ch 4 summary

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Chapter 4 summary
๏‚ท Position vector: ( ๐‘Ÿโƒ— =๐‘ฅ๐‘–ฬ‚ + ๐‘ฆ๐‘—ฬ‚ + ๐‘ง๐‘˜ฬ‚ )
๏‚ท Displacement : change in position vector ( โˆ†๐‘Ÿโƒ— =โˆ†๐‘ฅ๐‘–ฬ‚ + โˆ†๐‘ฆ๐‘—ฬ‚ + โˆ†๐‘ง๐‘˜ฬ‚ )
โˆ†๐‘Ÿโƒ—
๏‚ท
Average velocity ( ๐‘ฃโƒ— avg =
๏‚ท
Instantaneous velocity ( ๐‘ฃ
๏‚ท
Average velocity ( ๐‘Ž
โƒ— avg =
๏‚ท
Instantaneous acceleration (a =
โˆ†๐‘ก
)
๐‘‘๐‘Ÿโƒ—
= ๐‘‘๐‘ก )
โƒ—โƒ—
โˆ†๐‘ฃ
โˆ†๐‘ก
)
โƒ—โƒ—
๐‘‘๐‘ฃ
๐‘‘๐‘ก
)
Exercises:
1. Find the position vector ( 2m ,-1m,-3m )
2. A particle moves through displacement
[(2๐‘š)๐‘–ฬ‚ + (6๐‘š )๐‘—ฬ‚ − (4๐‘š)๐‘˜ฬ‚] in (2 s ) find the average
velocity ??
3. The x and y component of an object moving in xy plane
are given as( x=2t3 – 2t2 +t) m and (y=3t2+2t) m
determine :at ( t=2 s)
๏‚ท The position vector
๏‚ท Velocity vector
๏‚ท Acceleration vector
4. At t=0 the velocity of a particle is measured to be 3 m/s
along the negative y- axis and its acceleration is
a=( 2๐‘–ฬ‚ − 4๐‘—ฬ‚) (m/s2)
Find the velocity vector as a function of time ??
Projectile motion :
A projectile is :
๏‚ท A particle moving in the x-y plane
๏‚ท Have an initial velocity
๏‚ท It have the free fall acceleration (g)
Projectile motion : the motion of a projectile
๏‚ท Vi x = vi cos ๐œƒ ( โ€ซ =)ุซุงุจุชู‡ ุงู„ ุชุชุบูŠุฑ ุทูˆุงู„ ูุชุฑุฉ ุงู„ุญุฑูƒุฉโ€ฌvx at any time
๏‚ท Vi y = vi sinθ
๏‚ท Range : horizontal distance (R)
๏‚ท R=
๐‘ฃ๐‘– 2 ๐‘ ๐‘–๐‘› 2๐œƒ
๐‘”
๏‚ท Maximum range at ( ๐œƒ= 45°)
๏‚ท R max=
๐‘ฃ๐‘– 2
๐‘”
๏‚ท Maximum height ( H =
(๐‘ฃ๐‘– sin ๐œƒ)2
2๐‘”
โ€ซุฃูˆ ูŠู…ูƒู† ุงุณุชุฎุฏุงู… ู‚ูˆุงู†ูŠู† ุงู„ุญุฑูƒุฉโ€ฌ:
Vertical( โ€ซ↓)ุญุฑูƒุฉ ุนู…ูˆุฏูŠุฉ ู„ุฃู„ุณูู„โ€ฌ
Vertical( โ€ซ↑)ุญุฑูƒุฉ ุนู…ูˆุฏูŠุฉ ู„ุฃู„ุนู„ู‰โ€ฌ
Vf = vi+ gt
Vf = vi- gt
๐Ÿ
โˆ†๐’š =vi t + ๐Ÿgt2
Vf2 = vi2+ 2g โˆ†๐’š
๐Ÿ
โˆ†๐’š =๐Ÿ( vf+ vi ) t
๐Ÿ
โˆ†๐’š =vi t - ๐Ÿgt2
Vf2 = vi2- 2g โˆ†๐’š
๐Ÿ
โˆ†๐’š =๐Ÿ( vf+ vi ) t
๏‚ท Thrown horizontally ( โ€ซ = ๐œƒ )ู‚ุฐู ุงูู‚ูŠุงโ€ฌ0
๏‚ท โ€ซุนู†ุฏ ุงู‚ุตู‰ ุงุฑุชูุงุนโ€ฌvy =0 โ€ซู„ูƒู†โ€ฌ
๏‚ท vx
โ€ซู„ู‡ุง ู‚ูŠู…ุฉ ูˆู‡ูŠ ู†ูุณู‡ุง ุงู„ุณุฑุนุฉ ุงู„ุณูŠู†ูŠุฉ ุงุงู„ุจุชุฏุงุฆูŠุฉโ€ฌ
โ€ซ ููŠ ุญุฑูƒุฉ ุงู„ู…ู‚ุฐูˆูุงุช ุงุงู„ู„ุชุฒุงู… ุจุฃู…ุซู„ุฉ ุงู„ูƒุชุงุจโ€ฌ: โ€ซู…ุงู„ุญุธุฉโ€ฌ
Uniform circular motion :
๏‚ท it travels around a circle path
๏‚ท the tangential velocity of a particle has constant magnitude and
change direction continuously .
๏‚ท the center petal acceleration has constant magnitude and
towards at the center of the circle
๏‚ท the velocity and acceleration are perpendicular to each other
๏ƒฝ a=
๐‘ฃ2
๐‘Ÿ
a=center petal acceleration(m/s2)
v=velocity (m/s)
r= radius (m) ( โ€ซ) ู†ุตู ุงู„ู‚ุทุฑโ€ฌ
๏ƒฝ period of revolution : the time taken to complete one
revolution
2๐œ‹๐‘Ÿ
๐œ=
๐‘ฃ
๏ƒฝ Frequency :the number of revolution per second
1
๐‘“=
๐œ
๐‘“=
๐‘ฃ
2๐œ‹๐‘Ÿ
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