Question 513: Centre of Mass and Rotational Motion

Question

Two loops P and Q of radii $\mathrm{R}_1$ and $\mathrm{R}_2$ are made from uniform metal wire of same material. $I_p$ and $\mathrm{I}_{\mathrm{Q}}$ be the moment of inertia of loop P and Q respectively then ratio $R_1 / R_2$ is $\left(\right.$ Given $\left.I_P / I_Q=27\right)$

Accepted Answer

Two loops made from same material
$\mathrm{I}=\mathrm{MR}^2\quad 	ext{...(i)}$
$\mathrm{M=\sigma .2 \pi R}\quad$ 
...($\sigma$ is mass per unit length)
Substituting in (i),
$\mathrm{I}=\sigma .2 \pi \mathrm{R}^3\quad	ext{.... (ii)}$
$	herefore \quad \frac{\mathrm{I}_{\mathrm{p}}}{\mathrm{I}_{\mathrm{Q}}}=\frac{\sigma \cdot 2 \pi \mathrm{R}_1^3}{\sigma \cdot 2 \pi \mathrm{R}_2^3}\quad 	ext{...[From(ii)]}$
$$\begin{aligned}
\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{Q}}} & =\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}ight)^3 \
\frac{27}{1} & =\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}ight)^3 \quad \ldots\left(	ext { given, } \frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{Q}}}=27ight) \
	herefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2} & =\frac{3}{1}
\end{aligned}$$

Answer

$4: 1$

Answer

$3: 1$

Answer

$9: 1$

Answer

$6: 1$

Centre of Mass and Rotational Motion

Two loops P and Q of radii \mathrm{R}_1 and \mathrm{R}_2 are made from uniform metal wire of same material. I_p and \mathrm{I}_{\mathrm{Q}} be the moment of inertia of loop P and Q respectively then ratio R_1 / R_2 is \left(\right. Given \left.I_P / I_Q=27\right)

Choose the correct answer: