A sector cut from a uniform disk of radius 12 cm and a uniform rod of the same mass bent into shape

Question

A sector cut from a uniform disk of radius 12 cm and a uniform rod of the same mass bent into shape of an arc are arranged facing each other as shown in the figure. If center of mass of the combination is at the origin, what is the radius of the arc?

Accepted Answer

Sol. $x_{2} = \frac{2R_{2}}{	heta}\sin\frac{	heta}{2}$, $x_{1} = \frac{4R_{1}}{3	heta}\sin\frac{	heta}{2}$m1x1 = m2x2 ⇒ m. $\frac{4R_{1}}{3	heta}\sin\frac{\pi}{2} = m.\frac{2R_{2}}{	heta}\sin\frac{	heta}{2}$⇒ $R_{1} = \frac{3R_{2}}{2} \Rightarrow R_{2} = \frac{2R_{1}}{3}$⇒ R2 = 2 × $\frac{12}{3}$ = 8 cm

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