Question

a uniform thin rod of length L and mass M is hinged at one end and released on horizontal position as shown in the figure the angular acceleration of the rod about the hinge. When rod becomes vertical is​

Answer 1

We know that the moment of inertia of a rod of mass m and length l about its center of mass is ml² / 12.

So, by parallel axes theorem, moment of inertia of the rod about the hinge point O is,

I = ml² / 12 + ml² / 4

I = ml² / 3

And the torque acting on the rod about the hinge point is,

τ = mgl / 2,

because force acting on the rod about the center of mass is its weight and the perpendicular distance between the center of mass and axis of rotation is l / 2.

We have,

τ = Iα

α = τ / I

α = (mgl / 2) / (ml² / 3)

α = 3g / 2l

Hence the answer is (4).

Answer 2

Answer:

We know that the moment of inertia of a rod of mass m and length l about its center of mass is ml² / 12.

So, by parallel axes theorem, moment of inertia of the rod about the hinge point O is,

I = ml² / 12 + ml² / 4

I = ml² / 3

And the torque acting on the rod about the hinge point is,

τ = mgl / 2,

because force acting on the rod about the center of mass is its weight and the perpendicular distance between the center of mass and axis of rotation is l / 2.

We have,

τ = Iα

α = τ / I

α = (mgl / 2) / (ml² / 3)

α = 3g / 2l

Explanation: