/*! 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 556 A car is moving at a speed of \(... [FREE SOLUTION] | 91影视

91影视

A car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The angular retardation (伪) produced by the brakes is -29.52 \(rad/s^2\). So, the correct answer is: \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\)

Step by step solution

01

Converting linear speed to angular speed

We need to convert the linear speed (v) to the initial angular speed (蠅鈧). Use the formula: \(v = 蠅 脳 r\) where v is the linear speed, 蠅 is the angular speed, and r is the radius of the wheel. Given values: \(v = 72 \frac{km}{hr}\) \(r = 0.25 m\) Now, convert the linear speed to meters per second: \(v = 72 \frac{km}{hr} 脳 \frac{1,000 m}{1 km} 脳 \frac{1 hr}{3,600 s} = 20 m/s\) Using the formula: \(蠅_{1} = \frac{v}{r}\)
02

Calculate initial angular velocity (蠅鈧)

Plug the values into the formula for 蠅鈧: \(蠅_{1} = \frac{20}{0.25} = 80 rad/s\)
03

Calculate final angular velocity (蠅鈧)

If the wheels are stopped in 20 rotations (meaning the final angular velocity 蠅鈧 = 0), we have: \(蠅_{2} = 0 rad/s\)
04

Calculate angular displacement (胃)

Since there are 20 rotations, we need to convert rotations to radians. The angular displacement (胃) is the total angle covered during these 20 rotations. Recall that 1 rotation = \(2\pi\) radians, so: \(胃 = 20 脳 2\pi = 40\pi rad\)
05

Find the angular retardation (伪)

Now, use the relationship between angular velocities, angular displacement, and angular acceleration: \(蠅_{2}^2 = 蠅_{1}^2 + 2伪胃\) Plug in 蠅鈧, 蠅鈧, and 胃 to solve for 伪: \(0 = 80^2 + 2伪(40\pi)\) Solving for 伪, we get: \(伪 = -\frac{80^2}{80\pi} = -29.52 rad/s^2\) The angular retardation (伪) produced by the brakes is -29.52 \(rad/s^2\). So, the correct answer is: \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

angular velocity
Angular velocity is a measurement of how quickly an object rotates or spins around a central point or axis. It is generally denoted by the symbol \( \omega \) (omega) and is measured in radians per second (rad/s). When a wheel is in motion, the angular velocity tells us how fast it鈥檚 turning. Understanding angular velocity is crucial for solving problems in rotational motion.

To calculate angular velocity, you can use the formula \( \omega = \frac{v}{r} \), where \( v \) is the linear speed and \( r \) is the radius of the wheel. In this exercise, by converting the car's linear speed to angular speed, we have determined the initial angular velocity \( \omega_1 \) to be 80 rad/s. This conversion is essential because it helps relate linear and rotational motion, providing further insights into how fast the object is rotating.
angular displacement
Angular displacement is a way of describing how much an object has rotated or "displaced" from its starting position in terms of angle. It is measured in radians. In physics, angular displacement is used to determine the total rotation over a period of time.

When dealing with wheels or circular motion, the total amount of rotation in radians is given by the formula \( \theta = n \times 2\pi \), where \( n \) is the number of full rotations completed. In this exercise, the car's wheels make 20 full rotations before stopping, translating to an angular displacement of \( 40\pi \) radians. This value is critical for calculating other properties of rotational motion, such as angular retardation, as it directly affects how quickly an object comes to rest.
rotational motion
Rotational motion explains how objects move in a circular path. It involves concepts like angular velocity, angular displacement, and angular acceleration. It's a crucial part of mechanics that helps understand how wheels, gears, or any rotating systems operate.

One of the key things in rotational motion is understanding how the rotational aspects relate to their linear counterparts. Much like linear motion involves speed, acceleration, and distance, rotational motion involves angular velocity, angular acceleration, and angular displacement. By breaking down rotational motion into its fundamental elements, you can decode the behavior of rolling objects such as wheels, making it possible to analyze phenomena like braking physics, as demonstrated in the exercise.
Rotational motion not only explains how objects spin but also how they stop, intertwining with other concepts like braking physics.
linear speed conversion
Linear speed conversion is the process of translating a speed given in linear units, like kilometers per hour (km/hr), into angular speed expressed in radians per second (rad/s). This conversion is necessary in many physics problems involving wheels or circular paths, where understanding the rotation in terms of angular motion is needed.

