Answer:
11 radians or 1.7507 revolutions
9 rad/s
7.27565 revolutions
Explanation:
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
t = Time taken
Equation of rotational motion
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\times 2+\frac{1}{2}\times 3.5\times 2^2\\\Rightarrow \theta=11\ rad=\frac{11}{2\pi}=1.7507\ rev[/tex]
Angle the wheel rotates in the given time is 11 radians or 1.7507 revolutions
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=2+3.5\times 2\\\Rightarrow \omega_f=9\ rad/s[/tex]
The angular speed of the wheel at the given time is 9 rad/s
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{(9\times 2)^2-2^2}{2\times 3.5}\\\Rightarrow \theta=45.71428\ rad=\frac{45.71428}{2\pi}=7.27565\ rev[/tex]
The number of revolutions if the final angular speed doubles is 7.27565 revolutions
This question involves the concepts of the equations of motion for angular motion.
(a) The wheel rotates through the angle "11 rad" (OR) "1.75 rev".
(b) The angular speed at t = 2 s is "9 rad/s".
(c) The angular displacement for double speed is "45.71 rad" (OR) "7.28 rev".
(a)
The angular displacement can be found using the second equation of motion for angular motion.
[tex]\theta = \omega_it+\frac{1}{2}\alpha t^2[/tex]
where,
θ = angular displacement = ?
ωi = initial angular speed = 2 rad/s
t = time interval = 2 s
α = angular acceleration = 3.5 rad/s²
Therefore,
[tex]\theta = (2\ rad/s)(2\ s)+\frac{1}{2}(3.5\ rad/s^2)(2\ s)^2\\\theta = 4\ rad\ +\ 7\ rad[/tex]
θ = 11 rad
[tex]\theta = (11\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]
θ = 1.75 rev
(b)
The angular speed can be found using the first equation of motion for angular motion.
[tex]\omega_f = \omega_i+\alpha t\\\omega_f = 2\ rad/s + (3.5\ rad/s^2)(2\ s)\\[/tex]
ωf = 9 rad/s
(c)
The angular displacement can be found using the third equation of motion for angular motion after we double the value of ωf.
[tex]2\alpha \theta = \omega_f^2-\omega_i^2\\2(3.5\ rad/s^2)\theta = (18\ rad/s)^2-(2\ rad/s)^2\\\\\theta = \frac{320\ rad^2/s^2}{7\ rad/s^2}\\\\[/tex]
θ = 45.71 rad
[tex]\theta = (45.71\ rad)(\frac{1\ rev}{2\pi\ rad})[/tex]
θ = 7.28 rev
Learn more about the angular motion here:
brainly.com/question/14979994?referrer=searchResults
The attached picture shows the angular equations of motion.
