The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during the entire 20-s interval

Respuesta :

Answer:

50 revolutions

Explanation:

Data provided:

case I: From rest to top spin

The initial angular speed of the washer, ωi = 0 rev /s

Final angular speed of the washer ωf = 5 rev /s

Time taken, t₁ = 8 s

now,  

The angular displacement or the number of revolutions taken (θ₁) is calculated as:

  θ₁ = ωi t₁ + (1/2)α₁t₁²

where,

α is the angular acceleration

The angular acceleration can be calculated as:

  ωf - ωi = α₁t₁

on substituting the values, we get

8α₁ = 5 - 0

or

α₁ = 0.625 rev/s²

substituting the values in the equation for the number of revolutions, we get

θ₁ = 0 + (1/2) (0.625)(8)²

or

θ₁ = 20 revolutions

also,  

For the case II: From top spin to rest

we have

The initial angular speed, ωi = 5 rev /s

and the final angular speed, ωf = 0 rev /s

Total time taken, t₂ = 12 s

Now, angular acceleration for this case

  ωf - ωi = α₂t₂

on substituting the values, we have

  12α₂ = 0 - 5

α₂ = - 0.4166 rev/s²

Therefore, the number of revolutions ( i.e angular displacement  )

θ₂ = ωit₂ + (1/2)α₂t₂²

on substituting the values, we have

θ₂ = 5 × 12 + (1/2)(-0.4166)(12)²

or

θ₂ = 30 rev

Hence,

the total number of revolutions made by the washer during the 20s is  

θ = θ₁ + θ₂

or

θ = 20 rev + 30 rev

or

θ = 50 revolutions

The number of revolutions made by the tub during the entire time is 15 revolutions.

Angular acceleration of the tub

The angular acceleration of the tub is determined by applying the following kinematic equation as shown below;

[tex]\omega _f = \omega _i \ + \ a\lpha t\\\\[/tex]

where;

  • ωi is the initial angular velocity
  • ωf is the final angular velocity = 5 rev/s = 5 x 2π rad = 31.42 rad/s

[tex]31.42 = 0 \ + \ \alpha (8)\\\\\alpha = \frac{31.42}{8} \\\\\alpha = 3.93 \ rad/s^2[/tex]

Displacement of the tub after 20s

The displacement of the tub before it comes to rest is calculated as follows;

[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2\\\\\theta = 31.42(12) \ + \ 0.5(-3.93)(12)^2\\\\\theta = 377.04 \ - 282.96\\\\\theta = 94.1 \ rad\\\\\theta = 94.1\ rad \times \frac{1 \ rev}{2\pi \ rad} = 15 \ rev[/tex]

Thus, the number of revolutions made by the tub during the entire time is 15 revolutions.

Learn more about number of revolutions here: https://brainly.com/question/24463763