Respuesta :
Answer:
a)[tex]t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec[/tex]
b)[tex]\theta_1=\frac{w_1^2}{2\alpha}rad[/tex]
c)[tex]t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec[/tex]
Explanation:
1) Basic concepts
Angular displacement is defined as the angle changed by an object. The units are rad/s.
Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:
[tex]w=\frac{\Delat \theta}{\Delta t}[/tex]
Angular acceleration is the rate of change of the angular velocity respect to the time
[tex]\alpha=\frac{dw}{dt}[/tex]
2) Part a
We can define some notation
[tex]w_o=0\frac{rad}{s}[/tex],represent the initial angular velocity of the wheel
[tex]w_1=?\frac{rad}{s}[/tex], represent the final angular velocity of the wheel
[tex]\alpha[/tex], represent the angular acceleration of the flywheel
[tex]t_1[/tex] time taken in order to reach the final angular velocity
So we can apply this formula from kinematics:
[tex]w_1=w_o +\alpha t_1[/tex]
And solving for t1 we got:
[tex]t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec[/tex]
3) Part b
We can use other formula from kinematics in order to find the angular displacement, on this case the following:
[tex]\Delta \theta=wt+\frac{1}{2}\alpha t^2[/tex]
Replacing the values for our case we got:
[tex]\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2[/tex]
And we can replace [tex]t_1[/tex]from the result for part a, like this:
[tex]\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2[/tex]
Since [tex]\theta_o=0[/tex] and [tex]w_o=0[/tex] then we have:
[tex]\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}[/tex]
And simplifying:
[tex]\theta_1=\frac{w_1^2}{2\alpha}rad[/tex]
4) Part c
For this case we can assume that the angular acceleration in order to stop applied on the wheel is [tex]\alpha_1 =-5\alpha \frac{rad}{s}[/tex]
We have an initial angular velocity [tex]w_1[/tex], and since at the end stops we have that [tex]w_2 =0[/tex]
Assuming that [tex]t_2[/tex] represent the time in order to stop the wheel, we cna use the following formula
[tex]w_2 =w_1 +\alpha_1 t_2[/tex]
Since [tex]w_2=0[/tex] if we solve for [tex]t_2[/tex] we got
[tex]t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}[/tex]
And from part a) we can see that [tex]w_1=\alpha t_1[/tex], and replacing into the last equation we got:
[tex]t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec[/tex]
