Question 509: Centre of Mass and Rotational Motion

Question

Four particles each of mass M are placed at the corners of a square of side $L$. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

Accepted Answer

Let the four particles each of mass $M$ be placed at the corners of a square of side $L$.
Step 1: Calculate the distance of each particle from the centre
The centre of the square is equidistant from all corners.
The distance from the centre to a corner (letÃ¢ÂÂs call it $r$):
$$ r = 	ext{half the diagonal of the square} = \frac{L}{\sqrt{2}} $$
Step 2: Calculate the moment of inertia about the given axis
The axis is perpendicular to the square and passes through its centre.
Moment of inertia for each mass about the axis:
$$ I_{	ext{each}} = M r^2 = M \left(\frac{L}{\sqrt{2}}ight)^2 = M \cdot \frac{L^2}{2} $$
For 4 masses:
$$ I_{	ext{total}} = 4 \cdot \frac{M L^2}{2} = 2 M L^2 $$
Step 3: Total mass of the system
$$ M_	ext{total} = 4M $$
Step 4: Formula for radius of gyration $K$
The moment of inertia is also given by: $I = M_{	ext{total}} K^2$
So,
$$ I_{	ext{total}} = M_{	ext{total}} K^2 \implies 2 M L^2 = 4 M K^2 $$
Step 5: Solve for $K$
$$ 4 M K^2 = 2 M L^2 \ K^2 = \frac{2 M L^2}{4 M} \ K^2 = \frac{L^2}{2} \ K = \frac{L}{\sqrt{2}} $$
Correct option:
$\boxed{\frac{L}{\sqrt{2}}}$
So, the answer is
Option D
.

Answer

$\mathrm{L}$

Answer

$\frac{\mathrm{L}}{2}$

Answer

$\frac{\mathrm{L}}{4}$

Answer

$\frac{\mathrm{L}}{\sqrt{2}}$

Centre of Mass and Rotational Motion

Four particles each of mass M are placed at the corners of a square of side L. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

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