Question

A rod of length and mass m hinged at one end is released from rest from vertical position angular velocity of the rod is

Answer 1

Answer:

We know that Mg\frac { L }{ 2 } \quad =\quad \frac { 1 }{ 2 } I{ w }^{ 2 }

Where,

I\quad =\quad \frac { 1 }{ 3 } M{ L }^{ 2 }

{ w }^{ 2 }\quad =\quad \frac { 3g }{ L }

Average centripetal force = M\frac { L }{ 2 } { w }^{ 2 }\quad =\quad \frac { 3 }{ 2 } Mg

Net normal reaction at the hinge = weight + centripetal force

=\quad Mg\quad +\quad \frac { 3 }{ 2 } Mg\quad

=\quad \frac { 5 }{ 2 } Mg"

i Hope this will help uh

Answer 2

Answer:

w = 4.43/L

Explanation:

the other end falls freely and hence...

v^2 - u^2 = 2aS

v^2 - 0 = 2 × 9.8 × L

v = √(19.6L)

now...

v = r w

v = Lw

√(19.6L) = Lw

w = 4.43/L

Note :

w is the angular velocity