In the given figure,r_1 and r_2 are 10 cm an | vedclass

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(D) The net torque $	au$ acting on the wheel is the sum of the torques produced by the individual forces. Taking counter-clockwise as positive:$	au

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

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(D) The net torque $	au$ acting on the wheel is the sum of the torques produced by the individual forces. Taking counter-clockwise as positive:$	au = (5 \ N 	imes r_2) + (10 \ N 	imes r_2) - (15 \ N 	imes r_1)$Given $r_1 = 10 \ cm = 0.1 \ m$ and $r_2 = 20 \ cm = 0.2 \ m$,and $I = 1500 \ kg \cdot m^2$.$	au = (5 + 10) 	imes 0.2 - (15 	imes 0.1)$$	au = (15 	imes 0.2) - 1.5 = 3.0 - 1.5 = 1.5 \ N \cdot m$Using the relation $	au = I\alpha$:$\alpha = \frac{	au}{I} = \frac{1.5}{1500} = \frac{1.5}{1.5 	imes 10^3} = 10^{-3} \ rad/s^2$.

In the given figure,$r_1$ and $r_2$ are $10 \ cm$ and $20 \ cm$ respectively. If the moment of inertia of the wheel is $1500 \ kg \cdot m^2$,find the angular acceleration.

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