System Of Particles And Rotational Motion Ques 74
74. If the linear density (mass per unit length) of a rod of length $3$ $ m$ is proportional to $x$, where $x$ is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is
[2002]
(a) $2.5$ $ m$
(b) $1$ $ m$
(c) $1.5$ $ m$
(d) $2$ $ m$
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Answer:
Correct Answer: 74.(d)
Solution:
- (d) Consider an element of length $d x$ at a distance $x$ from end $A$.
Here, mass per unit length $\lambda$ of rod
$\lambda \propto x \Rightarrow \lambda=k x$
$\therefore d m=\lambda d x=k x d x$
Position of centre of gravity of rod from end $A$.
$ x _{C G}=\frac{\int_0^{L} xdm }{\int_0^{L} dm}$
$ \therefore x _{C G}=\frac{\int_0^{3} x(k x d x)}{\int_0^{3} k x d x}=\frac{[\frac{x^{3}}{3}]_0^{3}}{[\frac{x^{2}}{2}]_0^{3}}=\frac{\frac{(3)^{3}}{3}}{\frac{(3)^{3}}{2}}=2 $ $m $
BETA
