Question 56: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

The given situation is shown in figure.
Moment of inertia of the systems about an axis perpendicular to the square and passing through the centre is given as
$$I = m{\left( {{l \over {\sqrt 2 }}} ight)^2} + m{\left( {{l \over {\sqrt 2 }}} ight)^2} + m{\left( {{l \over {\sqrt 2 }}} ight)^2} + m{\left( {{l \over {\sqrt 2 }}} ight)^2}$$
$$ = 4{{m{l^2}} \over 2} = 2m{l^2}$$ ..... (i)
If $$k$$ be the radius of gyration, then
$$I=4mk^2$$ ..... (ii)
$$	herefore$$ From Eqs. (i) and (ii), we have
$$2ml^2=4mk^2$$
$$\Rightarrow k=\frac{l}{\sqrt2}$$

Answer

$$\frac{l}{\sqrt2}$$

Answer

$$\frac{l}{2}$$

Answer

$$l$$

Answer

$$\sqrt2l$$

Centre of Mass and Rotational Motion

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

Choose the correct answer: