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Chapter 8: Problem 21

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has a length of \(64 \mathrm{~cm}\) and is wound around the top at a spot where its radius is \(2.0 \mathrm{~cm}\). The thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of the string, thereby unwinding it and giving the top an angular acceleration of \(+12 \mathrm{rad} / \mathrm{s}^{2}\). What is the final angular velocity of the top when the string is completely unwound?

Short Answer

Expert verified
The final angular velocity is approximately 11.1 rad/s.

Step by step solution

01

Understanding the Problem

We need to find the final angular velocity of the top when the string is completely unwound. The given values are the length of the string (64 cm), the radius where it is wound around the top (2.0 cm), and the angular acceleration (12 rad/s虏).
02

Convert Units

Convert the length of the string from centimeters to meters to use SI units. The length of the string is 64 cm, which is equivalent to 0.64 meters.
03

Calculate Total Radians Wound

The length of the string divided by the circumference of the circle gives the total number of radians the string unwinds. The circumference of the circle is given by the formula \(C = 2\pi r\). Using the radius 2 cm (0.02 m): \[ C = 2\pi \times 0.02 = 0.04\pi \text{ meters} \]The total radians (\(\theta\)) is:\[ \theta = \frac{0.64}{0.04\pi} \approx 5.1\text{ radians} \]
04

Relate Angular Quantities

Using kinematic equations for rotational motion, find the final angular velocity \(\omega_f\) using:\[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]Where \(\omega_i\) (initial angular velocity) is 0 because it starts from rest, \(\alpha\) is the angular acceleration (12 rad/s虏), and \(\theta\) is approximately 5.1 radians.
05

Substitute and Solve

Substitute the known values into the equation:\[ \omega_f^2 = 0 + 2 \times 12 \times 5.1 \]\[ \omega_f^2 = 122.4 \]Finally, solve for \(\omega_f\):\[ \omega_f = \sqrt{122.4} \approx 11.1 \text{ rad/s} \]

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Key Concepts

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

Angular Velocity
Angular velocity describes how fast something spins around a central point or axis. It's about how quickly it's rotating. Angular velocity is symbolized by \( \omega \) and measured in radians per second (rad/s). For instance, if a top is spinning, its angular velocity tells us the rate of change of its angle over time.
  • Imagine you have a merry-go-round. As it spins, each part moves around its center. The speed of this rotation is its angular velocity.
  • Angular velocity is important because it helps us understand rotational movements, such as when a wheel turns, or in this case, a spinning top.
The final angular velocity is what an object like a toy top will reach once it stops accelerating and is constantly spinning. In our exercise, calculating the final angular velocity involves using rotational kinematic equations based on the initial conditions like starting from rest.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. When you start or stop something spinning, angular acceleration tells you how quickly the spin is speeding up or slowing down.
  • In simple terms, if a top starts from being still and gradually spins faster as you pull the string, the rate at which it speeds up is its angular acceleration.
  • It's symbolized by \( \alpha \) and measured in radians per second squared (rad/s虏).
In our scenario, the top was given an angular acceleration of \( +12 \text{ rad/s}^2 \). This indicates that as the string is pulled and the top spins faster, its rotational speed increases by 12 rad/s every second.
Kinematic Equations
Kinematic equations for rotational motion are the mathematical formulas that relate various aspects of motion like distance, velocity, time, and acceleration. They make it easier to figure out unknowns when you know certain variables, like angular acceleration and displacement. Similar to linear motion equations, but tailored for rotation.
  • For example, they can help us find the final angular velocity if we know the initial angular velocity, angular acceleration, and how much the object has rotated (angular displacement).
  • In rotational motion, we use equations like \( \omega_f^2 = \omega_i^2 + 2\alpha\theta \).
These equations are crucial in our exercise for determining the final velocity of the top as it unwinds. Knowing the initial velocity (zero, since the top was at rest), the amount of rotation, and the angular acceleration allows us to calculate the final angular speed.
Rotational Motion
Rotational motion is how an object rotates or spins around a central point or an axis. Unlike linear motion, which goes in a straight line, rotational motion involves circular paths.
  • An example is the Earth spinning on its axis, or a wheel turning as a car moves.
  • Understanding rotational motion is key to analyzing behaviors of many objects, from everyday toys to complex machinery.
In the context of our exercise, the toy top undergoes rotational motion as it spins around its own center, when the string is pulled. It starts from rest, gains speed due to angular acceleration, and continues spinning at a calculated angular velocity once the string is completely unwound.

