Science One Math March 11, 2019 Applications of Integration - PowerPoint PPT Presentation

Science One Math March 11, 2019

Applications of Integration • Computing areas • Computing changes • Computing volumes • Computing probabilities • Locating centre of mass of a lamina Key ideas ➥ “slicing, approximating, adding infinite contributions ” Today: Using integration to compute work done by a non constant force

From your Physics class (last Wednesday) + ⃗ ( 𝑒𝑡 ⃗ ) ∆𝑉 = −∫ 𝐺 change in potential energy (work done by 𝐺 , For a conservative force (when work is independent of path), we can define a potential 𝑉 such that ./ 𝐺 - = − .- This is the fundamental theorem of calculus!

Fundamental Theorem of Calculus is fundamental in Physics too! 5 FTC tells us ∫ 𝐺 𝑦 𝑒𝑦 = 𝑉 𝑐 − 𝑉(𝑏) where 𝑉 𝑦 = ∫ 𝐺 𝑦 𝑒𝑦 + 𝐷 . 6 ./ : FTC also tells us 𝐺 𝑦 = where 𝑉 𝑦 = ∫ 𝐺 𝑢 𝑒𝑢 . 6 .: 5 We call 𝑉 𝑦 the potential energy , then ∫ 𝐺 𝑦 𝑒𝑦 is Work 6 Convention: ./ 𝐺 𝑦 = .: for an external force (exerted on object) ./ 𝐺 𝑦 = − .: for a force exerted by the potential energy

+ ⃗ ( 𝑒𝑡 How to derive ∫ 𝐺 mathematically , (for straight paths) ⃗ is constant along the path ⇒ basic definition of work If 𝐺 ⃗ ( 𝑡 ⃗ = 𝐺 W = 𝐺 - ∆𝑡 ⃗ acting on a particle that moves along displacement 𝑡 ⃗ ) (work done by a constant force 𝐺

+ ⃗ ( 𝑒𝑡 How to derive ∫ 𝐺 mathematically , (for straight paths) ⃗ is constant along the path ⇒ basic definition of work If 𝐺 ⃗ ( 𝑡 ⃗ = 𝐺 W = 𝐺 - ∆𝑡 ⃗ acting on a particle that moves along displacement 𝑡 ⃗ ) (work done by a constant force 𝐺 ⃗ changes along the path ⇒ use Calculus! If 𝐺 • slice path into segments • assume force is constant along each segment ⇒ compute work done by constant ⃗ ( ∆𝑡 = 𝐺 force to move particle over the segment ∆W = 𝐺 - ∆𝑡 • add up all contributions to work + ⃗ ( 𝑒𝑡 • take the limit for an infinite number of segments ⇒ definite integral ∫ 𝐺 ,

Computing work done by a non constant force • “slice” path into n segments of length Δ𝑦 k -th segment is 𝑦 > ,𝑦 >@A • approximate force by a constant on each segment (possibly different for each segment) ∗ ≤ 𝑦 >@A ∗ ) be force component along k -th segment, for 𝑦 > ≤ 𝑦 > let 𝐺(𝑦 > • compute work done by (constant) force on each segment of path ∗ Δ𝑦 Δ𝑋 = 𝐺 𝑦 > • add up small amounts of work ⇒ Riemann Sum F ∗ )Δ𝑦 ∑ 𝐺(𝑦 > >GA • take the limit for 𝒐 → ∞ ⇒ a definite integral 5 F ∗ )Δ𝑦 = ∫ 𝐺 𝑦 𝑒𝑦 F→ N ∑ 𝑋 = lim 𝐺(𝑦 > >GA 6

A few examples of non constant forces: • elastic force • electric force • gravity on a point-like object of varying mass • gravity on a distributed mass • force exerted by (on) a gas during gas expansion

Work done on a spring Hooke’s law : force required to keep a spring compressed or stretched a distance 𝑦 is proportional to 𝑦 . Note: 𝑦 is measured from the natural length of spring.

Work done on a spring Problem : Compute the work done on the spring to compress it by 𝑌 . • “slice path”, each segment is Δ𝑦 long • approximate force as a constant on each segment, on 𝑗 -th segment, consider force of magnitude 𝐺 , = 𝑙𝑦 , • compute work to compress by Δ𝑦 , Recall: Force exerted on spring is in the same direction as displacement, 𝐺 , ( ∆𝑦 = 𝐺 , ∆𝑦 cos 0 = 𝐺 , ∆𝑦 ⇒ ∆𝑋 , = 𝑙𝑦 , Δ𝑦 • add up all contributions and take a limit ⇒ definite integral Z[ 𝑙𝑦 𝑒𝑦 = 1 2 𝑙 𝑌 Y 𝑋 = V \

