2.2 Dinamika — Masalalar

Jami: 35 ta | Yechim bilan: ✅

---

Asosiy Masalalar (1-12)

Masala 1 ⭐⭐

5 kg massali jismga 20 N kuch ta'sir etmoqda. Tezlanishni toping.

Yechim

$F = ma$

$a = \frac{F}{m} = \frac{20}{5} = 4$ m/s²

---

Masala 2 ⭐⭐

Avtomobil (1500 kg) 3 m/s² tezlanish bilan harakatlanyapti. Motor kuchini toping (ishqalanishni hisobga olmang).

Yechim

$F = ma = 1500 \times 3 = 4500$ N = 4.5 kN

---

Masala 3 ⭐⭐

10 kg jism gorizontal sirtda turibdi. Ishqalanish koeffitsienti $\mu = 0.3$. Jismni harakatga keltirish uchun minimal kuch?

Yechim

$N = mg = 10 \times 10 = 100$ N

$f_{max} = \mu N = 0.3 \times 100 = 30$ N

Minimal kuch: 30 N (statik ishqalanishni yengish uchun)

---

Masala 4 ⭐⭐

2 kg jism erkin tushmoqda. Havo qarshiligi 5 N. Tezlanishni toping.

Yechim

$W = mg = 2 \times 10 = 20$ N (pastga) $F_D = 5$ N (yuqoriga)

$F_{net} = W - F_D = 20 - 5 = 15$ N

$a = \frac{F_{net}}{m} = \frac{15}{2} = 7.5$ m/s²

---

Masala 5 ⭐⭐

Lift 800 kg, unda 70 kg odam turibdi. Lift 2 m/s² bilan yuqoriga ko'tarilmoqda. Arqon tarangligini toping.

Yechim

Umumiy massa: $m = 800 + 70 = 870$ kg

$T - mg = ma$

$T = m(g + a) = 870(10 + 2) = 870 \times 12 = 10440$ N ≈ 10.4 kN

---

Masala 6 ⭐⭐

30° qiya tekislikda 5 kg jism. Ishqalanish koeffitsienti 0.2. Tezlanishni toping.

Yechim

$N = mg\cos 30° = 5 \times 10 \times 0.866 = 43.3$ N

$f = \mu N = 0.2 \times 43.3 = 8.66$ N

$F_{parallel} = mg\sin 30° = 5 \times 10 \times 0.5 = 25$ N

$F_{net} = 25 - 8.66 = 16.34$ N

$a = \frac{F_{net}}{m} = \frac{16.34}{5} = 3.27$ m/s²

---

Masala 7 ⭐⭐

Ip bilan bog'langan 3 kg va 5 kg jismlar. 3 kg jismga 40 N kuch qo'yilgan. Ip tarangligini toping (ishqalanishsiz).

Yechim

Umumiy tezlanish: $a = \frac{F}{m_1 + m_2} = \frac{40}{3 + 5} = 5$ m/s²

5 kg jism uchun: $T = m_2 \times a = 5 \times 5 = 25$ N

---

Masala 8 ⭐⭐

Robot g'ildiragi 0.1 m radiusli. Motor 2 N·m moment beradi. Robot massasi 10 kg. Tezlanishni toping (ishqalanishsiz).

Yechim

G'ildirak kuchi: $F = \frac{\tau}{r} = \frac{2}{0.1} = 20$ N

Tezlanish: $a = \frac{F}{m} = \frac{20}{10} = 2$ m/s²

---

Masala 9 ⭐⭐

1000 kg avtomobil 20 m/s tezlikda ketayotib, 5 s da to'xtadi. Tormoz kuchini toping.

Yechim

$a = \frac{v - v_0}{t} = \frac{0 - 20}{5} = -4$ m/s²

$F = ma = 1000 \times 4 = 4000$ N = 4 kN

---

Masala 10 ⭐⭐

500 m radiusli egri yo'lda 30 m/s tezlikda harakatlanayotgan 1200 kg avtomobil. Markazga intilma kuchni toping.