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Most popular questions from this chapter

A quarterback throws a pass that is a perfect spiral. In other words, the football does not wobble, but spins smoothly about an axis passing through each end of the ball. Suppose the ball spins at \(7.7 \mathrm{rev} / \mathrm{s} .\) In addition, the ball is thrown with a linear speed of \(19 \mathrm{~m} / \mathrm{s}\) at an angle of \(55^{\circ}\) with respect to the ground. If the ball is caught at the same height at which it left the quarterback's hand, how many revolutions has the ball made while in the air?

A basketball player is balancing a spinning basketball on the tip of his finger. The angular velocity of the ball slows down from 18.5 to \(14.1 \mathrm{rad} / \mathrm{s} .\) During the slow-down, the angular displacement is 85.1 rad. Determine the time it takes for the ball to slow down.

(a) In general, is the direction of an object's average angular velocity \(\bar{\omega}\) the same as its initial angle \(\theta_{0}\), its final angle \(\theta\), or the difference \(\theta-\theta_{0}\) between its final and initial angles? (b) The table that follows lists four pairs of initial and final angles of a wheel on a moving car. Decide which pairs give a positive average angular velocity and which give a negative average angular velocity. Provide reasons for your answers. $$ \begin{array}{|c|c|c|} \hline & \text { Initial angle } \theta-\theta_{0} & \text { Final angle } \theta \\ \hline \text { (a) } & 0.45 \mathrm{rad} & 0.75 \mathrm{rad} \\ \hline \text { (b) } & 0.94 \mathrm{rad} & 0.54 \mathrm{rad} \\ \hline \text { (c) } & 5.4 \mathrm{rad} & 4.2 \mathrm{rad} \\ \hline \text { (d) } & 3.0 \mathrm{rad} & 3.8 \mathrm{rad} \\ \hline \end{array} $$ Problem The elapsed time for each pair of angles is \(2.0 \mathrm{~s}\). Review the concept of average angular velocity in Section 8.2 and then determine the average angular velocity (magnitude and direction) for each of the four pairs of angles in the table. Check to see that the directions (positive or negative) of the angular velocities agree with the directions found in the Concept Question.

The front and rear sprockets on a bicycle have radii of 9.00 and \(5.10 \mathrm{~cm}\), respectively The angular speed of the front sprocket is \(9.40 \mathrm{rad} / \mathrm{s}\). Determine (a) the linear speed (in \(\mathrm{cm} / \mathrm{s}\) ) of the chain as it moves between the sprockets and \((\mathrm{b})\) the centripetal acceleration (in \(\mathrm{cm} / \mathrm{s}^{2}\) ) of the chain as it passes around the rear sprocket.

A propeller is rotating about an axis perpendicular to its center, as the drawing shows. The axis is parallel to the ground. An arrow is fired at the propeller, travels parallel to the axis, and passes through one of the open spaces between the propeller blades. The vertical drop of the arrow may be ignored. There is a maximum value for the angular speed \(\omega\) of the propeller beyond which the arrow cannot pass through an open space without being struck by one of the blades. (a) If the arrow is to pass through an open space, does it matter if the arrow is aimed closer to or farther away from the axis (see points \(\mathrm{A}\) and \(\mathrm{B}\) in the drawing, for example)? Explain. (b) Does the maximum value of \(\omega\) increase, decrease, or remain the same with increasing arrow speed \(v ?\) Why? (c) Does the maximum value of \(\omega\) increase, decrease, or remain the same with increasing arrow length \(L\) ? Justify your answer.
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