Rotational Motion Question 207
Question: The moment of inertia of a uniform semicircular wire of mass m and radius r, about an axis passing through its centre of mass and perpendicular to its plane is $mr^{2}\left( 1-\frac{k}{{{\pi }^{2}}} \right)$ then find the value of k.
Options:
A) 2
B) 3
C) 4
D) 5
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Moment of inertia about z-axis,
$I _{z}=mr^{2}$
(about centre of mass) Applying parallel axes theorem,
$I _{z}=I _{cm}+mk^{2}$
$I _{cm}=I _{z}-m{{\left( \frac{2}{\pi }r \right)}^{2}}$
$=mr^{2}-\frac{m4r^{2}}{{{\pi }^{2}}}=mr^{2}\left( 1-\frac{4}{{{\pi }^{2}}} \right)$
i.e., k=4
