play phryzby with physics

Part of eval backend edge, agent version bdd945f

A Whirled Stone

solved UniformCircularMotion

Q. A stone on a string 80 cmrc=80cm long is whirled round a horizontal circle at a steady speed, 14nrev=14 times in 25 sΔt=25s. What is its acceleration, in m/s²?

r = 0.8 mvaωa = ?νTn = 14 turnsΔt = 25 s
Inputs
rc
80 cm
nrev
14
Δt
25.00 s
Outputs
accelerationac?

How it's solved

Solved with UniformCircularMotion

Given

  • rc=80cm
  • nrev=14
  • Δt=25s

Find ac.

Working
  1. rc=80cm=0.8m
  2. rounds over the time(round)ν=nrevΔt
  3. ν=1425
  4. ν=0.56Hz
  5. rounds a second(round)ν=1T
  6. 0.56=1T
  7. T=1.78571429s
  8. once round a period(round)ω=2⋅πT
  9. ω=2⋅3.141592651.78571429
  10. ω=3.51858377rad/s
  11. speed round the circle(round)v=ω⋅rc
  12. v=3.51858377⋅0.8
  13. v=2.81486702m/s
  14. towards the centre(round)ac=v2rc
  15. ac=2.8148670220.8
  16. ac=9.904m/s²
  17. ac≈9.90m/s²to 3 significant figures
Conclusion

The stone moves at a steady speed but its direction changes constantly, so it has a centripetal acceleration toward the centre.

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
1question :a_whirled_stone do2  kind :numerical3  describes "A stone on a string %{length} long is whirled round a horizontal circle at a steady speed, %{turns} times in %{time}. %{acceleration}, in m/s²?"4  given length: "80 cm"5  given turns: 14.06  given time: "25 s"7  asks acceleration: "What is its acceleration"8end
solution.rb
1solve :a_whirled_stone, using: UniformCircularMotion do2  objects :stone3  paths :circle4  stone.goes(:round, on: circle)5  take length: circle.radius, turns: stone.revolutions, time: round.time_taken6  acceleration = stone.centripetal_acceleration7  answer acceleration:, in: "m/s2", figures: 38  conclusion "The stone moves at a steady speed but its direction changes constantly, so it has a centripetal acceleration toward the centre."9end
playground.rb
1play_with :a_whirled_stone do2  title "A Whirled Stone"3end
The laws it drew on
quantitywrittenunitswithinfrom
piπ—0 … ∞UniformCircularMotion
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
radiusrm0 … ∞UniformCircularMotion
speedvm/s0 … ∞UniformCircularMotion
angular speedωrad/s—UniformCircularMotion
revolutionsnrev——UniformCircularMotion
frequencyνHz0 … ∞UniformCircularMotion
centripetal accelerationacm/s²—UniformCircularMotion
against gravityq——UniformCircularMotion
periodTs0 … ∞UniformCircularMotion
gravitygm/s²—UniformCircularMotion
massmkg0 … ∞UniformCircularMotion
centripetal forceFcN—UniformCircularMotion
displacement of stone during roundΔxm—
position of stone at end of roundx1m—
position of stone at start of roundx0m—
average velocity of stone during roundvavm/s—
time taken of roundΔts0 … 1e+05
average speed of stone during roundcavm/s—
path length of stone during rounddm—
speed of stone during roundvm/s0 … ∞
angular speed of stone during roundωrad/s—
radius of circlercm0 … ∞
period of roundTs0 … ∞
frequency of stone during roundνHz0 … ∞
revolutions of stone during roundnrev——
centripetal acceleration of stone during roundacm/s²—
against gravity of stone during roundq——
centripetal force on stone during roundFcN—
mass of stonemkg0 … ∞
x0≡xi(start of motion)
starting position
x≡xi(end of motion)
final position
di(whole)=di(part 1)+di(part 2)
path in parts
Δtw=Δtp+Δtp
time in parts
Δx=x1-x0
change of position(round)
vav=ΔxΔt
displacement a time(round)
cav=dΔt
path a time(round)
v=ω⋅rc
speed round the circle(round)
ω=2⋅πT
once round a period(round)
ν=1T
rounds a second(round)
ν=nrevΔt
rounds over the time(round)
ac=v2rc
towards the centre(round)
q=acg
measured in gs(round)
Fc=m⋅ac
mass times the turning(round)