2\] Substitute the known values to get: \[0.250 \cdot t = \frac{\pi}{2} + 0.005 \cdot t^2\]
05

Rearrange the equation into a quadratic form

Rearrange the equation: \[0.005t^2 - 0.250t + \frac{\pi}{2} = 0\] This is a standard quadratic equation of the form \(at^2 + bt + c = 0\) where \(a = 0.005\), \(b = -0.250\), and \(c = \frac{\pi}{2}\).
06

Solve the quadratic equation

Use the quadratic formula, \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], to solve for \(t\). Substitute the coefficients: \[t = \frac{-(-0.250) \pm \sqrt{(-0.250)^2 - 4 \cdot 0.005 \cdot \frac{\pi}{2}}}{2 \cdot 0.005}\]Calculate under the square root: \[t = \frac{0.250 \pm \sqrt{0.0625 - 0.0314}}{0.01}\] and simplify: \[t = \frac{0.250 \pm \sqrt{0.0311}}{0.01}\]\[t = \frac{0.250 \pm 0.176}\]Choose the positive solution for time: \[t = \frac{0.250 + 0.176}{0.01} = 42.6 \, \mathrm{s}\]
07

Verify the solution

Check that the time value obtained allows both the child and the horse to reach the same angular position.For the child: \[\theta_c = 0.250 \times 42.6 = 10.65 \, \mathrm{rad}\]For the merry-go-round including initial offset: \[\theta_m = \frac{\pi}{2} + 0.005 \times (42.6)^2 = 10.65 \, \mathrm{rad}\]Both outcomes are equal, confirming the solution is correct.

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Key Concepts

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

Angular Speed
Angular speed refers to how fast an object is rotating. It is the angle an object rotates through per unit time. In our exercise, the child running around the merry-go-round has a constant angular speed of 0.250 radians per second. This means that every second, the child covers an angular distance of 0.250 radians as he moves.

Understanding angular speed is crucial when analyzing rotational motion. Unlike linear speed, which is measured in distance per time, angular speed is measured in radians per second. This makes sense because, in circular motion, distance is considered in terms of the rotation of angles.
  • Angular speed is constant here, implying the child's rotational motion does not change in pace.
  • The child鈥檚 constant angular speed simplifies the calculations as each unit of time results in a predictable increase in the angular position.
Angular Acceleration
Angular acceleration is the rate of change of angular speed. It tells you how quickly the angular speed of an object increases or decreases. In the exercise, the merry-go-round starts with an angular acceleration of 0.0100 \ \( \mathrm{rad/s^2} \ \).

This means that every second, the merry-go-round's angular speed increases by 0.0100 radians per second. When the child spots the horse one-quarter turn away, the merry-go-round begins its acceleration, meaning it was initially at rest but starts spinning faster over time.
  • With angular acceleration, even if something starts from rest, it will gain speed rotationally over time.
  • In our problem, the merry-go-round moving with angular acceleration affects when or if the child catches the horse.
  • This quantity is important to understand how the spin rate changes and impacts the motion analysis.
Quadratic Equations
Quadratic equations are vital tools for solving problems involving rotational motion when variables change with time. In the given exercise, we derive a quadratic equation to determine when the child's and merry-go-round's angular positions match.

The form the equation \ \( 0.005t^2 - 0.250t + \frac{\pi}{2} = 0 \ \) comes from equating the angular positions of both the child and the merry-go-round. It's essential to know how to solve these equations since they often appear in physics when two opposing or related actions affect a system.
  • The solution involves substituting values into the quadratic formula: \ \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ \).
  • This helps to find the time when two moving entities are at the same position.
  • Quadratic equations often provide two solutions, but here the negative time isn't feasible, so only the positive is chosen.
Angular Position
Angular position is a way to describe the location of an object in its rotational path at a given time. The concept is crucial in tracking how far along it something is in its circular path. In our scenario, it's about how far the child and merry-go-round have traveled in their rotations.

The child鈥檚 angular position is modeled by \ \( \theta_c = \omega_c \cdot t \ \), assuming he continues at a constant speed. Meanwhile, the merry-go-round's angular position uses \ \( \theta_m = \frac{1}{2} \cdot \alpha \cdot t^2 \ \), acknowledging it speeds up over time.
  • This concept allows understanding of where rotational objects are located in relation to starting points.
  • Over time, tracking angular position can tell you when two rotating objects meet.
  • In this problem, equating child and merry-go-round's angular positions finds when they align, considering the initial distance gap.

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Most popular questions from this chapter

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 45 s, in order to buff an especially scuffed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Our sun rotates in a circular orbit about the center of the Milky Way galaxy. The radius of the orbit is \(2.2 \times 10^{30} \mathrm{m},\) and the angular speed of the sun is \(1.1 \times 10^{-15}\) rad/s. How long (in years) does it take for the sun to make one revolution around the center?

A spinning whecl on a fircworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of \(-4.00 \mathrm{rad} / \mathrm{s}^{2}\) . Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of \(-25.0 \mathrm{rad} / \mathrm{s}\) . While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

In \(9.5 \mathrm{~s}\) a fisherman winds \(2.6 \mathrm{~m}\) of fishing line onto a reel whose radius is \(3.0 \mathrm{~cm}\) (assumed to be constant as an approximation). The line is being reeled in at a constant speed. Determine the angular speed of the reel.

A dragster starts from rest and accelerates down a track. Each tire has a radius of 0.320 m and rolls without slipping. At a distance of 384 m, the angular speed of the wheels is 288 rad/s. Determine (a) the linear speed of the dragster and (b) the magnitude of the angular acceleration of its wheels.
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