Question 398: Centre of Mass and Rotational Motion

Question

Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ and moment of intertia $I_1$ and $I_2$, respectively. For $R_2=2 R_1$, the ratio of $I_1$ and $I_2$ would be $1 / x$, where $\mathrm{x}=$ _______ .

Accepted Answer

Let surface mass density = $\sigma$
So, $${M_1} = \sigma  	imes \pi R_1^2$$
$${M_2} = \sigma  	imes \pi R_2^2 = \sigma \pi {(2{R_1})^2}$$ (As $${R_2} = 2{R_1}$$
$$ = 4\sigma \pi R_1^2$$
$$ \Rightarrow {M_2} = 4{M_1}$$
$${{{I_1}} \over {{I_2}}} = {{{{{M_1}R_1^2} \over 2}} \over {{{{M_2}R_2^2} \over 2}}} = \left( {{{{M_1}} \over {{M_2}}}} ight){\left( {{{{R_1}} \over {{R_2}}}} ight)^2}$$
$$ \Rightarrow {{{I_1}} \over {{I_2}}} = \left( {{{{M_1}} \over {4{M_1}}}} ight){\left( {{{{R_1}} \over {2{R_1}}}} ight)^2} = {1 \over 4} 	imes {1 \over 4} = {1 \over {16}} = {1 \over x}$$
Hence, $$x = 16$$

Centre of Mass and Rotational Motion

Two iron solid discs of negligible thickness have radii R_1 and R_2 and moment of intertia I_1 and I_2, respectively. For R_2=2 R_1, the ratio of I_1 and I_2 would be 1 / x, where \mathrm{x}= _______ .

Integer / Numeric Answer Type