Question

Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point one-fourth of the length from one end

Answer 1

Answer:

Answered

Explanation:

a) Two balls are at a distance of L/2 from the axis of rotation and one block at the center. ( center of rod).

therefore,

$I=2\times m\frac{L}{2}^2$

$I= \frac{1}{2}mL^2$

b) two balls at a distance L/4 at the from the axis and 1 ball at a distance 3L/4 from the from the axis.

$I= 2\times m\times(L/4)^2 + m(\frac{3L}{4})^2$

= $\frac{1}{16} mL^2(2+9)= \frac{11}{16}mL^2$