Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Let's solve this problem step-by-step.
### (i) Calculate the angular acceleration
1. Initial Information:
- Initial speed ([tex]\(\omega_i\)[/tex]): [tex]\(1800 \, \text{rev/min}\)[/tex]
- Final speed ([tex]\(\omega_f\)[/tex]): [tex]\(1200 \, \text{rev/min}\)[/tex]
- Time ([tex]\(t\)[/tex]): [tex]\(20 \, \text{s}\)[/tex]
2. Convert speeds from revolutions per minute (rev/min) to radians per second (rad/s):
- [tex]\(1 \, \text{rev} = 2\pi \, \text{radians}\)[/tex]
- [tex]\(1 \, \text{min} = 60 \, \text{s}\)[/tex]
Therefore,
[tex]\[ \omega_i = 1800 \times \frac{2\pi}{60} \, \text{rad/s} = 1800 \times \frac{\pi}{30} \, \text{rad/s} = 60\pi \, \text{rad/s} \][/tex]
[tex]\[ \omega_f = 1200 \times \frac{2\pi}{60} \, \text{rad/s} = 1200 \times \frac{\pi}{30} \, \text{rad/s} = 40\pi \, \text{rad/s} \][/tex]
3. Calculate the angular acceleration ([tex]\(\alpha\)[/tex]):
[tex]\[ \alpha = \frac{\omega_f - \omega_i}{t} = \frac{40\pi - 60\pi}{20} = \frac{-20\pi}{20} = -\pi \, \text{rad/s}^2 \][/tex]
Hence, the angular acceleration is: [tex]\(\alpha = -\pi \, \text{rad/s}^2\)[/tex], which approximately equals [tex]\( -3.14 \, \text{rad/s}^2 \)[/tex].
### (ii) Calculate the number of revolutions made by the motor during this time
1. Use the equation for angular displacement ([tex]\(\theta\)[/tex]) during uniformly accelerated motion:
[tex]\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \][/tex]
2. Substitute the values:
[tex]\[ \theta = 60\pi \times 20 + \frac{1}{2} \times (-\pi) \times 20^2 \][/tex]
3. Simplify the expression:
[tex]\[ \theta = 1200\pi + \frac{1}{2} \times (-\pi) \times 400 = 1200\pi - 200\pi = 1000\pi \, \text{radians} \][/tex]
4. Convert the angular displacement from radians to revolutions:
[tex]\[ 1 \, \text{rev} = 2\pi \, \text{radians} \][/tex]
[tex]\[ \text{Number of revolutions} = \frac{1000\pi}{2\pi} = 500 \, \text{revs} \][/tex]
Hence, the motor makes [tex]\(500\)[/tex] revolutions during this time.
### (iii) Calculate the additional time required to come to rest with the same rate of deceleration
1. Use the final speed ([tex]\(\omega_f\)[/tex]) for the calculation (note that here, [tex]\(\omega_f\)[/tex] for this part will be zero since the motor comes to rest):
[tex]\[ \omega_f' = 0, \, \omega_i' = 1200 \times \frac{2\pi}{60} = 40\pi \, \text{rad/s}, \, \alpha = -\pi \, \text{rad/s}^2 \][/tex]
2. Using the formula for uniform angular deceleration:
[tex]\[ \alpha = \frac{\omega_f' - \omega_i'}{t'} \][/tex]
[tex]\[ -\pi = \frac{0 - 40\pi}{t'} \][/tex]
3. Solve for [tex]\(t'\)[/tex]:
[tex]\[ -\pi t' = -40\pi \][/tex]
[tex]\[ t' = 40 \, \text{s} \][/tex]
Hence, the additional time required for the motor to come to rest is [tex]\(40 \, \text{seconds}\)[/tex].
### Summary of Answers:
(i) Angular acceleration: [tex]\(-3.1 \, \text{rad/s}^2\)[/tex].
(ii) Number of revolutions: [tex]\(500 \, \text{revs}\)[/tex].
(iii) Additional time to come to rest: [tex]\(40.0 \, \text{s}\)[/tex].
