Question

dfuddatstsitstss885w5w8sts​

Answer 1

Answer:

thanks for free point thank you

Answer 2

Answer:

The loss in gravitational potential energy of the rod's center of mass is equal to the gain in rotational kinetic energy of the rod.

Hence mg

2

L

=

2

1

Iω

2

=

2

1

(

3

mL

2

)ω

2

⟹ω=

L

3g