Question

Four equal masses m are so small they can be treated as points, and they are equally spaced along a long, stiff wire of neglible mass. The distance between any two adjacent masses is ℓ . m m m m ℓ ℓ ℓ What is the rotational inertia Icm of this system about its center of mass?
A.) (1/2)ml^2
B.) ml^2
C.) 5ml^2
D.) 6ml^2
E.) 2ml^2
F.) 3ml^2
G.) 7ml^2
H.) 4ml^2.

Answer 1

Answer:C

Explanation:

Given

Four masses are attached to the wire such that distance between two mass is L

therefore the Length of wire is 4 L

and the center of mass is at 2L

moment of inertia is distribution of mass from its rotational axis

thus moment of Inertia I is given by

$I=m\times (\frac{3L}{2})^2+m\times (\frac{L}{2})^2+m\times (\frac{3L}{2})^2+m\times (\frac{L}{2})^2$

$I=2\times m\times (\frac{L}{2})^2+2\times m\times (\frac{3L}{2})^2$

$I=\frac{2mL^2}{4}+\frac{18mL^2}{4}$

$I=5mL^2$