Rotational Motion Question 214
Question: A particle of mass m is attached to q a thin uniform rod of length a and mass 4 m. The distance of the particle from the centre of mass of the rod is a/4. The moment of inertia of the combination about an axis passing through 0 normalto the rod is
Options:
A) $\frac{64}{48}ma^{2}$
B) $\frac{91}{48}ma^{2}$
C) $\frac{27}{48}ma^{2}$
D) $\frac{51}{48}ma^{2}$
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Moment of inertia $=m{{\left( \frac{3a}{4} \right)}^{2}}+m _{1}\frac{a^{2}}{3}$
For the centre of rod $\left( \frac{m _{1}a^{2}}{12}+\frac{m _{1}a^{2}}{4} \right)=\frac{m _{1}a^{2}}{3}$
$\therefore m _{1}=4m$
Total $I=m{{\left( \frac{3a}{4} \right)}^{2}}+\frac{4ma^{2}}{3}$ $=\frac{9ma^{2}}{16}+\frac{4ma^{2}}{3}$ $=\frac{(27+64)}{48}ma^{2}=\frac{91}{48}ma^{2}$
