Physics System of Particles and Rotational Motion

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New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New question posted

2 months ago

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New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

From conservation of momentum:
2*4 = 2v? + mv?
Given v? = 1 m/s (interpreted from intermediate steps)
8 = 2 (1) + mv?
mv? = 6 . (i)
From coefficient of restitution (e=1 for elastic collision):
e = (v? - v? )/ (u? - u? )
1 = (v? - v? )/ (4 - 0)
-1 = (v? - v? )/ (0 - 4)  (as written in the image)
⇒ 4 = v? - 1
⇒ v? = 5 . (ii)
Put (2) in (1), m (5) = 6
m = 1.2kg

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? * v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? * v? = 0
⇒ (10αt²î + 5β (t-5)? ) * (20αtî + 5β? ) = 0
⇒ 50αβt² (î*? ) + 100αβt (t-5) (? *î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

I? / I? = (MR²/2) / (mr²/4)
[ M = σπR², m = σπr² ]
= 2 (M/m) (R²/r²)
= 2 (σπR²/σπr²) (R²/r²)
= 2 (R²/r²)² = 2R? /r?

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

(a) I = ML²/12
(b) I = (2M)L²/3
(c) I = M (2L)²/12 = ML²/3
(d) I = (2M) (2L)²/3 = 8ML²/3

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2 months ago

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A
alok kumar singh

Contributor-Level 10

τ = Iα
I = ½MR² = ½ (10) (0.2)² = 0.2 kgm²
α = (ωf - ωi)/Δt = (0 - 600*2π/60)/10 = -2π rad/s²
τ = |Iα| = 0.2 * 2π = 0.4π = 4π * 10? ¹ Nm

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2 months ago

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New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

l 1 = 1 2 m r 2

l 2 = 1 2 m r 2 l 3 = 1 2 m r 2 l 4 = 2 5 m r 2

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

mg – T = ma. (1)

T * R = l . (2)

a = R. (3)

With the help of equations (1), (2) and (3), we get

a = m g m + l R 2

v = 2 a h = ω R

ω 2 = 2 m g h l + m R 2

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