/*! 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 27 A wheel rotating at a speed of \... [FREE SOLUTION] | 91Ó°ÊÓ

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A wheel rotating at a speed of \(600 \mathrm{rpm}\) (revolutions per minute) about its axis is brought to rest by applying a constant torque for 10 seconds. Find the angular deceleration and the angular velocity 5 seconds after the application of the torque.

Short Answer

Expert verified
Angular deceleration is \(-2\pi\) rad/s² and angular velocity after 5 seconds is \(10\pi\) rad/s.

Step by step solution

01

Convert Initial Angular Velocity to Radians Per Second

The problem gives the initial angular velocity as 600 rpm. To convert this to radians per second, we use the conversion factor: 1 revolution = \(2\pi\) radians and 1 minute = 60 seconds. Thus, the initial angular velocity \( \omega_0 \) is: \[ \omega_0 = 600 \times \frac{2\pi}{60} = 20\pi \text{ rad/s}. \]
02

Determine Angular Deceleration

The wheel is brought to rest in 10 seconds. We use the formula for angular deceleration \( \alpha \), given by \( \omega = \omega_0 + \alpha t \). Here, the final angular velocity \(\omega\) is 0, \(t\) is 10 seconds. Therefore, \[ 0 = 20\pi + (\alpha)(10). \] Solving for \(\alpha\), we get: \[ \alpha = -2\pi \text{ rad/s}^2. \]
03

Compute Angular Velocity After 5 Seconds

Using the formula \( \omega = \omega_0 + \alpha t \), where \(\alpha = -2\pi\) rad/s² and \(t = 5\) seconds, we find the angular velocity after 5 seconds: \[ \omega = 20\pi + (-2\pi)(5) = 20\pi - 10\pi = 10\pi \text{ rad/s}. \]

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Key Concepts

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

Understanding Angular Velocity
Angular velocity is the rate of change of angular displacement and is a fundamental concept in the study of rotational motion. It tells us how rapidly something is spinning around a fixed axis. Angular velocity is often expressed in units of radians per second (rad/s).
  • One revolution is a complete turn around a circle, which is equal to an angular displacement of 360 degrees or \(2\pi\) radians.
  • In the context of our problem, we start with a wheel rotating at 600 revolutions per minute (rpm).
  • This initial value of angular velocity needs to be converted to radians per second for better integration into physics equations.
Converting from rpm to rad/s is crucial for applying standard physics formulas. For example, if a wheel spins at 600 rpm, each revolution taking place is captured by multiplying the revolutions by \(2\pi\) for conversion from revolutions to radians. Then, because there are 60 seconds in one minute, you divide by 60 to get the rate per second. Thus, the formula \(\omega_0 = 600 \times \frac{2\pi}{60} = 20\pi\) rad/s is derived, showing that the wheel spins at \(20\pi\) radians per second initially.
Defining Angular Deceleration
Angular deceleration is the rate at which an object slows down its rotation. It is a measure of how much the angular velocity decreases over time. This is opposite to angular acceleration, which increases angular velocity. In our exercise, the wheel was brought to a stop over a span of 10 seconds by applying a constant negative torque.
  • The formula for angular velocity \( \omega = \omega_0 + \alpha t \) is used, where \(\omega_0\) is initial angular velocity, \(\alpha\) is angular acceleration (negative in deceleration), and \(t\) is time.
  • By setting the final angular velocity \( \omega \) to zero, because the wheel stops, we can solve for \( \alpha \).
  • In this case, plugging known values results in \( 0 = 20\pi + (\alpha)(10) \).
  • Solving for \(\alpha\), the calculation comes out as \(\alpha = -2\pi\) rad/s², signifying a decrease in speed at that rate each second.
Understanding angular deceleration is important for interpreting how quickly something can be brought to a halt. This is particularly useful in contexts like brakes in a vehicle or machinery slow-down mechanisms, where a rapid yet controlled reduction of speed is necessary.
Conversion Of Units in Rotational Motion
In physics, unit conversion is vital for solving problems accurately. Specifically in rotational dynamics, conversions like revolutions per minute to radians per second are frequent. These conversions harmonize our values with the standard units used in equations, enabling precise calculations.
  • A common conversion factor is \(1 \text{ revolution} = 2\pi \text{ radians}\) and 1 minute = 60 seconds.
  • Using these conversions, you can translate between systems, aiding in using consistent units such as converting \(\text{rpm to rad/s}\).
  • For example, converting \(600 \text{ rpm}\) to radians per second using \(\omega_0 = 600 \times \frac{2\pi}{60}\) simplifies to \(20\pi \) rad/s.
Through such conversions, one integrates rotational motion variables with general physics laws seamlessly, maintaining accuracy in calculations.
Converting units properly is crucial for understanding because it provides a common basis for comparing values and performing operations within different parts of a problem.

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Most popular questions from this chapter

A cylinder rotating at an angular speed of \(50 \mathrm{rev} / \mathrm{s}\) is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed?

Because of the friction between the water in oceans with the earth's surface, the rotational kinetic energy of the earth is continuously decreasing. If the earth's angular speed decreases by \(0-0016 \mathrm{rad} /\) day in 100 years, find the average torque of the friction on the earth. Radius of the earth is \(6400 \mathrm{~km}\) and its mass is \(6 \cdot 0 \times 10^{24} \mathrm{~kg}\).

A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella in his hand. The axis of the umbrella coincides with that of the platform. The moment of inertia of "the platform plus the boy system" is \(3 \cdot 0 \times 10^{-3} \mathrm{~kg}-\mathrm{m}^{2}\) and that of the umbrella is \(2 \cdot 0 \times 10^{-3} \mathrm{~kg}-\mathrm{m}^{2}\). The boy starts spinning the umbrella about the axis at an angular speed of \(2 \cdot 0\) rev/s with respect to himself. Find the angular velocity imparted to the platform.

A kid of mass \(M\) stands at the edge of a platform of radius \(R\) which can be freely rotated about its axis. The moment of inertia of the platform is \(I\). The system is at rest when a friend throws a ball of mass \(m\) and the kid catches it. If the velocity of the ball is \(v\) horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.

A metre stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end at the floor does not slip, find the angular speed of the rod when it hits the floor.

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