question_answer

A semicircular lamina of mass m, radius r and centre at C is shown in the figure. Its centre of mass is at a distance x from C. Its moment of inertia about an axis through its centre of mass and perpendicular to its plane is

A) \[\frac{1}{2}m{{r}^{2}}\]       

B) \[\frac{1}{4}m{{r}^{2}}\]

C) \[\frac{1}{2}m{{r}^{2}}+m{{x}^{2}}\]        

D) \[\frac{1}{2}m{{r}^{2}}-m{{x}^{2}}\]

Correct Answer: D

Solution :

[d]: We know, \[{{I}_{C}}=m{{r}^{2}}/2\]        Using parallel axes theorem,  \[{{I}_{C}}={{I}_{CM}}+m{{x}^{2}}\] \[\therefore \]\[{{I}_{CM}}={{I}_{C}}-m{{x}^{2}}=m{{r}^{2}}/2-m{{x}^{2}}\]