Xorazm ilm ziyo ma`ruza kinematika
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Kinematika Ma`ruza
- Bu sahifa navigatsiya:
- VI. O`rtacha tezlik. O`rtacha tezlik
- IX. Balandlikdan gorizontga burchak ostida otilgan jism harakati.
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C. Ikkita jism bir xil 0 ϑ
∆ vaqt interval bilan tik yuqoriga otildi.
Birinchi jismning otilgandan ikkinchi jism bilan uchrashgunga qadar ketgan vaqt: 2 0 1 t g t ∆ + = ϑ
Ikkinchi jism otilgandan keyin birinchi jism bilan uchrashgunga qadar ketgan vaqt: 2 0
t g t ∆ − = ϑ
Uchrashuv nuqtasida ikkala jismning tezliklari bir xil bo`ladi, ya`ni o`zaro tenglashadi. Jismlarning JUMANIYAZOV TEMUR
9 uchrashuv balandligini quyidagicha topamiz: 2 2 2 2 0 t g t h ⋅ − ⋅ = ϑ D. Jism
0 ϑ boshlang`ich tezlik bilan tik yuqoriga otildi. Bunda u ma`lum bir ϑ tezlikka ikki marta erishadi. Birinchi marta ko`tarilayotganda, ikkinchisi esa pastga qaytib tushayotganda. Bu nuqtalar bir xil balandlikda bo`ladi (B va C nuqtalar).
− − = 0 ϑ ϑ ;
g t AC 0 ϑ ϑ + =
Tik yuqoriga otilgan jism uchun harakat tenglamasi quyidagicha yoziladi:
2
0 2
y y t ϑ ⋅ = + ⋅ − Jism H balandlikdan tik yuqoriga 0 ϑ
tezlik bilan otildi.
2 2 0
g t H ⋅ + ⋅ − = ϑ ; bunda: t-jism otilgandan yerga tushguncha ketgan vaqt.
Istalgan vaqt momentidagi balandlikni quyidagi formuladan topish mumkin:
2 2 0
g t H ⋅ − ⋅ = ϑ G. Vertikal harakatda tezlik tenglamasi quyidagi ko`rinishda bo`ladi:
± = 0 ϑ ϑ . Harakat yuqoriga bo`lsa, tenglamada (-), pastga bo`lsa (+) qo`yiladi. Harakat tenglamasi esa quyidagicha bo`ladi: 2 2 0 t g t H ⋅ ± ⋅ = ϑ .
O`rtacha tezlik – butun yo`lni butun vaqtga nisbati. butun butun t S = ϑ Hususiy holatlar: 1. Jism yo`lning birinchi teng yarmini 1 ϑ
qolgan yarmini 2 ϑ tezlik bilan o`tdi. Ushbu holatda o`rtacha tezlikni hisoblaymiz:
1
1 2 1 2 1 2 2 2
S S S S S S S t t t ϑ ϑ ϑ ϑ ϑ = = = = = + + + ⋅ ⋅ ;
1 2 1 2 1 2 2 1 1 2 S S ϑ ϑ
ϑ ϑ ϑ ϑ ⋅ ⋅ = = + ⋅ + 2 1 2 1 2 ϑ ϑ ϑ ϑ ϑ + ⋅ ⋅ = 2. Jism vaqtning birinchi teng yarmini 1 ϑ
qolgan yarmini 2 ϑ tezlik bilan o`tdi. Ushbu holatda o`rtacha tezlikni hisoblaymiz:
2
2 2 1 2 1 2 2 1 1 2 1 ϑ ϑ ϑ ϑ ϑ ϑ ϑ + = ⋅ + ⋅ = ⋅ + ⋅ = + = = t t t t t t t S S t S ;
2 2 1 ϑ ϑ ϑ + =
3. Jism yo`lning birinchi uchdan bir qismini 1 ϑ
va qolgan qismini 2 ϑ tezlik bilan o`tdi. Ushbu holatda o`rtacha tezlikni hisoblaymiz:
JUMANIYAZOV TEMUR
10 1 2 1 2 1 2 1 2 1 2 2 1 2 3 3 3 S S S S S S S S S t t t S ϑ ϑ ϑ ϑ ϑ ϑ ϑ = = = = = = ⋅ + + + ⋅ + ⋅ ⋅
1 2 2 1 3 2 ϑ ϑ ϑ ϑ ⋅ ⋅ = + ⋅
4. Jism vaqtning birinchi uchdan bir qismini 1 ϑ
va qolgan qismini 2 ϑ tezlik bilan o`tdi. Ushbu holatda o`rtacha tezlikni hisoblaymiz:
3
3 2 3 2 1 2 1 2 2 1 1 2 1 ϑ ϑ ϑ ϑ ϑ ϑ ϑ ⋅ + = ⋅ + ⋅ = ⋅ + ⋅ = + = = t t t t t t t S S t S
Faraz qilaylik, jism 1
vaqtda 1 ϑ tezlik bilan 1
masofani, 2 t vaqtda
2 ϑ tezlik bilan 2 S masofani va 3
vaqtda esa 3 ϑ
3 S masofani bosib o`tdi. Ushbu holatda o`rtacha tezlikni hisoblaymiz:
3
1 3 3 2 2 1 1 3 2 1 3 2 1 t t t t t t t t t S S S t S + + ⋅ + ⋅ + ⋅ = + + + + = = ϑ ϑ ϑ ϑ
Tekis o`zgaruvchan harakatda o`rtacha tezlik boshlang`ich va oxirgi tezliklarning o`rta arifmetigidan topiladi: 2 0
ϑ ϑ + = .
