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An electric motor rotating a workshop grinding wheel at a rate of \(1.00 \times 10^{2}\) rev/min is switched off. Assume the wheel has a constant negative angular acceleration of magnitude \(2.00 \mathrm{rad} / \mathrm{s}^{2}\). (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in part (a)?

Short Answer

Expert verified
a) The time it takes for the grinding wheel to stop is 5.235 seconds. b) The wheel turned through approximately 27.63 radians before it stopped.

Step by step solution

01

Convert Revolutions per Minute to Radians per Second

Before starting calculations, convert the given angular speed from revolutions per minute to radians per second, which is the standard unit for angular speed. Use the conversion factor \(1 \: rev = 2\pi \: rad\) and \(1 \: min = 60 \: s\): \(1.00 \times 10^{2} \: rev/min \times \frac{2\pi \: rad}{1 \: rev} \times \frac{1 \: min}{60 \: s} = 10.47 \: rad/s.\)
02

Calculate the Time for the Motor to Stop

Now use the formula for stopping time with constant angular acceleration. The formula is:\[\Delta t = \frac{\omega_f - \omega_i}{\alpha},\]where \(\omega_f\) is the final angular velocity, \(\omega_i\) is the initial angular velocity, and \(\alpha\) is the angular acceleration. For this problem, \(\omega_f = 0 \: rad/s\) (because the wheel stops), \(\omega_i = 10.47 \: rad/s\) and \(\alpha = -2.00 \: rad/s^2\). Substituting these values:\[\Delta t = \frac{0 - 10.47}{-2} = 5.235 \: s.\] So, the wheel stops after approximately 5.235 seconds.
03

Calculate the Total Number of Radians that the Wheel Turned

Now calculate the total angle in radians that the wheel turned during this time period using the angular motion equation: \[\Delta \theta = \omega_i\Delta t + 0.5\alpha(\Delta t)^2.\]Here, \(\Delta \theta\) is the displacement in radians, \(\omega_i = 10.47 \: rad/s\) is the initial angular velocity, \(\Delta t = 5.235 \: s\) is the time, and \(\alpha = -2.00 \: rad/s^2\) is the angular acceleration. Plugging in these values:\[\Delta \theta = (10.47)(5.235) + 0.5(-2)(5.235)^2= \] \[27.63 \: rad.\] So, the wheel turned approximately 27.63 radians before it stopped.

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Key Concepts

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

Angular Motion Equations
Understanding the dynamics of rotational motion involves angular motion equations, which are analogous to the equations for linear motion but applied to rotational movement. Key equations describe the relationships between angular displacement (\( \theta \)), angular velocity (\( \omega \)), and angular acceleration (\( \alpha \)). One core equation is \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \), where \( \theta \) is the angular displacement, \( \omega_i \) is the initial angular velocity, and \( t \) is the time interval. Another important equation is \( \omega_f = \omega_i + \alpha t \), which relates the final angular velocity (\( \omega_f \)) to the initial angular velocity and the angular acceleration over time. These equations are powerful tools that can predict the future state of a rotating body using known initial conditions and forces acting upon it.

In our electric motor example, we used \( \Delta \theta = \omega_i\Delta t + 0.5\alpha(\Delta t)^2 \) to calculate the displacement in radians. This equation is a direct translation of the kinematic equation \( s = ut + \frac{1}{2}at^2 \) used for linear motion, showcasing the parallel between linear and angular motion concepts.
Converting Angular Speed
When working with angular motion, it's essential to use consistent units. Angular speed can be specified in units like revolutions per minute (rpm) or the standard unit of radians per second (rad/s). To convert from rpm to rad/s, use the conversion factors \( 1 \: rev = 2\pi \: rad \) and \( 1 \: min = 60 \: s \). The conversion formula looks like this: \( Angular\:speed\:(rad/s) = Angular\:speed\:(rpm) \times \frac{2\pi}{1} \times \frac{1}{60} \).

The clear understanding of these units is critical because equations of angular motions require consistency for accurate calculations. In our exercise, this conversion allowed us to express the initial angular velocity of the grinding wheel in the appropriate units before applying the motion equations.
Calculating Stopping Time
Calculating stopping time for a rotating object with constant angular acceleration is similar to finding the time it takes for an object to come to rest under constant linear deceleration. The formula used is \( \Delta t = \frac{\omega_f - \omega_i}{\alpha} \), where \( \Delta t \) is the stopping time, \( \omega_f \) and \( \omega_i \) are the final and initial angular velocities, respectively, and \( \alpha \) is the angular acceleration.

For an object coming to a standstill, \( \omega_f \) is zero, simplifying the formula to \( \Delta t = -\frac{\omega_i}{\alpha} \). It is important to consider the direction of acceleration; if it is opposite to the velocity, it should be taken as negative. In our problem, the angular acceleration was negative because it indicated a deceleration of the rotating wheel.
Displacement in Radians
Displacement in radians measures the angle through which a point or line has been rotated in a specified sense about a specified axis. It is a way to quantify the change in position of a rotating object and is particularly crucial in the study of circular motion. When calculating the displacement in radians, we look at the arc length on the path traveled by a point on the outer edge of the rotating object and the radius of its path. The formula for angular displacement is \( \Delta \theta = \omega_i\Delta t + 0.5\alpha(\Delta t)^2 \), where each symbol represents the same quantities as they do in the angular motion equations.

In the context of our example, after calculating the stopping time, we used the angular motion equation to obtain the total angular displacement. This resulted in the grinding wheel's total rotation before it ceased movement, expressed in radians, which is the standard unit of angular displacement and is crucial when analyzing rotational motion.

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Most popular questions from this chapter

ecp Because of Earth's rotation about its axis, a point on the equator has a centripetal acceleration of \(0.0340 \mathrm{~m} / \mathrm{s}^{2}\) whereas a point at the poles has no centripetal acceleration. (a) Show that, at the equator, the gravitational force on an object (the object's true weight) must exceed the object's apparent weight. (b) What are the apparent weights of a \(75.0\) -kg person at the equator and at the poles? (Assume Earth is a uniform sphere and take \(g=\) \(\left.9.800 \mathrm{~m} / \mathrm{s}^{2} .\right)\)

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