Two circular discs of radius 10 text{cm} eac | vedclass

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(B) The moment of inertia $I$ of the system about the axis $AB$ is the sum of the moments of inertia of the two discs and the rod.For each disc,the ax

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) The moment of inertia $I$ of the system about the axis $AB$ is the sum of the moments of inertia of the two discs and the rod.For each disc,the axis $AB$ is at a distance of $10 \ 	ext{cm}$ $(R)$ from the centre of the left disc and $20 \ 	ext{cm}$ $(2R)$ from the centre of the right disc.Using the parallel axis theorem,$I_{disc} = I_{cm} + md^2$.For the left disc: $I_1 = \frac{1}{4}mR^2 + mR^2 = \frac{5}{4}mR^2$.For the right disc: $I_2 = \frac{1}{4}mR^2 + m(2R)^2 = \frac{17}{4}mR^2$.For the rod of length $L = 3R = 30 \ 	ext{cm}$,the axis $AB$ passes through a point $10 \ 	ext{cm}$ from one end. The distance of the centre of the rod from $AB$ is $d = 5 \ 	ext{cm} = R/2$.$I_{rod} = I_{cm} + md^2 = \frac{mL^2}{12} + m(R/2)^2 = \frac{m(3R)^2}{12} + \frac{mR^2}{4} = \frac{9mR^2}{12} + \frac{mR^2}{4} = \frac{3mR^2}{4} + \frac{mR^2}{4} = mR^2$.Total $I = I_1 + I_2 + I_{rod} = \frac{5}{4}mR^2 + \frac{17}{4}mR^2 + mR^2 = \frac{22}{4}mR^2 + mR^2 = 5.5mR^2 + mR^2 = 6.5mR^2$.Given $m = 600 \ 	ext{g}$,$R = 10 \ 	ext{cm}$.$I = 6.5 	imes 600 	imes (10)^2 = 6.5 	imes 60000 = 390000 \ 	ext{g cm}^2 = 39 	imes 10^4 \ 	ext{g cm}^2$.Angular acceleration $\alpha = \frac{	au}{I} = \frac{43 	imes 10^5}{39 	imes 10^4} = \frac{430}{39} \approx 11.02 \ 	ext{rad/s}^2$.Thus,the angular acceleration is approximately $11 \ 	ext{rad/s}^2$.

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