A string is wound around the rim of a mounted | vedclass

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(C) Given: Mass of the disc $M = 20 \, kg$,Radius $R = 20 \, cm = 0.2 \, m$,Force $F = 25 \, N$.The torque $	au$ applied to the flywheel is given by

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$12.5$

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(C) Given: Mass of the disc $M = 20 \, kg$,Radius $R = 20 \, cm = 0.2 \, m$,Force $F = 25 \, N$.The torque $	au$ applied to the flywheel is given by $	au = F 	imes R$.$	au = 25 \, N 	imes 0.2 \, m = 5 \, N \cdot m$.The moment of inertia $I$ of a disc about its central axis is $I = \frac{1}{2} M R^2$.$I = \frac{1}{2} 	imes 20 	imes (0.2)^2 = 10 	imes 0.04 = 0.4 \, kg \cdot m^2$.Using the rotational analogue of Newton's second law,$	au = I \alpha$,where $\alpha$ is the angular acceleration:$5 = 0.4 	imes \alpha$.$\alpha = \frac{5}{0.4} = 12.5 \, rad/s^2$.

$A$ string is wound around the rim of a mounted flywheel (disc) of mass $20 \, kg$ and radius $20 \, cm$. $A$ steady pull of $25 \, N$ is applied on the cord. Neglecting friction and the mass of the string,the angular acceleration of the wheel in $rad/s^2$ is

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