Science One Math March 1, 2017 Applications of Integration - - PowerPoint PPT Presentation

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Science One Math March 1, 2017 Applications of Integration - - PowerPoint PPT Presentation

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Science One Math March 1, 2017 Applications of Integration Computing areas Computing changes Computing volumes Computing probabilities Locating centre of mass of an object Work done by a non constant force Application of

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Science One Math

March 1, 2017

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Applications of Integration

  • Computing areas
  • Computing changes
  • Computing volumes
  • Computing probabilities
  • Locating centre of mass of an object
  • Work done by a non constant force
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Application of Integration: Work

What is work?

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Application of Integration: Work

What is work? Work is change in energy when a force causes displacement. Basic formula: work = force x displacement valid if force is constant!

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Work done by a non-constant force

If F is constant, 𝑋 = 𝐺 $ 𝑦 If F is not constant, add up small amounts of work over short distances Δ𝑦 Key ideas: split displacement into small segments Δ𝑦, assume F to be constant over Δ𝑦 find Δ𝑋 work done by F over Δ𝑦

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Strategy for computing work done by varying F

  • divide displacement into n segments of length Δ𝑦
  • assume force is a constant over each segment (possibly different for each

segment), force resulting in k-th segment of displacement = 𝐺(𝑦(

βˆ—)

  • compute work done by constant force over short displacement Δ𝑋 = 𝐺 𝑦(

βˆ— Δ𝑦

  • add up small amounts of work, we get a Riemann Sum βˆ‘

𝐺(𝑦(

βˆ—)Δ𝑦

  • (./
  • take the limit for π‘œ β†’ ∞, we get a definite integral

𝑋 = lim

  • β†’ 6 βˆ‘

𝐺(𝑦(

βˆ—)Δ𝑦 = ∫ 𝐺 𝑦 𝑒𝑦 9 :

  • (./
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Fundamental Theorem of Calculus is fundamental in Physics too!

FTC tells us how to compute a definite integral 𝑋 = ∫ 𝐺 𝑦 𝑒𝑦 = 𝑉 𝑐 βˆ’ 𝑉(𝑏)

9 :

where 𝑉 𝑦 = ∫ 𝐺 𝑦 𝑒𝑦 + 𝐷. We call 𝑉 𝑦 the potential energy. FTC also tells us 𝐺 𝑦 =

AB AC (for an external force exerted on object).

Note for a force exerted by the potential energy we have 𝐺 𝑦 = βˆ’

AB AC

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SLIDE 9

Stretching and compressing a spring

Hooke’s law : force required to keep a spring compressed

  • r stretched a distance x is proportional to x.
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Stretching and compressing a spring

Hooke’s law : force required to keep a spring compressed

  • r stretched a distance x is proportional to x.

1) Force F exerted on spring is in the same direction as displacement (both positive when stretching, both negative when compressing). Work done to stretch/compress a spring by x is 𝑋 = ∫ 𝐺 𝑒 𝑒𝑒

C E

= ∫ 𝑙𝑒 𝑒𝑒 = /

G 𝑙 𝑦G C E

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SLIDE 11

Stretching and compressing a spring

Hooke’s law : force required to keep a spring compressed

  • r stretched a distance x is proportional to x.

1) Force F exerted on spring is in the same direction as displacement (both positive when stretching, both negative when compressing). Work done to stretch/compress a spring by x is 𝑋 = ∫ 𝐺 𝑒 𝑒𝑒

C E

= ∫ 𝑙𝑒 𝑒𝑒 = /

G 𝑙 𝑦G C E

2) Force 𝐺

H done by spring is opposite to displacement

Work done by a spring compressed by x, 𝑋 = ∫ βˆ’π‘™π‘’ 𝑒𝑒 = βˆ’ /

G 𝑙 𝑦G C E

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Pulling up a heavy rope

A 10-m long rope of density 2 kg/m is hanging from a wall which is 5 m high (so 5 m of rope runs down the length of the wall and the remaining 5 m is coiled at the bottom of the wall). How much work (in J) is required to pull the rope to the top of the wall? Let g be the acceleration due to gravity.

  • A. 𝑋 = 𝑕 ∫

2𝑧𝑒𝑧

/E E

B. 𝑋 = 50𝑕 + 𝑕 ∫ 2𝑧𝑒𝑧

Q E

C. 𝑋 = 50𝑕 + 𝑕 ∫ 2(10 βˆ’ 𝑧)𝑒𝑧

/E E

  • D. 𝑋 = 50𝑕 + 𝑕 ∫ 2(5 βˆ’ 𝑧)𝑧𝑒𝑧

Q E

E. 𝑋 = 50𝑕 + 𝑕 ∫ 2(5 βˆ’ 𝑧)𝑒𝑧

Q E

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Pulling up a heavy rope

A 10-m long rope of density 2 kg/m is hanging from a wall which is 5 m high (so 5 m of rope runs down the length of the wall and the remaining 5 m is coiled at the bottom of the wall). How much work (in J) is required to pull the rope to the top of the wall? Let g be the acceleration due to gravity. Force exerted = mg but mass depends of length of rope hanging

To pull up the first 5 m of rope we apply constant force of 2x5xg J To pull up the second 5 m of rope we apply a decreasing force because the mass of the rope is decreasing as more rope is pulled up 𝑋 = (50𝑕) + ∫ 2 5 βˆ’ 𝑧 𝑕 𝑒𝑧

Q E

force displacement

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What if different β€œparts” of an β€œobject” undergo different displacements?

