Question 673: Centre of Mass and Rotational Motion

Question

Four identical thin rods each of mass M and length $$l$$, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

Accepted Answer

As per theorem of parallel axes :
I = I
cm
+ Ma
2
where I
cm
= moment of inertia about an axis through centre of mass of rod
I = moment of inertia about parallel axis at distance Ã¢ÂÂaÃ¢ÂÂ
Now moment of inertia of thin rod of mass M and
length L about an axis through the midpoint of the
rod and perpendicular to its length is ML
2
/12.
The moment of inertia of a rod about parallel axis at
distance L/2 will be :
ML
2
/12 + M(L/2)
2
= ML
2
/3
As there are 4 rods in square frame, so moment of inertia of entire frame is 4 times the moment of inertia of 1 rod, so it will be (4/3) ML
2

Answer

$${2 \over 3}M{l^2}$$

Answer

$${{13} \over 3}M{l^2}$$

Answer

$${1 \over 3}M{l^2}$$

Answer

$${4 \over 3}M{l^2}$$

Centre of Mass and Rotational Motion

Four identical thin rods each of mass M and length $l$, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

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