System of Particles and Rotational Motion

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New answer posted

3 weeks ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

L ? = r ? * m v ?

= ( 3 i ^ ? j ^ ) * ( 3 j ^ + k ^ )

= 9 k ^ + 3 ( ? j ^ ) ? i ^

= ? i ^ ? 3 j ^ + 9 k ^

| L ? | = 1 + 9 + 8 1 = 9 1

New answer posted

3 weeks ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Use formula for M.I.

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

By conservation of momentum:

m ( 0 ) = 2 m 5 ( - v i ˆ ) + 2 m 5 ( - v j ˆ ) + m 5 v ? ' v ? ' = 2 v i ˆ + 2 v j ˆ v ' = ( 2 v ) 2 + ( 2 v ) 2 = 2 2 v

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

ω = ω 0 + α t α = ω - ω 0 t = ( 3120 - 1200 ) 16 s r p m = 1920 16 * 2 π 60 r a d / s 2 = 4 π r a d / s 2

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

k = 1 m

k 1 k 2 = I 1 I 2 = m R 2 / 2 m R 2 / 4 = 2 : 1

New answer posted

a month ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

X C M = 20 * 10 20 + 10 = 20 3 m

New answer posted

a month ago

0 Follower 2 Views

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

a month ago

0 Follower 2 Views

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

a month ago

0 Follower 3 Views

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

New answer posted

a month ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

I A B = 2 5 m a 2 * 2 + ( 2 5 m a 2 + m b 2 ) * 2 = 8 m a 2 5 + 2 m b 2

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