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Chapter 8: Problem 26

A gyroscope flywheel of radius \(2.83 \mathrm{~cm}\) is accelerated from rest at \(14.2 \mathrm{rad} / \mathrm{s}^{2}\) until its angular speed is 2760 rev/min. \((a)\) What is the tangential acceleration of a point on the rim of the flywheel? ( \(b\) ) What is the radial acceleration of this point when the flywheel is spinning at full speed? ( \(c\) ) Through what distance does a point on the rim move during the acceleration?

Short Answer

Expert verified
The tangential acceleration is approximately \(0.402 \, \mathrm{m/s^{2}}\), the radial acceleration at full speed is approximately \(291198.69 \, \mathrm{m/s^{2}}\), and the distance traveled by the point on the rim during the acceleration is approximately \(626.88 \, \mathrm{m}}.

Step by step solution

01

Find the Tangential Acceleration

Use the formula for tangential acceleration \(a_t\) to find its value. The formula is: \(a_t = r \cdot \alpha\), where \(r\) is the radius and \(\alpha\) is the angular acceleration. Take \(r = 2.83 \, \mathrm{cm} = 0.0283 \, \mathrm{m}\) (as we convert it into SI Units), and \(\alpha = 14.2 \, \mathrm{rad/s^{2}}\). Substitute these values into the formula to get \(a_t\).
02

Calculate the Radial Acceleration

After finding the tangential acceleration, you now need to find the radial acceleration. The formula for radial acceleration \(a_r\) is: \(a_r = r \cdot \omega^2\), where \(\omega\) is the angular speed. But the angular speed is not directly given in our problem, it is given in revolutions per minute, so first convert it into radians per second using the relation: \(\omega (\mathrm{rad/s}) = 2760 \, \mathrm{rev/min} \times \frac{2 \pi \, \mathrm{rad}}{1 \, \mathrm{rev}} \times \frac{1 \, \mathrm{min}}{60 \, \mathrm{s}}\). Once you have \(\omega\) in rad/s, substitute \(r\) and \(\omega\) into our formula to get \(a_r\).
03

Determine the Distance Traveled

To find the distance, \(s\), traveled by a point on the rim of the wheel, use the following formula: \(s = r \cdot \theta\), where \(\theta\) is the angular displacement. But \(\theta\) is not given directly; instead, you must use the formula for angular displacement when angular acceleration and final angular velocity are known: \(\theta = \frac{\omega_{final}^{2}}{2\alpha}\). Substitute the values of \(\omega_{final}\) (already converted in Step 2) and \(\alpha\) into the formula to get \(\theta\). Then substitute \(r\) and the derived \(\theta\) into \(s = r \cdot \theta\) to get the required distance, \(s\). Make sure all units are in the SI system.

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Most popular questions from this chapter

If an airplane propeller of radius \(5.0 \mathrm{ft}(=1.5 \mathrm{~m})\) rotates at 2000 rev/min and the airplane is propelled at a ground speed of \(300 \mathrm{mi} / \mathrm{h}(=480 \mathrm{~km} / \mathrm{h})\), what is the speed of a point on the tip of the propeller, as seen by \((a)\) the pilot and \((b)\) an observer on the ground? Assume that the plane's velocity is parallel to the propeller's axis of rotation.

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