Question 640: Centre of Mass and Rotational Motion

Question

A light rod of length $$l$$ has two masses m
1
and m
2
attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

Accepted Answer

Here, $${l_1} + {l_2} = l$$
Centre of mass of the system,
$${l_1} = {{{m_1} 	imes 0 + {m_2} 	imes l} \over {{m_1} + {m_2}}} = {{{m_2}l} \over {{m_1} + {m_2}}}$$
$${l_2} = l - {l_1} = {{{m_1}l} \over {{m_1} + {m_2}}}$$
Required moment of inertia of the system,
$$I = {m_1}l_1^2 + {m_2}l_2^2$$
$$ = \left( {{m_1}m_2^2 + {m_2}m_1^2} ight){{{l^2}} \over {{{\left( {{m_1} + {m_2}} ight)}^2}}}$$
$$ = {{{m_1}{m_2}\left( {{m_1} + {m_2}} ight){l^2}} \over {{{\left( {{m_1} + {m_2}} ight)}^2}}} = {{{m_1}{m_2}} \over {{m_1} + {m_2}}}{l^2}$$

Answer

$${{{m_1}{m_2}} \over {{m_1} + {m_2}}}{l^2}$$

Answer

$${{{m_1} + {m_2}} \over {{m_1}{m_2}}}{l^2}$$

Answer

$$\left( {{m_1} + {m_2}} \right){l^2}$$

Answer

$$\sqrt {{m_1}{m_2}} \,{l^2}$$

Centre of Mass and Rotational Motion

A light rod of length $l$ has two masses m 1 and m 2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

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