/*! 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 15 The drawing shows a device that ... [FREE SOLUTION] | 91Ó°ÊÓ

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The drawing shows a device that can be used to measure the speed of a bullet. The device consists of two rotating disks, separated by a distance of \(d=0.850 \mathrm{m},\) and rotating with an angular speed of \(95.0 \mathrm{rad} / \mathrm{s} .\) The bullet first passes through the left disk and then through the right disk. It is found that the angular displacement between the two bullet holes is \(\theta=0.240\) rad. From these data, determine the speed of the bullet.

Short Answer

Expert verified
The bullet's speed is approximately 336 m/s.

Step by step solution

01

Understand the Relationship Between Angular and Linear Quantities

The device uses two disks rotating with constant angular velocity. The bullet creates two holes at an angular displacement of \( \theta = 0.240 \) rad, which allows us to relate the bullet's linear speed with the angular speed of the disks, \( \omega = 95.0 \) rad/s, via the formula for angular displacement \( \theta = \omega \Delta t \) where \( \Delta t \) is the time the bullet takes to travel the distance \( d = 0.850 \) m.
02

Calculate the Time Interval for the Bullet's Travel

Rearrange the angular displacement formula to solve for the time interval, \( \Delta t \): \[ \Delta t = \frac{\theta}{\omega} = \frac{0.240 \text{ rad}}{95.0 \text{ rad/s}}. \]Calculate \( \Delta t \) to determine how long it took for the bullet to travel between the two disks.
03

Solve for the Bullet's Speed

With the time interval \( \Delta t \) from Step 2, use the linear speed formula \( v = \frac{d}{\Delta t} \) where \( d = 0.850 \) m is the distance between the disks. Substitute the value of \( \Delta t \) to find the bullet's speed \( v \).

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Key Concepts

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

Angular Displacement
Angular displacement refers to the angle through which a point or line has been rotated in a specified sense about a specific axis. In the case of our rotating disks, the angular displacement is the measure of how far the bullet has turned on the circle path between the two disks. Here, we use the symbol \( \theta \) to denote angular displacement. A key point to remember is that angular displacement is usually measured in radians.

In our exercise, the bullet causes an angular displacement of \( 0.240 \) rad upon passing through the two disks. Think of angular displacement as the angular equivalent of linear distance. It tells us how far around the circular path the bullet has moved, providing insight into the relationship between the bullet's straight-line speed and the rotational motion of the disks.

Angular displacement is an important concept because it helps relate linear quantities, such as speed, to angular quantities, such as rotational speed. By understanding \( \theta \), you can connect the dots between how fast something travels in a straight line and how it affects circular motion mechanics.
Angular Velocity
Angular velocity is the rate of change of angular displacement with time. It tells us how quickly an object rotates or revolves around an axis. It's measured in radians per second (rad/s). In this exercise, the disks are rotating at an angular velocity of \( 95.0 \) rad/s, which is constant.

The formula \( \omega = \frac{\theta}{\Delta t} \) connects angular displacement \( \theta \) and time \( \Delta t \). Here, \( \omega \) represents angular velocity. Given that the disks rotate at \( 95.0 \) rad/s and the bullet pierces through with an angular displacement of \( 0.240 \) rad, the formula also allows us to calculate how long it took for the bullet to travel this path. Each rotation helps us better understand the relationship between time, motion, and speed.

Understanding angular velocity is crucial when evaluating any scenarios involving rotational motion. It forms the bridge between rotational movement (like the spinning disks) and linear movement (such as the speed of the bullet passing through).
Linear Velocity
Linear velocity refers to the rate at which an object moves along a linear path. It's directly related to how fast something is moving from one point to another. This concept becomes especially relevant when translating angular motion into linear terms. In our case, it involves the bullet's speed as it travels through the disks.

In essence, linear velocity \( v \) can be found by dividing the distance \( d \) by the time it took to travel this distance \( \Delta t \). Using the formula \( v = \frac{d}{\Delta t} \), we can compute the bullet's speed. The challenge lies in translating the known angular quantities into this straightforward linear motion concept.

By applying this formula, you can determine the linear speed of the bullet, linking the rotating disk mechanics to the bullet's straightforward travel path. Linear and angular velocities are interrelated, with angular motion often serving as a catalyst for unveiling linear properties.
Rotational Motion
Rotational motion is the circular movement of an object around a center or axis of rotation. In this exercise, the rotating disks exhibit rotational motion, which is fundamental to understanding the speed measurement of the bullet.

Rotational motion encompasses several key concepts, such as angular displacement, angular velocity, and other rotational dynamics. Each rotating disk can be thought of as traveling around a rotational path, with every point on the disk executing circular motion. The study of this motion helps explain how the bullet's speed can be calculated from its interaction with the disks.

Given that the disks rotate at a constant angular speed, their rotational motion remains consistent and predictable. Observing this consistency allows for an accurate measurement of other motion-related parameters, such as the bullet's speed. Therefore, understanding the principles of rotational motion opens the door to exploring the fascinating interplay between circular and linear dynamics, significantly impacting measurements and calculations in varied engineering and physics applications.

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Most popular questions from this chapter

An automobile, starting from rest, has a linear acceleration to the right whose magnitude is 0.800 m/s2 (see the figure). During the next 20.0 s, the tires roll and do not slip. The radius of each wheel is 0.330 m. At the end of this time, what is the angle through which each wheel has rotated?

A motorcycle, which has an initial linear speed of \(6.6 \mathrm{m} / \mathrm{s}\), decelerates to a speed of \(2.1 \mathrm{m} / \mathrm{s}\) in \(5.0 \mathrm{s}\). Each wheel has a radius of \(0.65 \mathrm{m}\) and is rotating in a counterclockwise (positive) direction. What are (a) the constant angular acceleration (in \(\mathrm{rad} / \mathrm{s}^{2}\) ) and (b) the angular displacement (in rad) of each wheel?

The drive propeller of a ship starts from rest and accelerates at \(2.90 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2}\) for \(2.10 \times 10^{3} \mathrm{s} .\) For the next \(1.40 \times 10^{3} \mathrm{s}\) the propeller rotates at a constant angular speed. Then it decelerates at \(2.30 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2}\) until it slows (without reversing direction) to an angular speed of \(4.00 \mathrm{rad} / \mathrm{s}\). Find the total angular displacement of the propeller.

The wheels of a bicycle have an angular velocity of \(+20.0 \mathrm{rad} / \mathrm{s}\). Then, the brakes are applied. In coming to rest, each wheel makes an angular displacement of +15.92 revolutions. (a) How much time does it take for the bike to come to rest? (b) What is the angular acceleration (in \(\mathrm{rad} / \mathrm{s}^{2}\) ) of each wheel?

Suppose you are driving a car in a counterclockwise direction on a circular road whose radius is \(r=390 \mathrm{m}\) (see the figure). You look at the speedometer and it reads a steady \(32 \mathrm{m} / \mathrm{s}\) (about \(72 \mathrm{mi} / \mathrm{h}\) ). Concepts: (i) Does an object traveling at a constant tangential speed (for example, \(\left.v_{\mathrm{T}}=32 \mathrm{m} / \mathrm{s}\right)\) along a circular path have an acceleration? (ii) Is there a tangential acceleration \(\overrightarrow{\mathbf{a}}_{\mathrm{T}}\) when the angular speed of an object changes (e.g., when the car's angular speed decreases to \(4.9 \times 10^{-2} \mathrm{rad} / \mathrm{s}\) )? Calculations: (a) What is the angular speed of the car? (b) Determine the acceleration (magnitude and direction) of the car. (c) To avoid a rear-end collision with the vehicle ahead, you apply the brakes and reduce your angular speed to \(4.9 \times 10^{-2} \mathrm{rad} / \mathrm{s}\) in a time of 4.0 s. What is the tangential acceleration (magnitude and direction) of the car?

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