Two masses each of mass M are attached to the end of a rigid massless rod of length L. The moment of inerita of the system about an axis passing centre of mass and perpendicular to its length is
Answers
Answered by
1
Answer:
Answer -
Length = \frac{L}{4}
4
L
Let the axis at length L/4 be AB.
We know that the moment of inertia of uniform rod with an axis passing through its middle is \frac{ {ML}^{2}}{12}.
12
ML
2
.
So, I_{AB} = I_{CD} + {ML}^{2}I
AB
=I
CD
+ML
2
I_{AB}= \frac{ {ML}^{2}}{12} +\frac{ {ML}^{2}}{16}I
AB
=
12
ML
2
+
16
ML
2
I_{AB}= \frac { {4ML}^{2} + {3ML}^{2}}{48}I
AB
=
48
4ML
2
+3ML
2
\boxed {I_{AB}= \frac {{7ML}^{2}}{48}}
I
AB
=
48
7ML
2
Explanation:
Please make me a brainlist answer and like and follow please
Similar questions