Answer :
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].
