Question 210: Centre of Mass and Rotational Motion

Question

The linear mass density of a thin rod AB of length L varies from A to B as
$$\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right)$$,  where
x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing
through A and perpendicular to the rod is :

Accepted Answer

dm = $$\lambda $$dx
= $${\lambda _0}\left( {1 + {x \over L}} \right)$$dx
Integrating both side, we get
$$\int\limits_0^M {dm}  = \int\limits_0^L {{\lambda _0}\left( {1 + {x \over L}} \right)} dx$$
$$ \Rightarrow $$ M = $${\lambda _0}L + {{{\lambda _0}{L^2}} \over {2L}}$$ = $${{3{\lambda _0}L} \over 2}$$
$$ \Rightarrow $$ $${\lambda _0} = {{2M} \over {3L}}$$ .....(1)
Moment of inertia of small part dx is
dI = dmx
2
Integrating both side, we get
$$\int\limits_0^I {dI}  = \int\limits_0^L {dm{x^2}} $$
$$ \Rightarrow $$ I = $$\int\limits_0^L {{\lambda _0}\left( {1 + {x \over L}} \right)} dx\,{x^2}$$
= $${\lambda _0}\int\limits_0^L {\left( {{x^2} + {{{x^3}} \over L}} \right)} dx$$
= $${\lambda _0}\left[ {{{{L^3}} \over 3} + {{{L^3}} \over 4}} \right]$$
= $${{7{L^3}{\lambda _0}} \over {12}}$$
Here by putting the value of $$\lambda $$
0
form (1)
I = $${{7{L^3}} \over {12}}\left( {{{2M} \over {3L}}} \right)$$ = $${7 \over {18}}M{L^2}$$

Answer

$${2 \over 5}M{L^2}$$

Answer

$${5 \over {12}}M{L^2}$$

Answer

$${7 \over {18}}M{L^2}$$

Answer

$${3 \over 7}M{L^2}$$

Centre of Mass and Rotational Motion

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