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

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 (A) (m r2/2) (B) m r2 (C) m r2(1-(4/π2)) (D) m r2(1+(4/π2))

Tardigrade Admin · Accepted Answer

m r2=Ic m+m((2 r/π))2 Ic m=m r2[1-(4/π2)]

Q. 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

$\frac{m r ^{2}}{2}$

$m r^{2}$

$m r^{2} (1 - \frac{4}{π ^{2}})$

$m r^{2} (1 + \frac{4}{π ^{2}})$

$m r^{2} = I_{c m} + m (\frac{2 r}{π})^{2}$
$I_{c m} = m r^{2} [1 - \frac{4}{π ^{2}}]$

Q. 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

2mr2​

mr2

mr2(1−π24​)

mr2(1+π24​)