Rotational Mech

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A thin uniform hemispherical bowl of mass m and radius R is lying on a smooth horizontal surface,A horizontal force F is applied perpendicular to the rim of the bowl,what is the instantaneous angular acceleration of the bowl?

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6 Answers

tapanmast Vora ·2009-03-14 20:47:57

RF = 3/5mR² * Î±

ther4 Î±=5mF/(3R)

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Vivek ·2009-03-14 20:52:32

how did u get I = 3/5 mr²

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tapanmast Vora ·2009-03-14 20:55:04

is da ans crrct..........

thats da MOI of holo hemisfere

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Vivek ·2009-03-14 20:57:53

yeah correct

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Vivek ·2009-03-14 20:59:31

and the MOI is for the base ,right? i was trying to find COM and then find moi,didn't know about this formula

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tapanmast Vora ·2009-03-14 21:16:17

HMMMMMMMMM jus find MOI abt the base perpendi to plane

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√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

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+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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Rotational Mech

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