# Three identical thin rods each of length L and mass M are joined together to form a letter H. The moment of inertia of the system about one of the outher sides of H is :-

Kuldeep Pal
53 Points
7 years ago
Divide The Three rods Of H as:
H=I1  +  --2  + I3.
We Assume That The Moment Of Inertia Is About I1 ,
Now,
The Moment Of Inertia Of I1=0
By Applying Theorem Of Perpendicular Axes , The Moment Of Inertia Of  --= (ml2)/3.
By Applying Theorem Of Parallel Axes, The Moment Of inertia Of I3 =  0(for I1) + ml2(for I3 at lenght L)  = ml2.
So, I(Moment of Inertia of H) = 0 + (ml2)/3 + ml2 = (4/3)ml2.
3 years ago
Dear student,

Divide The Three rods Of H as:
H = I1  + I2  + I3.
We Assume That The Moment Of Inertia Is About I1 ,
Now,
The Moment Of Inertia Of I1 = 0
By Applying Theorem Of Perpendicular Axes , The Moment Of Inertia Of  I2 = (mL2)/3.
By Applying Theorem Of Parallel Axes, The Moment Of inertia Of I3 =  0(for I1) + mL2(for I3 at lenght L)  = mL2.
So, I(Moment of Inertia of H) = 0 + (mL2)/3 + mL2 = (4/3)mL2.

Thanks and regards,
Kushagra