Dinamika rotasi - masalah lan solusi

1. Gaya F ditrapake ing tali sing dililit ing katrol silinder. torsi is 2 N m lan wayahe inersia is 1 kg m2, opo iku percepatan sudut saka silinder kasebut.

Dinamika rotasi – masalah lan solusi 1Dikenal:

Torsi (τ) = 2 Nm

Momen inersia (I) = 1 kg m2

Dikarepake: Percepatan sudut silinder

Solusi:

Στ = Aku α

Στ = torsi bersih, I = momen inersia, α = percepatan sudut

Percepatan sudut silinder:

α = Στ / I = 2/1 = 2 rad/s2

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2. Gaya F ditrapake ing tali sing dililit ing katrol silinder. Gedhene gaya kasebut yaiku 10 N, radius silinder yaiku 0.2 m lan momen inersia yaiku 1 kg m2, WApa percepatan sudut silinder kasebut?

Dinamika rotasi – masalah lan solusi 2Dikenal:

Gaya (F) = 10 N

Radius silinder (R) = 0.2 m

Momen inersia (I) = 1 kg m2

Dikarepake: Percepatan sudut silinder kasebut.

Solusi:

τ = FR

τ = torsi, F = gaya, R = radius silinder

Torsi:

τ = FR = (10 N)(0.2 m) = 2 N m

Στ = Aku α

Στ = torsi bersih, I = momen inersia, α = percepatan sudut

Percepatan sudut silinder:

α = Στ / I = 2 / 1 = 2 rad/s2

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3. Gaya F ditrapake ing tali sing dililit ing katrol silinder. Gedhene gaya kasebut yaiku 10 N, radius silinder yaiku 0.2 m lan massa silinder yaiku 20 kg m2,. Wiku percepatan sudut silinder.

Dinamika rotasi – masalah lan solusi 3Dikenal:

Gaya (F) = 10 N

Radius silinder (R) = 0.2 m

Massa silinder (M) = 20 kg

Dikarepake: Percepatan sudut silinder

Solusi:

τ = FR = (10 N)(0.2 m) = 2 N m

Momen inersia:

Aku = 1⁄2 MR2 = 1⁄2 (20)(0.2)2 = 1⁄2 (20)(0.04) = 0.4 kg m²2

Percepatan sudut silinder:

α = Στ / I = 2 / 0.4 = 5 rad/s2

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4. Balok 1 kg digantung saka tali sing dililit ing katrol silinder. Momen inersia katrol kasebut yaiku 1 kg m2 lan radius katrol yaiku 0.2 m. Pira percepatan sudut katrol kasebut. Akselerasi amarga gravitasi yaiku 10 m/s2.

Dinamika rotasi – masalah lan solusi 4Dikenal:

Momen inersia katrol (I) = 1 kg m2

Massa saka blok (m) = 1 kg

Akselerasi amarga gravitasi (g) = 10 m/s2

bobot (w) = mg = (1 kg)(10 m/s)2) = 10 kg m/s2 = 10 N

Radius katrol (R) = 0.2 m

Dikarepake: Akselerasi sudut

Solusi:

Torsi:

τ = FR = w R = (10 N)(0.2 m) = 2 N m

Momen inersia:

Aku = 1 kg/m2

Percepatan sudut:

α = Στ / I = 2 / 1 = 2 rad/s2

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5. Balok 1 kg digantung saka tali sing dililit ing katrol silinder. Massa katrol kasebut yaiku 20 kg lan radius katrol kasebut yaiku 0,2 m. Pira percepatan sudut katrol lan tiba gratis percepatan balok kasebut. Percepatan amarga gravitasi yaiku 10 m/s2.

Dinamika rotasi – masalah lan solusi 5Dikenal:

Massa katrol (M) = 20 kg

Radius katrol (R) = 0,2 m

Massa balok (m) = 1 kg

Akselerasi amarga gravitasi (g) = 10 m/s2

Bobot (w) = mg = (1 kg)(10 m/s)2) = 10 kg m/s2 = 10 N

Dikarepake: percepatan sudut katrol lan percepatan jatuh bebas balok.

Solusi:

Torsi kasebut:

τ = FR = w R = (10 N)(0.2 m) = 2 N m

Momen inersia katrol silinder:

Aku = 1⁄2 MR2 = 1⁄2 (20)(0.2)2 = (10)(0.04) = 0.4 kg m²2

Percepatan sudut katrol:

α = Στ / I = 2 / 0.4 = 5 rad / s2

Percepatan jatuh bebas balok:

a = R α = (0.2)(5) = 1 m/s2

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