Question 641: Centre of Mass and Rotational Motion

Question

A light rod of length $$l$$ has two masses m1 and m2 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 m1 and m2 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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