Rotational Motion Question 174
Question: If the linear density (mass per unit length) of a rod of length 3m 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
Options:
A) $ 2.5 $ m
B) $ 1 $ m
C) $ 1.5 $ m
D) $ 2 $ m
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Consider an element of length $d x$ at a distance $x$ from end $\mathrm{A}$.
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 x d m}{\int_0^L d m} $
$ \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}^3}{\frac{x^2}{2}_0^3}=\frac{\frac{(3)^3}{3}}{\frac{(3)^3}{2}}=2 m $
BETA
