Question

A semicircular plate of mass $m$ has radius $r$ and centre $c$. The centre of mass of the plate is at a distance $x$ from its centre $c$. Its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is

• A. $\frac{mr^2}{2}$
• B. $\frac{mr^2}{4}$
• C. $\frac{mr^2}{2} + mx^2$
• D. $\frac{mr^2}{2} - mx^2$

Answer 1

Answer: (D)

$\frac{mr^2}{2} - mx^2$

Answer 2

Given,
mass of a semicircular plate $=m$
radius of semicircular plate $=r$
According to the question we can drawn the following diagram,

Now, from the parallel axis theorem,
$I_{c} =I_{ cm }+m x^{2}$
$I_{ cm } =I_{c}-m x^{2}$
$I_{c} =\frac{m r^{2}}{2}$
Hence, moment of inertia of semicircular plate about an axis passing through its centre of mass and perpendicular to it's plane is
$I_{ cm }=\frac{m r^{2}}{2}-m x^{2}$