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Chapter 5: Problem 63

A rotating flywheel slows down at a constant rate due to friction in its bearings. After 1 min, its angular velocity has diminished to 0.80 of its initial value \(\alpha\). At the end of the third minute, what is the angular velocity in terms of the initial value?

Short Answer

Expert verified
Answer: The angular velocity of the flywheel at the end of the third minute in terms of the initial value is \(\omega_f = \omega_i - 15\alpha\).

Step by step solution

01

Determine the constant rate of deceleration

Since the flywheel decelerates at a constant rate, we can use the formula for angular velocity and angular acceleration: \(\omega_f = \omega_i + \alpha t\), where \(\omega_f\) is the final angular velocity, \(\omega_i\) is the initial angular velocity, \(\alpha\) is the angular acceleration, and \(t\) is the time. We know that after one minute, the flywheel's angular velocity is 0.80 times its initial value. So, we can set up the equation: \(0.80\alpha = \alpha - \alpha t\). Now, we should solve for the acceleration, \(\alpha\).
02

Solve for angular acceleration

The equation from step 1 can be rewritten as: \(0.80\alpha = \alpha - \alpha t\). We know that \(t = 1\) min, so we can substitute the value to get: \(0.80\alpha = \alpha - \alpha(1)\). Simplifying the equation, we get \(0.20\alpha = \alpha\). Divide both sides by 0.20 to solve for \(\alpha\): \(\alpha = 5\alpha\). Now, we have found the angular deceleration rate is \(5\alpha\).
03

Calculate the angular velocity at the end of the third minute

Now that we have the angular deceleration rate, we can use the formula from step 1 again to find the angular velocity at the end of the third minute. We need to solve for the final angular velocity, \(\omega_f\), when \(t = 3\) min: \(\omega_f = \omega_i + \alpha t\). Substitute the angular deceleration rate we found in step 2 and the time: \(\omega_f = \omega_i - 5\alpha(3)\). Simplifying the equation, we get \(\omega_f = \omega_i - 15\alpha\). Thus, the angular velocity of the flywheel after the third minute in terms of the initial value will be \(\omega_f = \omega_i - 15\alpha\).

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