Yechim

$F_c = \frac{mv^2}{r} = \frac{1200 \times 900}{500} = \frac{1080000}{500} = 2160$ N

---

Masala 11 ⭐⭐

Quadcopter massasi 2 kg. Havoda qo'zg'almasdan turishda har bir motor qancha kuch berishi kerak? (4 ta motor)

Yechim

Umumiy ko'tarish kuchi = og'irlik $F_{total} = mg = 2 \times 10 = 20$ N

Har bir motor: $F_{motor} = \frac{20}{4} = 5$ N

---

Masala 12 ⭐⭐

Prujina bikrligi k = 500 N/m. 10 cm cho'zilgan. Prujina kuchini va agar 2 kg jism ulangan bo'lsa, tezlanishni toping.

Yechim

$F = kx = 500 \times 0.1 = 50$ N

$a = \frac{F}{m} = \frac{50}{2} = 25$ m/s²

---

O'rtacha Masalalar (13-25)

Masala 13 ⭐⭐⭐

Dron (1.5 kg) vertikal 3 m/s² tezlanish bilan ko'tarilmoqda. Propellerlar umumiy itarish kuchini toping.

Yechim

$F - mg = ma$

$F = m(g + a) = 1.5(10 + 3) = 1.5 \times 13 = 19.5$ N

---

Masala 14 ⭐⭐⭐

Atwood mashinasi: 8 kg va 6 kg jismlar. Tezlanish va ip tarangligini toping.

Yechim

$a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(8-6) \times 10}{14} = \frac{20}{14} = 1.43$ m/s²

$T = \frac{2m_1 m_2 g}{m_1 + m_2} = \frac{2 \times 8 \times 6 \times 10}{14} = \frac{960}{14} = 68.6$ N

---

Masala 15 ⭐⭐⭐

45° qiya tekislikda 10 kg jismni bir tekisda yuqoriga tortish uchun qanday kuch kerak? $\mu = 0.25$

Yechim

$N = mg\cos 45° = 100 \times 0.707 = 70.7$ N

$f = \mu N = 0.25 \times 70.7 = 17.7$ N (pastga, harakatga qarshi)

Tekislik bo'ylab gravitatsiya: $mg\sin 45° = 70.7$ N (pastga)

Bir tekisda tortish ($a = 0$): $F = mg\sin 45° + f = 70.7 + 17.7 = 88.4$ N

---

Masala 16 ⭐⭐⭐

60 kg parashyutchi terminal tezlikka erishdi. Havo qarshiligi kuchini va $C_D A = 1.2$ m² bo'lsa, terminal tezlikni toping. ($\rho = 1.2$ kg/m³)

Yechim

Terminal tezlikda: $F_D = mg = 60 \times 10 = 600$ N

$F_D = \frac{1}{2}\rho v^2 C_D A$

$600 = \frac{1}{2} \times 1.2 \times v^2 \times 1.2$

$v^2 = \frac{600}{0.72} = 833.3$

$v_t = \sqrt{833.3} = 28.9$ m/s ≈ 104 km/h

---

Masala 17 ⭐⭐⭐

Robot (15 kg) 20° qiyalikdan ko'tarilmoqda. Motor kuchi 80 N. Ishqalanish $\mu = 0.1$. Tezlanishni toping.

Yechim

$N = mg\cos 20° = 150 \times 0.94 = 141$ N

$f = \mu N = 0.1 \times 141 = 14.1$ N

Gravitatsiya (pastga): $mg\sin 20° = 150 \times 0.342 = 51.3$ N

$F_{net} = F - mg\sin 20° - f = 80 - 51.3 - 14.1 = 14.6$ N

$a = \frac{14.6}{15} = 0.97$ m/s²

---

Masala 18 ⭐⭐⭐

100 m radiusli egri yo'lda avtomobil ($\mu = 0.7$) maksimal qanday tezlikda burilishi mumkin?

Yechim

Markazga intilma kuch = ishqalanish kuchi

$\frac{mv^2}{r} = \mu mg$

$v^2 = \mu gr = 0.7 \times 10 \times 100 = 700$

$v_{max} = \sqrt{700} = 26.5$ m/s ≈ 95 km/h

---

Masala 19 ⭐⭐⭐

3 ta jism ip bilan bog'langan (gorizontal): 2 kg, 3 kg, 5 kg. 2 kg jismga 50 N kuch qo'yildi. Iplar tarangligini toping.

Yechim

Umumiy tezlanish: $a = \frac{50}{2+3+5} = 5$ m/s²

2-3 orasidagi ip ($T_1$): 3 kg va 5 kg ni tortishi kerak $T_1 = (3+5) \times 5 = 40$ N

3-5 orasidagi ip ($T_2$): faqat 5 kg ni tortadi $T_2 = 5 \times 5 = 25$ N

---

Masala 20 ⭐⭐⭐

Prujinada (k = 200 N/m) 0.5 kg jism osilgan. Muvozanat holatidan 5 cm pastga tortib qo'yib yuborildi. Maksimal tezlanishni toping.

Yechim

Maksimal cho'zilishda (5 cm qo'shimcha) maksimal kuch: $F_{max} = k \times x_{max} = 200 \times 0.05 = 10$ N

Lekin muvozanatda prujina allaqachon cho'zilgan: $x_{eq} = \frac{mg}{k} = \frac{0.5 \times 10}{200} = 0.025$ m

Maksimal tezlanish (5 cm pastda): $F_{spring} = k(x_{eq} + 0.05) = 200 \times 0.075 = 15$ N $F_{net} = F_{spring} - mg = 15 - 5 = 10$ N

$a_{max} = \frac{10}{0.5} = 20$ m/s²

---

Masala 21 ⭐⭐⭐

Quadcopter (2 kg) 30° burilgan holda gorizontal uchmoqda. Propellerlar itarish kuchini va gorizontal tezlanishni toping.

Yechim

Vertikal muvozanat: $F\cos 30° = mg$ $F = \frac{mg}{\cos 30°} = \frac{20}{0.866} = 23.1$ N

Gorizontal kuch: $F_h = F\sin 30° = 23.1 \times 0.5 = 11.55$ N

Gorizontal tezlanish: $a_h = \frac{F_h}{m} = \frac{11.55}{2} = 5.78$ m/s²

---

Masala 22 ⭐⭐⭐

0.5 m uzunlikdagi ipda 1 kg jism konussimon mayatnik sifatida 30° burchak ostida aylanmoqda. Ip tarangligi va aylanish davri?

Yechim

Vertikal: $T\cos 30° = mg$ $T = \frac{10}{0.866} = 11.5$ N

Gorizontal (markazga intilma): $T\sin 30° = \frac{mv^2}{r}$

Aylanish radiusi: $r = L\sin 30° = 0.5 \times 0.5 = 0.25$ m

$11.5 \times 0.5 = \frac{1 \times v^2}{0.25}$ $v^2 = 1.44$ $v = 1.2$ m/s

Davr: $T_{period} = \frac{2\pi r}{v} = \frac{2\pi \times 0.25}{1.2} = 1.31$ s

---

Masala 23 ⭐⭐⭐

Raketa (500 kg) vertikal ko'tarilmoqda. Dvigatel 8000 N itarish beradi. 10 s dagi balandlik va tezlik?

Yechim

$F_{net} = F - mg = 8000 - 5000 = 3000$ N

$a = \frac{3000}{500} = 6$ m/s²

$v = at = 6 \times 10 = 60$ m/s

$h = \frac{1}{2}at^2 = 0.5 \times 6 \times 100 = 300$ m

---

Masala 24 ⭐⭐⭐

Differensial drive robot (10 kg). Chap motor 5 N, o'ng motor 7 N kuch beradi. G'ildiraklar orasidagi masofa 0.4 m. Chiziqli va burchak tezlanishni toping.

Yechim

Umumiy kuch: $F = F_L + F_R = 5 + 7 = 12$ N

Chiziqli tezlanish: $a = \frac{F}{m} = \frac{12}{10} = 1.2$ m/s²

Aylanuvchi moment (massa markaziga nisbatan): $\tau = \frac{(F_R - F_L) \times L}{2} = \frac{(7-5) \times 0.4}{2} = 0.4$ N·m

Burchak tezlanish (taxminan, $I \approx \frac{1}{12}mL^2$ disk sifatida): $I \approx \frac{1}{12} \times 10 \times 0.4^2 = 0.133$ kg·m²

$\alpha = \frac{\tau}{I} = \frac{0.4}{0.133} = 3$ rad/s²

---

Masala 25 ⭐⭐⭐

Havo sharida (V = 2 m³) 80 kg yuk. Havo zichligi 1.2 kg/m³, geliy 0.17 kg/m³. Vertikal tezlanish?

