physics 81
DESCRIPTION
Physics 81TRANSCRIPT
11/20/2015
1
Post
Laboratory
Discussion PHYSICS 81
EXERCISE 7: CONSERVATION OF ENERGY
AND LINEAR MOMENTUM
)()()()( edgeBedgeAedgeBhiA
edgehi
PEKEPEPE
MEME
RESULTS
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edgeBedgeAedgeA
edgeacedgebc
KEKEKE
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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