VII. Balandlikdan gorizontal otilgan jism harakati. (
= 0
) Ushbu harakatda jism gorizontal yo`nalishda tekis va vertikal yo`nalishda esa tekis tezlanuvchan harakat qiladi. Natijada ikkala harakat uyg`unlashib parabolik trayektoriya hosil bo`ladi.
Bunda:
H -jism otilgan balandlik; S -uchish uzoqligi; ϑ -yerga urilishdagi tezlik; L -jism otilgan nuqtadan yerga urilish nuqtasi orasidagi eng kichik masofa; y ϑ - tezlikning vertikal tashkil etuvchisi; α -yerga urilish burchagi. 2 2 t g H ⋅ = ; t S ⋅ = 0 ϑ ; g H ⋅ − = 2 2 0 2 ϑ ϑ ;
H t ⋅ = 2 ⇒
g H S ⋅ ⋅ = 2 0 ϑ ; 2 2 2
H L + = t g y ⋅ = ϑ ;
ϑ ϑ α 0 cos
= ;
ϑ ϑ ϑ α t g y ⋅ = = sin
; 0 0 ϑ ϑ ϑ α t g tg y ⋅ = =
2 2 2 0 2 2 0 t g y ⋅ + = + = ϑ ϑ ϑ ϑ - istalgan vaqt momentidagi tezlikni topish. Tik pastga boshlang`ich tezliksiz tashlangan va gorizontal yo`nalishda otilgan jismlar bir xil vaqtda tushadilar.
3
1 t t t = = Gorizontal uchayotgan samolyotdan yuk yoki bomba tashlansa, bunda yuk yoki bombaning trayektoriyasi parabolik bo`ladi: g H S ⋅ ⋅ = 2 ϑ
Balandlikdan gorizontal otilgan jism uchun harakat tenglamalari quyidagicha:
0 x t ϑ = ⋅ ; 2 2 g t y ⋅ = . 2 2 0 2
y ϑ ⋅ = ⋅ -trayektoriya tenglamasi. JUMANIYAZOV TEMUR
11 0
t ϑ = ⋅ ; 2 0 2
y y ⋅ = − .
2 0 2 0 2
y y ϑ ⋅ = − ⋅ -trayektoriya tenglamasi. VIII. Gorizontga burchak ostida otilgan jism harakati.
Gorizontga burchak ostida otilgan jism gorizontal yo`nalishda tekis va vertikal yo`nalishda esa tekis o`zgaruvchan harakatlanadi. Natijada parabolik trayektoriya hosil bo`ladi. Jism qanday boshlang`ich va burchak ostida otilsa, huddi shunday tezlik va burchak ostida yerga qaytib tushadi.
α
ϑ cos
0 ⋅ = x - tezlikning gorizontal tashkil etuvchisi, uchish davomida o`zgarmas qiymatda saqlanadi: const x = ϑ . α ϑ ϑ sin
0 0 ⋅ = y - boshlang`ich holatdagi tezlikning vertikal tashkil etuvchisi bo`lib, u uchish davomida quyidagi qonun bo`yicha o`zgaradi: 0 0
y y g t gt ϑ ϑ ϑ α = − ⋅ = ⋅ − . Eng yuqori nuqtada tezlik faqat gorizontal yo`nalishda yo`naladi va bu tezlik uchish davomidagi eng kichik (minimal) tezlik bo`ladi: x ϑ ϑ = min
. Jismning uchish vaqtini quyidagicha hisoblaymiz: Aytish kerakki, ko`tarilish vaqti va tushish vaqti o`zaro teng bo`lgan kattaliklardir: 0 0
y k t g g ϑ ϑ α = = ⇒ g t t t k α ϑ sin 0 ⋅ = =
g t uchish α ϑ sin 2 0 ⋅ ⋅ = - jismning uchish vaqti ya`ni otilgandan yerga qaytib tushguncha ketgan vaqti. Uchish uzoqligi formulalari:
⋅ ⋅ = ⋅ = α ϑ ϑ cos 0 ; 0 0 2 sin cos
S g ϑ α ϑ α = ⇒ g S α ϑ 2 sin
2 0 ⋅ = ; bunda
α - jismning otilish burchagi. Maksimal ko`tarilish balandligi formulalari: 0 2 max 2