Work done by a spring Problem : Compute work done by the spring when compressed by 𝑌 . • approximate force as a constant on each segment, on 𝑗 -th segment, consider force of magnitude 𝐺 , = 𝑙𝑦 , • compute work when compressed by Δ𝑦 , Recall: Force exerted by spring is opposite to displacement, 𝐺 , ( ∆𝑦 = 𝐺 , ∆𝑦 cos 𝜌 = −𝐺 , ∆𝑦 ⇒ ∆𝑋 , = −𝑙𝑦 , Δ𝑦 add up all contributions and take a limit ⇒ definite integral • Z[ A Y 𝑙 𝑌 Y 𝑋 = − ∫ 𝑙𝑦 𝑒𝑦 = − \

Work done on a point charge Problem : Find the electric potential energy between two charges a distance 𝑠 apart. Recall: When a conservative force acts on a particle that moves from 𝑏 to 𝑐 , the change in potential energy is the negative work done by conservative force, 𝑉 5 − 𝑉 6 = −𝑋. Strategy : Compute work done on 𝑟 A by the electric force exerted by a second (stationary) charge 𝑟 Y when 𝑟 A moves from very far ( ∞ ) to 𝑠 .

Work done on a point charge Problem : Find the electric potential energy between two charges a distance 𝑠 apart. • “slice” path, each segment is ∆𝑠 long >a b a c • approximate force, on the 𝑗 -th segment consider 𝐺 , = (d e ) c • compute work done to move 𝑟 A by ∆𝑠 Recall: Electric force is in the same direction as the displacement >a b a c ⇒ ∆𝑋 , = 𝐺 , ( ∆𝑠 = 𝐺 , ∆𝑠 , = (d e ) c ∆𝑠 , d >a b a c Add up all contributions and take a limit ⇒ W = ∫ 𝑒𝑨 improper integral f c N d >a b a c >a b a c >a b a c >a b a c >a b a c i d ∆𝑉 = −∫ 𝑒𝑨 = − lim h→N − h = − lim = f c N f d h d h→N

A leaky bucket… A 2 kg bucket and a light rope are used to draw water from a well that is 40 m deep. The bucket is filled with 20 kg of water and is pulled up at 0.5 m/s, but water leaks out of a hole in the bucket at 0.1 kg/s. Find the work done in pulling the bucket to the top of the well.

A leaky bucket… A 2 kg bucket and a light rope are used to draw water from a well that is 40 m deep. The bucket is filled with 20 kg of water and is pulled up at 0.5 m/s, but water leaks out of a hole in the bucket at 0.1 kg/s. Find the work done in pulling the bucket to the top of the well. Force acting on bucket changes as water leaks out ⇒ need to integrate! Solution • slice up path into segments ∆𝑧 long • on the 𝑗 -th segment consider 𝐺 , = 𝑛 , 𝑕 • work to lift water by ∆𝑧 is ∆𝑋 = 𝑛 , 𝑕∆𝑧 ⇒ 𝑛 , = 𝑛(𝑧 , ) , 𝑛 is a function of 𝑧 .n .o = − \.A \.p , 𝑛 0 = 20 ⇒ 𝑛 𝑧 = −0.2𝑧 + 20 t\ 𝑋 q6rsd = ∫ (−0.2𝑧 + 20 )𝑕 𝑒𝑧 𝑋 5uv>sr = 𝑛𝑕 ( 40 = 80𝑕 (constant 𝑛 ) \ 𝑋 ryr6z = 𝑋 q6rsd + 𝑋 5uv>sr

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 10-m long rope of density 2 kg/m is hanging from a wall which A heavy rope… 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. Rope has distributed mass, NOT point-like object ☛ portion of rope near the top undergoes small displacement, ☛ portion of rope near the ground undergoes bigger displacement Mass is distributed uniformly along rope ⇒ force is constant Displacement changes ⇒ need to integrate! Strategy: “slice” rope into small segments of mass ∆𝒏. • Compute work to lift each segment to the top of wall •

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. What’s the work ∆𝑋 to lift a segment of rope ∆𝑧 long from a height 𝑧 to the top of the wall? A. ∆𝑋 = 2∆𝑧 𝑕 𝑧 B. ∆𝑋 = 2∆𝑧 𝑕(5 − 𝑧) C. ∆𝑋 = 2∆𝑧 𝑕(10 − 𝑧) D. ∆𝑋 = 2∆𝑧 𝑕(5 + 𝑧) E. ∆𝑋 = 2∆𝑧 𝑕(10 + 𝑧)

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. What’s the work ∆𝑋 to lift a segment of rope ∆𝑧 long from a height 𝑧 to the top of the wall? A. ∆𝑋 = 2∆𝑧 𝑕𝑧 B. B. ∆𝑿 = 𝟑∆𝒛 𝒉(𝟔 − 𝒛) C. ∆𝑋 = 2∆𝑧 𝑕(10 − 𝑧) D. ∆𝑋 = 2∆𝑧 𝑕(5 + 𝑧) E. ∆𝑋 = 2∆𝑧 𝑕(10 + 𝑧)

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. p A. 𝑋 = ∫ 2𝑕 5 − 𝑧 𝑒𝑧 \ A\ B. 𝑋 = ∫ 2𝑕 5 − 𝑧 𝑒𝑧 \ p C. 𝑋 = ∫ 2𝑕 5 − 𝑧 𝑒𝑧 + 50𝑕 \