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.

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

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

Answer 2

The moment of inertia of the system about an axis perpendicular to the rod and passing through the center of the rod is 1/2mL².

How to calculate the moment of inertia?

It should be noted that the moment of inertia of the system about an axis perpendicular to the rod and passing through the center of the rod will be calculated thus:

I = 1/2 × mL²/2

I = 1/2mL²

Also, the moment of inertia of the system about an axis perpendicular to the rod and passing through a point one-fourth of the length from one end will be:

I = 2 × m × (L/4)² + m(3L/4)²

I = (2 + 9)1/16mL²

I = 11/16mL²

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