Strategy:

  • slice object into n layers of thickness Δ𝑧
  • assume force on each layer is constant

(possibly different for each layer)

  • find work done on each layer by

multiplying force by displacement undergone by layer

  • add up all amounts of work.

Ξ”y

y H

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What if different β€œparts” of an object undergo different displacements?

Strategy:

  • slice object into n layers of thickness Δ𝑧
  • assume force on each layer is constant

(possibly different for each layer)

  • find work done on each layer by

multiplying force by displacement undergone by layer

  • add up all amounts of work.

Ξ”y

y H 𝐺

j:klm = (π‘₯π‘“π‘—π‘•β„Žπ‘’)j:klm

Δ𝑋 = (Δ𝑛j:klm 𝑕) $ π‘’π‘—π‘‘π‘žπ‘šπ‘π‘‘π‘“π‘›π‘“π‘œπ‘’ Δ𝑛j:klm = π‘’π‘“π‘œπ‘‘π‘—π‘’π‘§ βˆ†π‘Š

j:klm , Ξ”π‘Š j:klm = 𝐡j:klmΔ𝑧

total work 𝑋 = ∫ 𝜍 𝐡j:klm 𝑧 𝑕 𝑒𝑧

| E

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Building a pile of sand

How much work must be done in producing a conical heap of sand of base radius R and height H? The specific weight of sand is 𝜍. You may assume that all the sand is taken from the surface of the earth (that is, from height 0). Work to lift a layer of mass m up a height y from the ground force displacement

Δ𝑋 = (𝑛j:klm $ 𝑕) $ 𝑧

𝑛j:klm = 𝜍H:-A $ π‘Š

j:klm

π‘Š

j:klm = 𝜌 𝑠GΔ𝑧 = 𝜌 𝑆 βˆ’ € | 𝑧 G

Δ𝑧 𝑋 = β€’ 𝜍H:-A

| E

𝑕 𝜌 𝑆 βˆ’ 𝑆 𝐼 𝑧

G

𝑧 𝑒𝑧

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Digging a hole

Consider two workers digging a hole. How deep should the first worker dig so that each does the same amount of work? Let the weight density

  • f the dirt be the constant 𝜍, the depth of the hole is D, and the cross-

sectional area of the hole is A (so we assume that the hole does not get any wider or narrower as the workers dig).

  • A. The first worker should dig to a depth 𝐸/2
  • B. The first worker should dig to a depth 𝐸/4
  • C. The first worker should dig to a depth

† G

  • D. The first worker should dig to a depth 2 𝐸
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Digging a hole

Consider two workers digging a hole. How deep should the first worker dig so that each does the same amount of work? Let the weight density

  • f the dirt be the constant 𝜍, the depth of the hole is D, and the cross-

sectional area of the hole is A (so we assume that the hole does not get any wider or narrower as the workers dig). 𝑋

‑ˆ‑ = β€’ 𝜍 𝐡 𝑕 † E

𝑧 𝑒𝑧 = πœπ΅π‘• 𝐸G 2 Let x be the depth of the hole the first worker digs, then we must solve the following equation for x, ∫ 𝜍 𝐡 𝑕 𝑧𝑒𝑧 =

/ G C E

πœπ΅π‘•

†‰ G , which yields 𝑦 = † G .

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SLIDE 19

Pumping out fluid from a tank

A cylindrical tank with a length of L m and a radius of R m is on its side and half-full of gasoline. How much work is done to empty the tank through an outlet pipe at the top of the tank? Let ρ be density of gasoline, and A be the cross-sectional area

  • f a layer at height y,
  • A. ∫

𝜍 𝑕 𝐡 2𝑆 βˆ’ 𝑧 𝑒𝑧

G€ E

  • B. ∫

𝜍 𝑕 𝐡 𝑆 βˆ’ 𝑧 𝑒𝑧

G€ E

  • C. ∫ 𝜍 𝑕 𝐡 2𝑆 βˆ’ 𝑧 𝑒𝑧

€ E

  • D. ∫ 𝜍 𝑕 𝐡 𝑧 𝑒𝑧

€ E

  • E. ∫ 𝜍 𝑕 𝐡 𝑒𝑧

€ E

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Pumping out fluid from a tank

A cylindrical tank with a length of L m and a radius of R m is on its side and half-full of gasoline. How much work is done to empty the tank through an outlet pipe at the top of the tank? Let ρ be density of gasoline, and A be the cross-sectional area

  • f a layer at height y,
  • A. ∫

𝜍 𝑕 𝐡 2𝑆 βˆ’ 𝑧 𝑒𝑧

G€ E

  • B. ∫

𝜍 𝑕 𝐡 𝑆 βˆ’ 𝑧 𝑒𝑧

G€ E

  • C. ∫ 𝜍 𝑕 𝐡 2𝑆 βˆ’ 𝑧 𝑒𝑧

€ E

  • D. ∫ 𝜍 𝑕 𝐡 𝑧 𝑒𝑧

€ E

  • E. ∫ 𝜍 𝑕 𝐡 𝑒𝑧

€ E