Yechim

Ko'taruvchi kuch (Arximed): $F_b = \rho_{havo} \times V \times g = 1.2 \times 2 \times 10 = 24$ N

Shar og'irligi: $W_{He} = \rho_{He} \times V \times g = 0.17 \times 2 \times 10 = 3.4$ N

Yuk og'irligi: $W_{yuk} = 80 \times 10 = 800$ N

Jami og'irlik: $W = 3.4 + 800 = 803.4$ N

$F_{net} = F_b - W = 24 - 803.4 = -779.4$ N (pastga)

Umumiy massa: $m = 0.34 + 80 = 80.34$ kg

$a = \frac{-779.4}{80.34} = -9.7$ m/s² (tushadi)

Shar juda kichik — ko'tara olmaydi!

---

Murakkab Masalalar (26-35)

Masala 26 ⭐⭐⭐⭐

Quadcopter (m = 2 kg, I = 0.05 kg·m²) roll burchagi 20° ga o'zgarishi kerak. Propellerlar orasidagi masofa 0.4 m. Tezlik farqi qancha bo'lishi kerak agar 0.5 s da burilishi kerak bo'lsa?

Yechim

Burchak tezlanish (20° = 0.35 rad, 0.5 s da): $\theta = \frac{1}{2}\alpha t^2$ $\alpha = \frac{2\theta}{t^2} = \frac{2 \times 0.35}{0.25} = 2.8$ rad/s²

Kerakli moment: $\tau = I\alpha = 0.05 \times 2.8 = 0.14$ N·m

Kuch farqi ($\tau = \Delta F \times d/2$, d = 0.4 m): $\Delta F = \frac{2\tau}{d} = \frac{2 \times 0.14}{0.4} = 0.7$ N

---

Masala 27 ⭐⭐⭐⭐

Raketa massasi vaqt bo'yicha: $m(t) = 1000 - 5t$ kg. Itarish kuchi doimiy F = 15000 N. t = 100 s dagi tezlanish?

Yechim

$t = 100$ s da: $m = 1000 - 500 = 500$ kg

$a(t) = \frac{F - m(t)g}{m(t)} = \frac{15000 - 500 \times 10}{500} = \frac{10000}{500} = 20$ m/s²

---

Masala 28 ⭐⭐⭐⭐

Robot qo'li (2 bo'g'inli) vertikal tekislikda. $L_1 = 0.5$ m, $m_1 = 2$ kg, $L_2 = 0.3$ m, $m_2 = 1$ kg. Gorizontal holatda ($\theta_1 = \theta_2 = 0$) birinchi bo'g'in motoriga kerak moment?

Yechim

Birinchi bo'g'in massa markazi: $L_1/2 = 0.25$ m Ikkinchi bo'g'in massa markazi: $L_1 + L_2/2 = 0.65$ m

Gravitatsiya momentlari (birinchi motor atrofida): $\tau_1 = m_1 g \times (L_1/2) = 2 \times 10 \times 0.25 = 5$ N·m

$\tau_2 = m_2 g \times (L_1 + L_2/2) = 1 \times 10 \times 0.65 = 6.5$ N·m

Jami: $\tau_{total} = 5 + 6.5 = 11.5$ N·m

---

Masala 29 ⭐⭐⭐⭐

Yer sun'iy yo'ldoshi 400 km balandlikda. Orbital tezlik va markazga intilma tezlanish? (Yer R = 6371 km, g = 9.81 m/s²)

Yechim

Orbital radius: $r = 6371 + 400 = 6771$ km

Gravitatsiya bu balandlikda: $g' = g \times (\frac{R}{r})^2 = 9.81 \times (\frac{6371}{6771})^2 = 9.81 \times 0.885 = 8.68$ m/s²

Orbital tezlik: $\frac{v^2}{r} = g'$ $v = \sqrt{g' \times r} = \sqrt{8.68 \times 6.771 \times 10^6} = 7670$ m/s

Markazga intilma tezlanish = $g' = 8.68$ m/s²

---

Masala 30 ⭐⭐⭐⭐

Avtomobil (1500 kg) 30° bankli yo'lda 25 m/s tezlikda burilmoqda. Yo'l radiusini toping (ishqalanishsiz).