For example, in this exercise, the linear speed of the car is 72 km/hr. First, it鈥檚 converted to meters per second by using the conversion factor: \( 1 \text{ km/hr} = \frac{1,000 \text{ m}}{3,600 \text{ s}} \). Hence, 72 km/hr becomes 20 m/s.

Once you have the speed in meters per second, you can then calculate the angular speed using \( \omega = \frac{v}{r} \), where \( v = 20 \text{ m/s} \) and \( r = 0.25 \text{ m} \), resulting in an initial angular speed of 80 rad/s. This detailed conversion is vital as it bridges the gap between linear motion experienced by the car and the rotational behavior of its wheels.
braking physics
Braking physics is all about understanding how forces slow down a moving object, such as a car in this exercise. It involves aspects such as angular retardation, which describes how quickly the wheels lose their angular velocity and eventually come to a stop.

In our exercise, when brakes are applied, the wheels decelerate, reducing their angular velocity to zero. Here, the concept of angular retardation (or angular acceleration, with a negative value) comes into play. It's computed using the formula \( \alpha = \frac{\omega_2^2 - \omega_1^2}{2\theta} \). As a result, the angular retardation produced during this process is \(-29.52 \text{ rad/s}^2\).
Braking physics not only helps in understanding how braking affects rotational motion but also assists in designing efficient braking systems, ensuring that vehicles can stop safely and within a reasonable distance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

A mass \(\mathrm{m}\) is moving with a constant velocity along the line parallel to the \(\mathrm{x}\) -axis, away from the origin. Its angular momentum with respect to the origin \(\\{\mathrm{A}\\}\) Zero \\{B \\} remains constant \(\\{\mathrm{C}\\}\) goes on increasing \\{D\\} goes on decreasing

A binary star consist of two stars \(\mathrm{A}(2.2 \mathrm{Ms})\) and \(\mathrm{B}\) (mass \(11 \mathrm{Ms}\) ) where \(\mathrm{Ms}\) is the mass of sun. They are separated by distance \(\mathrm{d}\) and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of \(\operatorname{star} B\). about the centre of mass is \(\\{\mathrm{A}\\} 6\) \(\\{\mathrm{B}\\} \overline{(1 / 4)}\) \(\\{C\\} 12\) \(\\{\mathrm{D}\\}(1 / 2)\)

In a bicycle the radius of rear wheel is twice the radius of front wheel. If \(\mathrm{r}_{\mathrm{F}}\) and \(\mathrm{r}_{\mathrm{r}}\) are the radius, \(\mathrm{v}_{\mathrm{F}}\) and \(\mathrm{v}_{\mathrm{r}}\) are speed of top most points of wheel respectively then... \(\\{\mathrm{A}\\} \mathrm{v}_{\mathrm{r}}=2 \mathrm{v}_{\mathrm{F}}\) \(\\{\mathrm{B}\\} \mathrm{v}_{\mathrm{F}}=2 \mathrm{v}_{\mathrm{r}} \quad\\{\mathrm{C}\\} \mathrm{v}_{\mathrm{F}}=\mathrm{v}_{\mathrm{r}}\) \(\\{\mathrm{D}\\} \mathrm{v}_{\mathrm{F}}>\mathrm{v}_{\mathrm{r}}\)

Two bodies of mass \(1 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) have position vector \((\mathrm{i} \wedge+2 \mathrm{j} \wedge+\mathrm{k} \wedge)\) and \((-3 \mathrm{i}-2 \mathrm{j}+\mathrm{k})\) respectively the centre of mass of this system has a position vector ..... \(\\{\mathrm{A}\\}-2 \mathrm{i} \wedge+2 \mathrm{k} \wedge\) \(\\{B\\}-2 i \wedge-j \wedge+k \wedge\) \(\\{C\\} 2 i \wedge-j \wedge-k \wedge\) \(\\{\mathrm{D}\\}-\mathrm{i} \wedge+\mathrm{j} \wedge+\mathrm{k} \wedge\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.