The exercise is solved based on the consideration of Torque through the moment of inertia and angular acceleration, that is,
[tex]T= I\alpha[/tex]
Where ,
I= Moment of inertia
[tex]\alpha =[/tex] Angular acceleration.
Our torque is 1.2mN, so we can find easily the Angular Acceleration.
But first we will find the Inertia for a solid cylinder given by,
[tex]I= \frac{1}{2}mR^2[/tex]
Where m is the mass of 4.8Kg and R the radius 0.071m
[tex]I= \frac{1}{2} (4.8)(0.0710)^2[/tex]
[tex]I = 0.012098Kgm^2[/tex]
Replacing in the first equation we have,
[tex]-1.2 = 0.012098\alpha\\\alpha = -99.1899rad/s^2[/tex]
A) With the acceleration found now we can know how many revolutions the body had, through the equation of rotational dynamics when time is not counted,
[tex]w_f^2-w_i^2= 2\alpha\theta[/tex]
[tex]0^2- 10300rpm(2\pi/60rpm)^2 = -2*99.1899\theta[/tex]
[tex]\theta = \frac{1163407}{198.37}[/tex]
[tex]\theta = 5864.83 revolutions.[/tex]
B) Using a similar formula of rotational dynamics we can find the time, that is,
[tex]w_f=w_i+\alpha t[/tex]
[tex]0= 10300rpm(2\pi/60rpm)-99.1899t[/tex]
[tex]t=\frac{1078.613}{99.1899t}[/tex]
[tex]t=10.87s[/tex]
Answer:
Explanation:
Revolution of Rotor N = [tex]10300rpm[/tex]
Angular speed w = [tex]\frac{2\pi N}{60}[/tex]
[tex]w = \frac{2*3.14*10300}{60}\\\\w = 1078.58rad/s[/tex]
mass of rotor m = [tex]4.8kg[/tex]
radius of rotor r = [tex]0.071m[/tex]
MOI of rotor I = [tex]\frac{mr^2}{2}[/tex]
[tex]I = \frac{4.8}{2}*(0.071)^2\\\\I = 0.012kg-m^2[/tex]
Torque on rotor T = [tex]1.2N-M[/tex]
[tex]T = I\alpha\\\\\alpha = \frac{T}{I}\\\\\alpha = \frac{1.2}{0.012}\\\\\alpha = 100[/tex]
using equation of angular motion
[tex]w = w_o - \alpha t[/tex]
where w = 0
[tex]\alpha t = w_o\\\\t = \frac{w_o}{\alpha}\\\\t = \frac{1078.58}{100}\\\\t = 10.7858sec[/tex]
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