Yechim

Bankli yo'lda ishqalanishsiz: $\tan\theta = \frac{v^2}{rg}$

$r = \frac{v^2}{g\tan\theta} = \frac{625}{10 \times 0.577} = \frac{625}{5.77} = 108$ m

---

Masala 31 ⭐⭐⭐⭐

Shar (m = 0.5 kg, r = 0.1 m) qiya tekislikdan (30°) dumalab tushmoqda. Tezlanishni toping. ($I = \frac{2}{5}mr^2$)

Yechim

Dumalashda siljish + aylanish: $mg\sin\theta = ma + \frac{I\alpha}{r} = ma + \frac{I \cdot a/r}{r} = ma(1 + \frac{I}{mr^2})$

$mg\sin\theta = ma(1 + \frac{2}{5})$

$a = \frac{g\sin 30°}{1.4} = \frac{10 \times 0.5}{1.4} = 3.57$ m/s²

(Sirpanib tushganda: $a = g\sin 30° = 5$ m/s² — tezroq)

---

Masala 32 ⭐⭐⭐⭐

Quadcopter 4 ta propeller bilan. Har biri F itarish va $\tau$ reaktiv moment beradi. Yaw burchak tezlanish 2 rad/s² bo'lishi uchun (I = 0.1 kg·m²), diagonal propellerlar qanday farq bilan aylanishi kerak?

Yechim

Yaw momenti: $\tau_{yaw} = I\alpha = 0.1 \times 2 = 0.2$ N·m

Quadcopter'da diagonal propellerlar bir xil yo'nalishda aylanadi. 1-3 soat strelkasi bo'yicha, 2-4 soat strelkasiga qarshi.

Yaw uchun: $\tau_{yaw} = 2(\tau_1 - \tau_2)$ (har bir juft)

$\Delta\tau = \frac{0.2}{2} = 0.1$ N·m farq kerak

---

Masala 33 ⭐⭐⭐⭐

Magnit levitatsiya poezdi (500000 kg) 3 m/s² bilan tezlanmoqda. Havo qarshiligi $F_D = 0.5v^2$ N. 100 m/s dagi motor kuchi?

Yechim

$v = 100$ m/s da: $F_D = 0.5 \times 10000 = 5000$ N

$F - F_D = ma$ $F = ma + F_D = 500000 \times 3 + 5000 = 1505000$ N = 1.505 MN

---

Masala 34 ⭐⭐⭐⭐

Kosmik kemada (massa M = 10000 kg) astronavt (m = 80 kg) 2 m/s tezlik bilan sakradi. Kema tezligi qanday o'zgaradi?

Yechim

Impuls saqlanishi (umumiy boshlang'ich impuls = 0): $0 = mv_{astronavt} + Mv_{kema}$

$v_{kema} = -\frac{m \times v}{M} = -\frac{80 \times 2}{10000} = -0.016$ m/s

Kema teskari yo'nalishda 0.016 m/s = 1.6 cm/s tezlik oladi.

---

Masala 35 ⭐⭐⭐⭐

Hexacopter (6 motor) massasi 5 kg. Bir motor buzildi. Qolgan 5 motor qanday kuch berishi kerak va dron barqaror qola oladimi?

Yechim

Normal holatda har bir motor: $F = \frac{mg}{6} = \frac{50}{6} = 8.33$ N

Bir motor buzilganda:

• Ko'tarish uchun: $5F' = mg$, $F' = 10$ N har biri
• Lekin muvozanat buziladi!

Buzilgan motor qarama-qarshisidagi motor ham o'chirilishi kerak (yoki kamaytirilishi):

• 4 ta motor bilan: $F'' = \frac{50}{4} = 12.5$ N

Dron barqaror qolishi qiyin — avtopilot kompensatsiya algoritmi kerak.

---

✅ Tekshirish Ro'yxati

• 1-12: Asosiy Nyuton qonunlari
• 13-25: O'rtacha masalalar
• 26-35: Murakkab masalalar

---

Keyingi Qadam

🔬 Amaliyot — Python simulyatsiya!