The moment of inertia of a solid flywheel abo | vedclass

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(B) Let $M = 2\,kg$ be the mass suspended,$R = 0.1\,m$ be the radius of the flywheel,and $I = 0.1\,kg\cdot m^2$ be its moment of inertia.Let $a$ be th

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$16.3$

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(B) Let $M = 2\,kg$ be the mass suspended,$R = 0.1\,m$ be the radius of the flywheel,and $I = 0.1\,kg\cdot m^2$ be its moment of inertia.Let $a$ be the linear acceleration of the mass and $T$ be the tension in the string.The equation of motion for the falling mass is: $Mg - T = Ma$  ...$(1)$The torque $	au$ applied to the flywheel is $	au = T \cdot R = I \alpha$,where $\alpha$ is the angular acceleration.Since the string does not slip,$a = R \alpha$,so $T = \frac{I \alpha}{R}$  ...$(2)$Substituting $T$ and $a$ into equation $(1)$:$Mg - \frac{I \alpha}{R} = M(R \alpha)$$Mg = \alpha \left( \frac{I}{R} + MR ight) = \alpha \left( \frac{I + MR^2}{R} ight)$$\alpha = \frac{MgR}{I + MR^2}$Substituting the values: $\alpha = \frac{2 	imes 9.8 	imes 0.1}{0.1 + 2 	imes (0.1)^2} = \frac{1.96}{0.1 + 0.02} = \frac{1.96}{0.12} \approx 16.33\,rad/s^2$.

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