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Post

Laboratory

Discussion PHYSICS 81

EXERCISE 7: CONSERVATION OF ENERGY

AND LINEAR MOMENTUM

)()()()( edgeBedgeAedgeBhiA

edgehi

PEKEPEPE

MEME

RESULTS

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)(

'

)()(

'

)(

'

)()(

)()(

edgeBedgeAedgeA

edgeBedgeAedgeA

edgeacedgebc

KEKEKE

ppp

pp

Law of Conservation of Mechanical Energy

Law of Conservation of Linear Momentum

EXERCISE 8: TORQUE

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𝜏 ∝ 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑟𝑚

𝜏 ∝ 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 𝑓𝑜𝑟𝑐𝑒

Torque produced by the bar is not

included in the summation of torque

produced in Part I and II since

_______________.

The normal force from the iron stand is

not included on the forces that produce

torque because ___________________.

RESULTS

Part I

Displacement Method

𝐹𝐵 = 𝑊𝑓𝑑 = 𝑚𝑓𝑑𝑔

using Archimedes’ principle

FBD Method

𝐹𝐵 = 𝑊𝑙𝑜𝑠𝑠 = 𝑊𝑎𝑖𝑟 −𝑊𝑓𝑙𝑢𝑖𝑑

Part II: U-tube manometer

𝑃𝑤𝑎𝑡𝑒𝑟 = 𝑃𝑜𝑖𝑙(𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒)

EXERCISE 9: BUOYANT FORCE AND

SPECIFIC GRAVITY

Part I:

𝐹𝐵(𝑤𝑎𝑡𝑒𝑟) > 𝐹𝐵(𝑜𝑖𝑙)

𝐹𝐵 ∝ 𝜌 𝑠𝑖𝑛𝑐𝑒 𝜌(𝑤𝑎𝑡𝑒𝑟) > 𝜌(𝑜𝑖𝑙)

FBD method

𝐹 = 𝑇 + 𝐹𝐵 − 𝑊 = 0

𝐹𝐵 = 𝑊 − 𝑇

𝐹𝐵 = 𝑊𝑎𝑖𝑟 −𝑊𝑓𝑙𝑢𝑖𝑑

RESULTS

Part II:

𝑃𝑤𝑎𝑡𝑒𝑟 = 𝑃𝑜𝑖𝑙 𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒

𝜌𝑤𝑔 ℎ𝑤 − ℎ𝑖 = 𝜌𝑜𝑖𝑙𝑔 ℎ𝑜𝑖𝑙 − ℎ𝑖

𝜌𝑜′ = 𝜌𝑜𝑖𝑙𝜌𝑤𝑎𝑡𝑒𝑟

=(ℎ𝑤−ℎ𝑖)(ℎ𝑜𝑖𝑙−ℎ𝑖)

𝜌 1/∝ ℎ

RESULTS

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𝐷𝑒𝑟𝑖𝑣𝑒: 𝑇 = 2𝜋 𝑙𝑔

𝐹 = 𝐹𝑠 = −𝑘𝑥 = 𝑚𝑔𝑠𝑖𝑛𝜃

since 𝑠 = 𝑙𝜃 and sinθ ≈ 𝜃 −𝑘𝑥 = −𝑘𝑠 = 𝑚𝑔𝜃

𝑘 = −𝑚𝑔𝜃𝑠=𝑚𝑔𝜃𝑙𝜃= 𝑚𝑔𝑙

𝑇 = 2𝜋 𝑚𝑘= 2𝜋 𝑚

𝑚𝑔/𝑙= 2𝜋 𝑙

𝑔

EXERCISE 10: SIMPLE HARMONIC

MOTION RESULTS

y = 3.9359x + 0.1006

R² = 0.997

0

1

2

3

4

5

6

0 0.5 1 1.5

T2

length (m)

%25.2%

/03.10

/9359.3

4

44

2

2

2

22

error

smg

msg

mg

gm

Part I: Finding the

specific heat

capacity of an

unknown metal

Part II: Finding the

latent heat of

fusion of ice

EXERCISE 11: CALORIMETRY

Part I:

−𝑄𝑙𝑜𝑠𝑠 = 𝑄𝑔𝑎𝑖𝑛

−𝑄𝑚𝑒𝑡𝑎𝑙 = 𝑄𝑐 + 𝑄𝑤 + 𝑄𝑠

Part II:

𝑄𝑙𝑜𝑠𝑠 = 𝑄𝑔𝑎𝑖𝑛

−(𝑄𝑐+𝑄𝑤 + 𝑄𝑠) = 𝑄𝑖 (𝑇𝑖 𝑡𝑜 0) + 𝑄𝑝𝑐 + 𝑄𝑖 (0 𝑡𝑜 𝑇𝑓)

RESULTS