H g ϑ = ⇒ g H ⋅ ⋅ = 2 sin 2 2 0 max α ϑ ; 2 2 max ( ) 2 2 2 uchish k t g gt H = = ⇒ 8 2 max uchish t g H ⋅ = ; g g H x ⋅ − = ⋅ − = 2 2 2 min
2 0 2 2 0 max ϑ ϑ ϑ ϑ . max H va
S orasidagi bog`lanish quyidagicha: 2 0 sin 2 S g ϑ α = ; 2 2 0 max sin 2
g ϑ α = ; 2 2 0 0 ϑ ϑ = ⇒
⇒ max
2 2 sin 2 sin gH gS α α = ⇒ max 2 2 2 sin cos sin
H S α α α = ⇒ ⇒ max
4 sin
cos H S α α = ⇒
S H tg max
4 ⋅ = α . Jism gorizontga α burchak ostida otilganda qandaydir 1 t va
2 t vaqtlardan so`ng jismning tezlik yo`nalishi gorizont bilan β burchak tashkil qiladi: 0 0 0 sin
cos y y x x gt gt tg ϑ ϑ ϑ α β ϑ ϑ ϑ α − − = = = ⇒ ⇒ 0 0 0 sin sin cos
cos ( ) cos gt tg tg α ϑ α ϑ α β ϑ
α β α = − = −
( ) 0 1 cos t tg tg g ϑ α α β ⋅ = ⋅ − ( ) 0 0 2 1 2 sin cos uchish t t t tg tg g g ϑ α ϑ α α β ⋅ = − = − ⋅ − ⇒ ⇒ ( ) 0 0 2 2 sin cos t tg tg g g ϑ α ϑ α α β ⋅ = − ⋅ − = 0 cos
sin (2 ) cos tg tg g ϑ α α α β α ⋅ = − +
( ) 0 2 cos
t tg tg g ϑ α α β ⋅ = ⋅ + JUMANIYAZOV TEMUR
12 Istalgan vaqt momentidagi jism tezligini quyidagicha topamiz: 2 2
x ϑ ϑ ϑ + = ; Ushbu harakat uchun harakat tenglamalarini quyidagicha yozish mumkin: 0 cos x x t t S ϑ ϑ α = ⋅ = ⋅ ⋅ =
; 0 2 2 0 sin 2 2
g t g t y t t h ϑ ϑ α ⋅ ⋅ = ⋅ −
= ⋅ ⋅ − =
Bu yerda: S- t vaqtdagi jismning gorizontal ko`chishi; h- t vaqtdagi ko`tarilgan balandlik.
otilgan jism harakati. α ϑ ϑ cos
0 ⋅ = x - tezlikning gorizontal tashkil etuvchisi, uchish davomida o`zgarmas qiymatda saqlanadi: const x = ϑ . α ϑ ϑ sin
0 0 ⋅ = y - boshlang`ich holatdagi tezlikning vertikal tashkil etuvchisi bo`lib, u uchish davomida quyidagi qonun bo`yicha o`zgaradi: 0 0
y y g t gt ϑ ϑ ϑ α = − ⋅ = ⋅ − . Eng yuqori nuqtada tezlik faqat gorizontal yo`nalishda yo`naladi va bu tezlik uchish davomidagi eng kichik (minimal) tezlik bo`ladi: x ϑ ϑ = min
.
Jism uchun harakat tenglamalari quyidagicha: 0 cos
x x t t ϑ ϑ α = ⋅ = ⋅ ⋅ ; 0 2 2 0 0 sin 2 2 y g t g t y y t t ϑ ϑ α ⋅ ⋅ = + ⋅ − = ⋅ ⋅ − yoki
0 cos
x S t t ϑ ϑ α = ⋅ = ⋅ ⋅ -gorizontal ko`chish. 0 2 2 0 sin 2 2
g t g t h H t H t ϑ ϑ α ⋅ ⋅ = + ⋅ − = + ⋅ ⋅ − -istalgan vaqtdagi balandlik. Agarda jism yerga kelib urilsa, h=0 yoki y=0 ekanligidan: 0 2 2 uchish y uchish g t H t ϑ ⋅ = − ⋅ + yoki
2 0 sin 2 uchish uchish g t H t ϑ α ⋅ = − ⋅
⋅ +
max
max H H h = + -yerdan hisoblaganda maksimal ko`tarilish balandligi. 2 2
max sin
2 h g ϑ α ⋅ = ⋅ ; 2 max 2 k g t h ⋅ = ; 2 2 2 2 0 0 min
max 2 2 x h g g ϑ ϑ ϑ ϑ − − = = ⋅ ⋅ . Istalgan vaqt momentidagi jism tezligini quyidagicha topamiz: 2 2
x ϑ ϑ ϑ + = . Download 1.11 Mb. Do'stlaringiz bilan baham: 
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