# a uniform rod of length l,hinged at lower end is free to rotate in the vertical plane.if the rod is held vertically in the beginning and then released,the angular acceleration of the rod when it makes and angle 45 deg. with horizontal is?

148 Points
12 years ago

Dear tushar

When it makes 45 degree with the horizontal then at this position torque acting on rod is =mgl/2 cos45

=mgl/2√2

We know that

Torque=moment of inertia * angular acceleration

mgl/2√2  =ml2/3  * angular acceleration         where m is the mass of rod

angular acceleration =3g/2√2 l

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Rishi Sharma
one year ago
Dear Student,

When it makes 45 degree with the horizontal then at this position torque acting on rod is
=mgl/2 cos45
=mgl/2√2
We know that
Torque=moment of inertia * angular acceleration
= mgl/2√2
= ml2/3 * angular acceleration
where m is the mass of rod
angular acceleration =3g/2√2 l

Thanks and Regards