In the following figure,r_1 = 5 , cm and r_2 | vedclass

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(B) Given: $r_1 = 0.05 \, m$,$r_2 = 0.30 \, m$,$I = 5100 \, kg \cdot m^2$.Forces acting on the wheel:$1$. $A$ force of $10 \, N$ acts tangentially at

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$10^{-3} \, rad/s^2$

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(B) Given: $r_1 = 0.05 \, m$,$r_2 = 0.30 \, m$,$I = 5100 \, kg \cdot m^2$.Forces acting on the wheel:$1$. $A$ force of $10 \, N$ acts tangentially at $r_2$ (clockwise torque).$2$. $A$ force of $9 \, N$ acts tangentially at $r_2$ (clockwise torque).$3$. $A$ force of $12 \, N$ acts at $r_1$ at an angle of $30^\circ$ with the horizontal. The perpendicular component is $12 \sin(30^\circ) = 12 	imes 0.5 = 6 \, N$ (counter-clockwise torque).Net torque $	au_{net} = (10 	imes 0.30) + (9 	imes 0.30) - (6 	imes 0.05) = 3.0 + 2.7 - 0.3 = 5.4 \, N \cdot m$.Using $	au_{net} = I \alpha$:$5.4 = 5100 	imes \alpha$$\alpha = \frac{5.4}{5100} = \frac{54}{51000} \approx 1.058 	imes 10^{-3} \, rad/s^2$.Given the options,the closest value is $10^{-3} \, rad/s^2$.

In the following figure,$r_1 = 5 \, cm$ and $r_2 = 30 \, cm$. If the moment of inertia of the wheel is $5100 \, kg \cdot m^2$,then its angular acceleration will be:

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