Question

A thin wire of uniform density is bend in the form of an arc of a circle which subtends an angle $\alpha$ at the centre of the circle as shown in the figure. Where is the centre of mass of this circular arc located?

• A. $(0,0)$
• B. $(\frac{2R\sin \frac{\alpha }{2}}{\alpha },0)$
• C. $(\frac{2R\cos \frac{\alpha }{2}}{\alpha },\frac{2R\cos \frac{\alpha }{2}}{\alpha })$
• D. $(0,\frac{2R\sin \frac{\alpha }{2}}{\alpha })$

Answer 1

The correct option is D $(0,\frac{2R\sin \frac{\alpha }{2}}{\alpha })$

The wire is symmetric about the $Y-$ axis.
Hence $X_{COM}$ lies on the centre.

Let the mass of the wire be $M$.
The ratio of the mass of the element of the wire and the length it subtends on the arc remains constant.
Let $dm$ be the mass of the wire that an angle $d\theta$ subtends on the arc.
Then, we get, $\frac{dM}{Rd\theta }=\frac{M}{R\alpha }$
$\Rightarrow dM=\frac{Md\theta }{\alpha }$
The $y$ co-ordinate of $dm=R\sin \theta$.
$\therefore$ We get, $Y_{COM}=\frac{\int ydm}{\int dm}$
$\Rightarrow Y_{COM}=\frac{\int R\sin \theta .\frac{Md\theta }{\alpha }}{M}$
$\Rightarrow Y_{COM}=\frac{R}{\alpha }\times \underset{90-\frac{\alpha }{2}}{\overset{90+\frac{\alpha }{2}}{\int }}\sin \theta d\theta$
$\Rightarrow Y_{COM}=\frac{R}{\alpha }{[-\cos \theta ]}_{90-\frac{\alpha }{2}}^{90+\frac{\alpha }{2}}$
$\Rightarrow Y_{COM}=\frac{R}{\alpha }[-(\cos 90+\frac{\alpha }{2}-\cos 90-\frac{\alpha }{2})]$
$\Rightarrow Y_{COM}=\frac{R}{\alpha }[-(-\sin \frac{\alpha }{2}-\sin \frac{\alpha }{2})]$
$\Rightarrow Y_{COM}=\frac{R}{\alpha }\times 2\sin \frac{\alpha }{2}$
$\Rightarrow Y_{COM}=\frac{2R\sin \frac{\alpha }{2}}{\alpha }$
Hence the correct option is (d).