Respuesta :
Answer:
(A)Angular acceleration, [tex]\alpha =446.01\ rad/s^2[/tex]
(B) Number of revolution is 8.87.
Explanation:
Given ,time taken to reach 2300 rpm speed from 0 is ,t = 0.54 s.
2300 rpm = 240.85 rad/s
(A) Applying equation of motion to find the drill's angular acceleration
[tex]\omega=\omega_o+\alpha\times t[/tex]
[tex]\omega_o=0[/tex]
[tex]\alpha =\dfrac{\omega}{t}[/tex]
[tex]\alpha =\dfrac{240.85}{0.54}[/tex]
[tex]\alpha =446.01\ rad/s^2[/tex]
(B) Let there are [tex]\theta[/tex] number of revolutions it turn during this first 0.50 s. It can be calculated using second equation of motion as :
[tex]\theta=\omega_ot+\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta=\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta=\dfrac{1}{2}\times 446.01\times (0.5)^2[/tex]
[tex]\theta=55.75\ rad[/tex]
Radian can be converted to revolution as :
[tex]55.75\ rad=8.87\ revolution[/tex]
Therefore, this is the required solution.
The angular acceleration of the drill is 446.1 rad/s².
The number of revolutions of the drill in 0.5 second is 19.2 revolutions.
The given parameters;
- speed of the drill, ω = 2300 rpm
- time of motion, t = 0.54 s
The angular speed in radian per second is calculated as follows;[tex]\omega _f = 2300 \ \frac{rev }{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} = 240.89 \ rad/s[/tex]
The angular acceleration of the drill is calculated as follows;
[tex]\omega _f = \omega _i + \alpha t\\\\\alpha = \frac{\omega _f - \omega _i }{t} \\\\ \alpha = \frac{240.89 -0 }{0.54} = 446.1 \ rad/s^2[/tex]
The number of revolutions of the drill in 0.5 second is calculated as follows;
[tex]N = 2300 \times \frac{rev}{\min} \times \frac{1 \min}{60 \ s} \times 0.5 \ s = 19.2 \ rev[/tex]
Learn more here:https://brainly.com/question/15661545
