Physics System of Particles and Rotational Motion

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New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

L_i = L_f
mvL = Iω
mvL = (ML³/3 + mL²)ω


Before collision
After collision
0.1 * 80 * 1 = (0.9 * 1²)/3 + 0.1 * 1²)ω
8 = (3/10 + 1/10)ω 8 = (4/10)ω ω = 20 rad/sec

New answer posted

5 months ago

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R
Raj Pandey

Contributor-Level 9

L o i = L o f

m v 2 * l 2 = 4 m l 2 12 + m l 2 4 * ω ω = 6 v 7 2 l = 3 2 v 7 l

 

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

m g h = 1 2 m v 2 + 1 2 l ω 2

v = ω R (no slipping)

m g h = 1 2 m ω 2 R 2 + 1 2 m R 2 ω 2

m g h = 3 4 m ω 2 R 2

ω = 4 g h 3 R 2 = 1 R 4 g h 3

 

 

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

Take 1 k g  mass at origin

X c m = 1 * 0 + 1.5 * 3 + 2.5 * 0 5 = 0.9 c m

Y c m = 1 * 0 + 1.5 * 0 + 2.5 * 4 5 = 2 c m

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

M l 2 12 + M l 4 = M K 2

l 2 12 + l 2 16 = K 2

K = 7 48 l

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5 months ago

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R
Raj Pandey

Contributor-Level 9

( 0,3 )

r ? c m = 1 * i ˆ 2 + j ˆ + 1 * i ˆ + 5 j ˆ 2 2 r ? c m = 3 4 i ˆ + 7 4 j ˆ

( 0,0 )

New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

Moment of inertia of ring

d l = d m r 2

I = 0 a ? ( A + B r ) 2 π r d r π r 2

= 2 π A 0 a ? r 3 d r + 2 π B 0 a ? r 4 d r

= 2 π A a 4 4 + B a 5 5

= 2 π a 4 A 4 + B a 5

 

New answer posted

5 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

5 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

5 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

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