Respuesta :
Judging by the data delivered corresponding to the change of speeds and the number of revolutions, it is possible to complement the question by deducing that the angular acceleration is sought. Then to solve this problem we will apply the kinematic equations of angular motion, for which the square of the velocity change is described, in proportion to twice the product between the angular acceleration and the angle. That is to say,
[tex]\omega_f^2-\omega_i^2 = 2\alpha \theta[/tex]
Where,
[tex]\omega_{f,i}[/tex] = Initial and final angular velocity
[tex]\alpha[/tex]= Angular acceleration
[tex]\theta[/tex] = Angular displacement
Now convert the revolutions in radians we have,
[tex]\theta = 2.3rev (\frac{2\pi rad}{1rev})[/tex]
[tex]\theta = 14.45rad[/tex]
Replacing and solving to find alpha we have
[tex]\omega_f^2-\omega_i^2 = 2\alpha \theta[/tex]
[tex]\alpha = \frac{\omega_f^2-\omega_i^2 }{2\theta}[/tex]
[tex]\alpha = \frac{(28)^2-(15)^2}{2(14.45)}[/tex]
[tex]\alpha = 19.34rad/s^2[/tex]
Therefore the angular acceleration is [tex]19.34rad/s^2[/tex]
Answer:
Angular acceleration α = 20.34 rads/s^2
Complete Question:
The angular speed of a propeller on a boat increases with constant acceleration from 14 rad/s to 28 rad/s in 2.3 revolutions.
What is the acceleration of the propeller?
Explanation:
When a particle moves on a circular path(revolution) with uniformly increasing angular velocity(constant acceleration) its motion follows the following equations of kinematics
The angular velocity at time t is given by
ω=ω0+αt............1
Angular displacement at time t
θ=ω0t+0.5αt. .....2
angular velocity after covering angular displacement is given by
ω^2=(ω0)^2+2αθ .......3
α = [ω^2 - (ω0)^2]/2θ. ......4
Where
θ = angular displacement = 2.3 rev = 2.3(2π) rads
t = time
ω0 = initial angular velocity = 14 rads/s
ω = final angular velocity = 28 rads/s
α = angular acceleration
Substituting the values into equation 4.
α = [28^2 - 14^2]/[2(2.3×2π)]
α = 20.34 rads/s^2