play phryzby with physics

A Motor Wheel Speeds Up

solved RotationalKinematics

Q. The angular speed of a motor wheel is increased from 1200 rpmω0=1200rpm to 3120 rpmω1=3120rpm in 16 sΔt=16s. Assuming the angular acceleration is uniform, how many revolutions does the engine make during this time?

ωω = 126 → 327 rad/sαθn = ?Δt = 16 s
Inputs
ω0
1200 rpm
ω1
3120 rpm
Δt
16.00 s
Outputs
turnsnrev?

How it's solved

Solved with RotationalKinematics

Given

  • ω0=1200rpm
  • ω1=3120rpm
  • Δt=16s

Find nrev.

Working
  1. ω0=1200rpm=125.663706rad/s
  2. ω1=3120rpm=326.725636rad/s
  3. angular velocity after a time(speeding up)ω1=ω0+αr⋅Δt
  4. 326.725636=125.663706+αr⋅16
  5. αr=12.5663706rad/s²
  6. angle after a time(speeding up)θ=ω0⋅Δt+αr⋅Δt22
  7. θ=125.663706⋅16+12.5663706⋅1622
  8. θ=3619.11474
  9. turns in an angle(speeding up)nrev=θ2⋅π
  10. nrev=3619.114742⋅3.14159265
  11. nrev=576
  12. nrev≈576to 3 significant figures
Conclusion

With a steady angular acceleration the wheel turns as if at the mean of its two speeds for the whole time.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb
1require_relative "../physics"23question :a_motor_wheel_speeds_up do4  kind :numerical5  describes "The angular speed of a motor wheel is increased from %{start} to %{finish} in %{time}. Assuming the " \6            "angular acceleration is uniform, %{turns}?"7  given start: "1200 rpm"8  given finish: "3120 rpm"9  given time: "16 s"10  asks turns: "how many revolutions does the engine make during this time"11end
solution.rb
1require_relative "../questions/a_motor_wheel_speeds_up"2require_relative "../science/systems_of_particles_and_rotational_motion/rotational_kinematics"34solve :a_motor_wheel_speeds_up, using: RotationalKinematics do5  objects :wheel6  wheel.goes(:speeding_up)78  take start: wheel.initial_angular_velocity9  take finish: wheel.final_angular_velocity10  take time: speeding_up.time_taken1112  turns = wheel.turns13  answer turns:, figures: 31415  conclusion "With a steady angular acceleration the wheel turns as if at the mean of its two speeds " \16             "for the whole time."17end
playground.rb
1require_relative "../solutions/a_motor_wheel_speeds_up"23play_with :a_motor_wheel_speeds_up do4  title "A Motor Wheel Speeds Up"5end
The laws it drew on
quantitywrittenunitswithinfrom
piπ—0 … ∞RotationalKinematics
positionxm—Velocity
displacementΔxm—Velocity
path lengthdm—Velocity
time takenΔts0 … 1e+05Velocity
average velocityvavm/s—Velocity
average speedcavm/s—Velocity
starting positionx0m—Velocity
final positionxm—Velocity
angular velocityωrad/s—RotationalKinematics
angular accelerationαrrad/s²—RotationalKinematics
angle turnedθ——RotationalKinematics
turnsnrev——RotationalKinematics
starting angular velocityω0rad/s—RotationalKinematics
final angular velocityωrad/s—RotationalKinematics
displacement of wheel during speeding upΔxm—
position of wheel at end of speeding upx1m—
position of wheel at start of speeding upx0m—
average velocity of wheel during speeding upvavm/s—
time taken of speeding upΔts0 … 1e+05
average speed of wheel during speeding upcavm/s—
path length of wheel during speeding updm—
angular velocity of wheel at end of speeding upω1rad/s—
angular velocity of wheel at start of speeding upω0rad/s—
angular acceleration of wheel during speeding upαrrad/s²—
angle turned of wheel during speeding upθ——
turns of wheel during speeding upnrev——
x0≡xi(start of motion)
starting position
x≡xi(end of motion)
final position
ω0≡ωi(start of motion)
starting angular velocity
ω≡ωi(end of motion)
final angular velocity
di(whole)=di(part 1)+di(part 2)
path in parts
Δtw=Δtp+Δtp
time in parts
Δx=x1-x0
change of position(speeding up)
vav=ΔxΔt
displacement a time(speeding up)
cav=dΔt
path a time(speeding up)
ω1=ω0+αr⋅Δt
angular velocity after a time(speeding up)
θ=ω0⋅Δt+αr⋅Δt22
angle after a time(speeding up)
ω12=ω02+2⋅αr⋅θ
angular velocity after an angle(speeding up)
nrev=θ2⋅π
turns in an angle(speeding up)