Physics System of Particles and Rotational Motion

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New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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a month ago

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V
Vishal Baghel

Contributor-Level 10

Position vector about O (r_o) = 5i + 5√3j

Force vector (F) = 4i - 3j

Torque about O (τ_o) = r_o * F
τ_o = (5i + 5√3j) * (4i - 3j)
τ_o = -15k - 20√3k = (-15 - 20√3)k

Position vector about Q (r_q) = -5i + 5√3j

Torque about Q (τ_q) = r_q * F
τ_q = (-5i + 5√3j) * (4i - 3j)
τ_q = 15k - 20√3k = (15 - 20√3)k

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

dm = λdx = λ? (1 + x/L)dx
M = ∫? λ? (1 + x/L)dx = λ? [L + L²/2L] = 3λ? L/2
dI = dmx² = λ? (1 + x/L)dx * x²
I = λ? ∫? (x² + x³/L)dx = λ? [L³/3 + L? /4L]
I = (7λ? L³)/12 = (7/12) * (2M/3L) * L³ = (7/18)ML²

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Vishal Baghel

Contributor-Level 10

l? = 0.6M
l? = 0.8M
√l? ² + l? ² = 1
MI about 0 = M/12 (l? ² + l? ²)
I? = M/12

MI about O' = M/12 (l? ² + l? ²) + M (l? ²+l? ²)/4
I? = M/12 + M/4 = (4m)/12; I? /I? = 1/4

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a month ago

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Vishal Baghel

Contributor-Level 10

I = 0.8 kgM², |μ? | = 20Am²
U? = -μ? ⋅ B? = 0
Uf = -μBcos (30°) = -20 x 4 x √3/2
U? – Uf = 40√3 = ½ I ω² = 0.4ω²
ω² = 100√3; ω = 10 (3¹/? )

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V
Vishal Baghel

Contributor-Level 10

I? ∝ R? ² & I? ∝ R? ²
I? /I? = (R? /R? ) ² = 1/α² = 1/16
α = 4

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a month ago

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alok kumar singh

Contributor-Level 10

Angular momentum conservation:
⇒ I? ω? + I? ω? = (I? + I? )ωf
⇒ (MR²/2)ω? = (MR²/2 + MR²/8)ωf
⇒ ωf = 4/5 ω?
⇒ KEfinal = ½ (I? + I? )ωf² = (MR²ω? ²)/5
⇒ KEinitial = ½I? ω? ² = (MR²ω? ²)/4
⇒ % loss ⇒ 20%

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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