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Chapter 9: Problem 37

You are holding the axle of a bicycle wheel with radius \(35.0 \mathrm{~cm}\) and mass \(1.00 \mathrm{~kg}\). You get the wheel spinning at a rate of 75.0 rpm and then stop it by pressing the tire against the pavement. You notice that it takes \(1.20 \mathrm{~s}\) for the wheel to come to a complete stop. What is the angular acceleration of the wheel?

Short Answer

Expert verified
Answer: The angular acceleration of the bicycle wheel is approximately -6.545 rad/s^2.

Step by step solution

01

Convert given units

Before we proceed with the computation, we need to convert the given rpm (revolutions per minute) to rad/s (radians per second). 1 revolution = \(2\pi\) radians 1 minute = 60 seconds Initial angular velocity, \(\omega_0 = 75.0 \,\text{rpm} \times \frac{2\pi \,\text{rad}}{\text{rev}} \times \frac{1\,\text{min}}{60\,\text{s}}\)
02

Calculate the initial angular velocity in radians per second

Now, compute the value of the initial angular velocity \(\omega_0 = 75.0 \times \frac{2\pi}{60} \approx 7.854 \,\text{rad/s}\)
03

Use kinematic equation for rotational motion

We will use the kinematic equation, which relates final angular velocity (\(\omega\)), initial angular velocity (\(\omega_0\)), angular acceleration (\(\alpha\)), and time (\(t\)): \(\omega = \omega_0 + \alpha t\) Since the wheel comes to a complete stop, the final angular velocity (\(\omega\)) will be 0. We can now plug in the values and solve for the angular acceleration (\(\alpha\)). \(0 = 7.854 + \alpha (1.2)\)
04

Solve for the angular acceleration

Rearrange the equation and solve for \(\alpha\): \(\alpha = \frac{-7.854}{1.2} \approx -6.545 \,\text{rad/s}^2\) The angular acceleration of the wheel is approximately \(-6.545\,\text{rad/s}^2\). The negative sign indicates that the acceleration is acting in the opposite direction of the initial angular velocity, which makes sense since the wheel is slowing down.

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