A uniform circular disc of radius 50, cm at r | vedclass

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(D) Given: Radius $r = 50\, cm = 0.5\, m$,angular acceleration $\alpha = 2.0\, rad\, s^{-2}$,initial angular velocity $\omega_0 = 0$,and time $t = 2.0

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$8$

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(D) Given: Radius $r = 50\, cm = 0.5\, m$,angular acceleration $\alpha = 2.0\, rad\, s^{-2}$,initial angular velocity $\omega_0 = 0$,and time $t = 2.0\, s$.First,calculate the angular velocity $\omega$ at $t = 2.0\, s$ using $\omega = \omega_0 + \alpha t = 0 + (2.0)(2.0) = 4.0\, rad\, s^{-1}$.The tangential acceleration is $a_t = r\alpha = 0.5 	imes 2.0 = 1.0\, m\, s^{-2}$.The radial (centripetal) acceleration is $a_r = \omega^2 r = (4.0)^2 	imes 0.5 = 16 	imes 0.5 = 8.0\, m\, s^{-2}$.The net acceleration $a$ is given by $a = \sqrt{a_t^2 + a_r^2} = \sqrt{1.0^2 + 8.0^2} = \sqrt{1 + 64} = \sqrt{65} \approx 8.06\, m\, s^{-2}$.Rounding to the nearest value,the net acceleration is approximately $8.0\, m\, s^{-2}$.

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