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Chapter 9: Problem 60

Gear \(A\), with a mass of \(1.00 \mathrm{~kg}\) and a radius of \(55.0 \mathrm{~cm}\) is in contact with gear \(\mathrm{B}\), with a mass of \(0.500 \mathrm{~kg}\) and a radius of \(30.0 \mathrm{~cm} .\) The gears do not slip with respect to each other as they rotate. Gear A rotates at 120. rpm and slows to 60.0 rpm in \(3.00 \mathrm{~s}\). How many rotations does gear B undergo during this time interval?

Short Answer

Expert verified
Answer: Gear B undergoes approximately 7.87 rotations during the 3-second time interval.

Step by step solution

01

Convert angular speeds to rad/s

To convert rpm to rad/s, we can use the conversion factor: 1 rpm = \((2\pi \mathrm{~rad})/(\mathrm{60~s})\). For gear A, \(\omega_\mathrm{Ai} = 120\, \mathrm{rpm} \times \frac{2\pi \mathrm{~rad}}{\mathrm{60~s}} = 12.566\,\mathrm{rad/s}\) (initial angular speed) \(\omega_\mathrm{Af} = 60\, \mathrm{rpm} \times \frac{2\pi \mathrm{~rad}}{\mathrm{60~s}} = 6.283\, \mathrm{rad/s}\) (final angular speed)
02

Calculate the angular acceleration of gear A

We can calculate the angular acceleration (\(\alpha_\mathrm{A}\)) of gear A using the equation: \(\alpha_\mathrm{A} = \frac{\omega_\mathrm{Af} - \omega_\mathrm{Ai}}{t}\), where \(t = 3.00\, \mathrm{s}\). \(\alpha_\mathrm{A} = \frac{6.283 - 12.566}{3.00} = -2.094\, \mathrm{rad/s^2}\)
03

Find the angular acceleration of gear B

Given the radii of gears A and B, since they are in contact without slipping, their accelerations are related by: \(\alpha_\mathrm{B} = \frac{r_\mathrm{A}}{r_\mathrm{B}} \times \alpha_\mathrm{A}\) \(\alpha_\mathrm{B} = \frac{55.0}{30.0} \times (-2.094) = -3.824\, \mathrm{rad/s^2}\)
04

Calculate the final angular speed of gear B

We know the initial angular speed of gear B can be found from the gears' initial speeds: \(\omega_\mathrm{Bi} = \frac{r_\mathrm{A}}{r_\mathrm{B}} \times \omega_\mathrm{Ai}\) \(\omega_\mathrm{Bi} = \frac{55.0}{30.0} \times 12.566 = 23.045\, \mathrm{rad/s}\) Now we can calculate the final angular speed (\(\omega_\mathrm{Bf}\)) of gear B: \(\omega_\mathrm{Bf} = \omega_\mathrm{Bi} + \alpha_\mathrm{B} \times t\) \(\omega_\mathrm{Bf} = 23.045 - 3.824 \times 3.00 = 11.573\, \mathrm{rad/s}\)
05

Find the angular displacement of gear B

We can find the angular displacement (\(\theta_\mathrm{B}\)) covered by gear B during the time interval using the equation: \(\theta_\mathrm{B} = \omega_\mathrm{Bi} \times t + 0.5 \times \alpha_\mathrm{B} \times t^2\) \(\theta_\mathrm{B} = 23.045 \times 3.00 + 0.5 \times (-3.824) \times 3.00^2 = 49.455\, \mathrm{rad}\)
06

Convert angular displacement to number of rotations

To find the number of rotations that gear B underwent, divide the angular displacement by \(2\pi\). Number of rotations of gear B = \(\frac{\theta_\mathrm{B}}{2\pi}\) Number of rotations of gear B = \(\frac{49.455}{6.283} \approx 7.87\) Therefore, gear B undergoes approximately 7.87 rotations during the time interval.

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