Question 244: Centre of Mass and Rotational Motion

Question

The radius of gyration of a uniform rod of length $$l$$, about an axis passing through a
point $${l \over 4}$$ away from the centre of the rod,
and perpendicular to it, is :

Accepted Answer

I = $${{M{l^2}} \over {12}} + M{\left( {{l \over 4}} \right)^2}$$
$$ \Rightarrow $$ I = $${{7M{l^2}} \over {48}}$$
$$ \Rightarrow $$ MK
2
= $${{7M{l^2}} \over {48}}$$
$$ \Rightarrow $$ K = $$\sqrt {{7 \over {48}}} l$$

Answer

$${1 \over 8}l$$

Answer

$${1 \over 4}l$$

Answer

$$\sqrt {{7 \over {48}}} l$$

Answer

$$\sqrt {{3 \over 8}} l$$

Centre of Mass and Rotational Motion

The radius of gyration of a uniform rod of length $l, about an axis passing through a point {l \over 4}$ away from the centre of the rod, and perpendicular to it, is :

Choose the correct answer: