/*! 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 13 ssm The drawing shows a device t... [FREE SOLUTION] | 91Ó°ÊÓ

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ssm 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 \(\theta=0.240 \mathrm{rad}\). From these data, determine the speed of the bullet.

Short Answer

Expert verified
The speed of the bullet is approximately 336.48 m/s.

Step by step solution

01

Understanding the Problem

We need to find the speed of a bullet using a device with two rotating disks. The disks are separated by a distance of \(d=0.850 \text{ m}\) and rotate at \(95.0 \text{ rad/s}\). The angular displacement \(\theta\) between the holes made by the bullet on the disks is \(0.240 \text{ rad}\).
02

Time Taken to Cover the Distance

The time \(t\) it takes for the bullet to travel from the first to the second disk can be determined from the angular displacement \(\theta\) divided by the angular velocity \(\omega\): \[ t = \frac{\theta}{\omega}. \] Plugging in the values, we have \( \omega = 95.0 \text{ rad/s} \) and \( \theta = 0.240 \text{ rad} \), thus: \[ t = \frac{0.240}{95.0} \approx 0.002526 \text{ s}. \]
03

Calculate the Speed of the Bullet

The speed \(v\) of the bullet is the distance \(d\) divided by the time \(t\) it takes to travel that distance. Using the distance \(d = 0.850 \text{ m}\) and the time found in the previous step, we calculate: \[ v = \frac{d}{t} = \frac{0.850}{0.002526} \approx 336.48 \text{ m/s}. \]
04

Conclusion

With the given values and using the derived time, the speed of the bullet is found to be approximately \(336.48 \text{ m/s}\).

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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 an object moves on a circular path. In the context of the bullet speed measurement, it's the angle at which the bullet holes appear to have shifted as the bullet passes through the two rotating disks. This angular measurement, denoted as \(\theta\), is crucial for determining the bullet's speed because it helps us understand how far along the rotation the bullet has moved during its travel from one disk to the next.

Think of angular displacement as the arc of a circle that gets traced out between two points rather than a straight line. In this problem, \(\theta = 0.240 \, \text{rad}\) tells us the angle between the two bullet marks due to the disk's rotation. Understanding this concept helps to relate linear motion (bullet's path) with rotational motion (disk's spinning).
Angular Velocity
Angular velocity, represented by \(\omega\), is the rate at which an object rotates or spins around a central point. In our case, it describes how fast the disks rotate. Angular velocity is typically measured in radians per second (\(\text{rad/s}\)). For the bullet speed measurement device, \(\omega = 95.0 \, \text{rad/s}\), indicating the speed at which the disks are spinning when the bullet passes through.

Understanding angular velocity is key for calculating the bullet's speed, because it allows us to find out how much time the disks took to rotate over the angle \(\theta\), while the bullet moved between them. This relay of information from rotational to linear is fundamental to solving the problem, making angular velocity an essential component.
Rotational Motion
Rotational motion is the movement of an object around a central axis or point. In this scenario, the disks are undergoing rotational motion. This motion is characterized by parameters such as angular velocity and angular displacement, which describe how fast and how far the disk rotates.

To solve the bullet speed problem, we leverage the rotational motion of the disks to determine the time elapsed between the bullet passing through the two disks. This motion helps bridge the gap between the rotational aspect (disks spinning) and the bullet's linear motion (straight-line path through the disks). By analyzing the rotational motion, we can convert angular information into a parameter needed for linear speed calculations, thereby linking the bullet's passage with the rotating disks.
Speed Calculation
Speed calculation involves determining how fast something is moving along its path. In this exercise, we are tasked to find the speed of the bullet using the rotational data from the disks. The key to calculation is using the time \(t\) it takes for the bullet to travel between the disks, which we've deduced through angular displacement and angular velocity.

The straightforward formula to find speed is \(v = \frac{d}{t}\), where \(d\) is the distance between the disks, and \(t\) is the time calculated from the rotational motion data. With \(d = 0.850 \, \text{m}\) and \(t = 0.002526 \, \text{s}\), the bullet's speed is calculated to be approximately \(336.48 \, \text{m/s}\). This speed calculation seamlessly integrates the rotational and linear motion concepts to give us the bullet’s actual speed as it travels from one disk to the next.

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Most popular questions from this chapter

A gymnast is performing a floor routine. In a tumbling run she spins through the air, increasing her angular velocity from \(3.00\) to \(5.00\) rev \(/ \mathrm{s}\) while rotating through one-half of a revolution. How much time does this maneuver take?

Two Formula One racing cars are negotiating a circular turn, and they have the same centripetal acceleration. However, the path of car A has a radius of \(48 \mathrm{~m}\), while that of car \(\mathrm{B}\) is \(36 \mathrm{~m}\). Determine the ratio of the angular speed of car \(\mathrm{A}\) to that of car \(\mathrm{B}\).

An electric circular saw is designed to reach its final angular speed, starting from rest, in \(1.50 \mathrm{~s}\). Its average angular acceleration is \(328 \mathrm{rad} / \mathrm{s}^{2}\). Obtain its final angular speed.

The earth orbits the sun once a year \(\left(3.16 \times 10^{7} \mathrm{~s}\right)\) in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{~m} .\) With respect to the sun, determine (a) the angular speed of the earth, (b) the tangential speed of the earth, and (c) the magnitude and direction of the earth's centripetal acceleration.

Refer to Interactive Solution \(\underline{8.49}\) at in preparation for this problem. A car is traveling with a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. The wheels have a radius of \(0.300 \mathrm{~m} .\) If the car speeds up with a linear acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{~s},\) find the angular displacement of each wheel during this period.

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