Question

5. A uniform rod of length '6L' and mass '8m' is pivoted at its centre 'C'. Two masses 'm' and 2m with speed 2v and v (as shown) strike the rod and stick to the rod. Initially the rod is at rest. Due to impact, if it rotates with angular velocity o' then 'o' will be [MHT-CET 2019)​

Answer 1

Answer:

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Step-by-step explanation:

As F

ext

=0, so linear momentum of the system is conserved, Thus

−2mv+m×2v+0=(2m+m+8m)v

c

or

v

c

=0

Now as τ

ext

=0, angular momentum of the system is also conserved, Thus

m

1

v

1

r

1

+m

2

v

2

r

2

=(I

1

+I

2

+I

3

)ω

or

2mva+m(2v)(2a)=(2ma

2

+m(2a)

2

+

12

8m×(6a)

2

)ω

or

6mva=30mωa

2

or

ω=

5a

v

Now as the system has no translatory motion but only rotatory motion we have

E=

2

1

Iω

2

=

2

1

(30ma

2

)(

2a

v

)

2

=

5

3

mv

2