### (i) Calculate the angular acceleration
1. Initial Information:
- Initial speed ([tex]\(\omega_i\)[/tex]): [tex]\(1800 \, \text{rev/min}\)[/tex]
- Final speed ([tex]\(\omega_f\)[/tex]): [tex]\(1200 \, \text{rev/min}\)[/tex]
- Time ([tex]\(t\)[/tex]): [tex]\(20 \, \text{s}\)[/tex]
2. Convert speeds from revolutions per minute (rev/min) to radians per second (rad/s):
- [tex]\(1 \, \text{rev} = 2\pi \, \text{radians}\)[/tex]
- [tex]\(1 \, \text{min} = 60 \, \text{s}\)[/tex]
Therefore,
[tex]\[ \omega_i = 1800 \times \frac{2\pi}{60} \, \text{rad/s} = 1800 \times \frac{\pi}{30} \, \text{rad/s} = 60\pi \, \text{rad/s} \][/tex]
[tex]\[ \omega_f = 1200 \times \frac{2\pi}{60} \, \text{rad/s} = 1200 \times \frac{\pi}{30} \, \text{rad/s} = 40\pi \, \text{rad/s} \][/tex]
3. Calculate the angular acceleration ([tex]\(\alpha\)[/tex]):
[tex]\[ \alpha = \frac{\omega_f - \omega_i}{t} = \frac{40\pi - 60\pi}{20} = \frac{-20\pi}{20} = -\pi \, \text{rad/s}^2 \][/tex]
Hence, the angular acceleration is: [tex]\(\alpha = -\pi \, \text{rad/s}^2\)[/tex], which approximately equals [tex]\( -3.14 \, \text{rad/s}^2 \)[/tex].
### (ii) Calculate the number of revolutions made by the motor during this time
1. Use the equation for angular displacement ([tex]\(\theta\)[/tex]) during uniformly accelerated motion:
[tex]\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \][/tex]
2. Substitute the values:
[tex]\[ \theta = 60\pi \times 20 + \frac{1}{2} \times (-\pi) \times 20^2 \][/tex]
3. Simplify the expression:
[tex]\[ \theta = 1200\pi + \frac{1}{2} \times (-\pi) \times 400 = 1200\pi - 200\pi = 1000\pi \, \text{radians} \][/tex]
4. Convert the angular displacement from radians to revolutions:
[tex]\[ 1 \, \text{rev} = 2\pi \, \text{radians} \][/tex]
[tex]\[ \text{Number of revolutions} = \frac{1000\pi}{2\pi} = 500 \, \text{revs} \][/tex]
Hence, the motor makes [tex]\(500\)[/tex] revolutions during this time.
### (iii) Calculate the additional time required to come to rest with the same rate of deceleration
1. Use the final speed ([tex]\(\omega_f\)[/tex]) for the calculation (note that here, [tex]\(\omega_f\)[/tex] for this part will be zero since the motor comes to rest):
[tex]\[ \omega_f' = 0, \, \omega_i' = 1200 \times \frac{2\pi}{60} = 40\pi \, \text{rad/s}, \, \alpha = -\pi \, \text{rad/s}^2 \][/tex]
2. Using the formula for uniform angular deceleration:
[tex]\[ \alpha = \frac{\omega_f' - \omega_i'}{t'} \][/tex]
[tex]\[ -\pi = \frac{0 - 40\pi}{t'} \][/tex]
3. Solve for [tex]\(t'\)[/tex]:
[tex]\[ -\pi t' = -40\pi \][/tex]
[tex]\[ t' = 40 \, \text{s} \][/tex]
Hence, the additional time required for the motor to come to rest is [tex]\(40 \, \text{seconds}\)[/tex].
### Summary of Answers:
(i) Angular acceleration: [tex]\(-3.1 \, \text{rad/s}^2\)[/tex].
(ii) Number of revolutions: [tex]\(500 \, \text{revs}\)[/tex].
(iii) Additional time to come to rest: [tex]\(40.0 \, \text{s}\)[/tex].
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
