Question 489: 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

Step 1: Recall definition of radius of gyration
Radius of gyration is defined by:
$ k = \sqrt{\frac{I}{M_{	ext{total}}}} $
where $I$ = moment of inertia of the system about the given axis, and $M_{	ext{total}}$ is the total mass.
Step 2: Distance of each mass from axis
The axis is perpendicular to the square and passes through its centre.
So, the distance of each mass (corner of square) from centre = half of the diagonal =
$ r = \frac{\sqrt{2}}{2} L = \frac{L}{\sqrt{2}} $
Step 3: Moment of inertia
Each mass contributes $M r^2$ where $r = L/\sqrt{2}$.
So,
$ I = 4 \cdot M \left(\frac{L}{\sqrt{2}}ight)^2 = 4M \cdot \frac{L^2}{2} = 2 M L^2 $
Step 4: Total mass
$ M_{	ext{total}} = 4M $
Step 5: Radius of gyration
$ k = \sqrt{ \frac{I}{M_{	ext{total}}} } = \sqrt{ \frac{2 M L^2}{4M} } = \sqrt{\frac{L^2}{2}} = \frac{L}{\sqrt{2}} $
Ã¢ÂÂ Final Answer:
$ \boxed{\frac{L}{\sqrt{2}}} $
That corresponds to
Option B
.

Answer

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

Answer

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

Answer

$\quad 2 \mathrm{~L}$

Answer

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

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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