contestada

The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle through which the rotor turns. (b) What is the magnitude of the angular acceleration?

Respuesta :

Explanation:

Using the circular equations of motion,

dw = dθ/dt

θ = (wi - wo)*t/2

Where,

θ = angular displacement

wi = final angular velocity

= 1420 rad/s

wo = initial angular velocity

= 420 rad/s

t = time

= 5s

θ = (1420 - 420)*5/2

= 2500 rad.

2*pi rad = 1 rev

= 397.89 rev

wi = wo + αt

Where,

α = angular accelaration

1420 = 420 + 5α

α = 1000/5

= 200 rad/s^2.

Lanuel

a. The angle through which the rotor turns is equal to 4600 rads.

b. The magnitude of the angular acceleration is equal to 200 [tex]rad/s^2[/tex]

Given the following data:

  • Initial angular speed = 420 rad/s
  • Final angular speed = 1420 rad/s
  • Time = 5.00 seconds

a. To find the angle through which the rotor turns, we would use the second equation of kinematics for rotational motion:

[tex]\theta = \frac{1}{2} (w_f + w_o)t[/tex]

Where:

  • [tex]\theta[/tex] is the angle of rotation.
  • [tex]w_f[/tex] is the final angular speed.
  • [tex]w_i[/tex] is the initial angular speed.
  • t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

[tex]\theta = \frac{1}{2} (1420 + 420)5\\\\\theta = \frac{1}{2} (1840)5\\\\\theta = 920 \times 5\\\\\theta = 4600 \;rads[/tex]

b. To find the magnitude of the angular acceleration, we would use the following formula:

[tex]\alpha = \frac{w_f \; -\;w_i}{t} \\\\\alpha = \frac{ 1420\; -\;420}{5}\\\\\alpha = \frac{ 1000}{5}\\\\\alpha = 200\;rad/s^2[/tex]

Read more: https://brainly.com/question/14703849

Otras preguntas