As shown in the figure,a mass m = 500 g hang | vedclass

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(D) Given: $m = 0.5 \ kg$,$r = 0.2 \ m$,$s = 2.0 \ m$,$t = 8 \ s$,$g = 10 \ m/s^2$. First,calculate the acceleration $a$ of the falling mass using the

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$3.18$

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(D) Given: $m = 0.5 \ kg$,$r = 0.2 \ m$,$s = 2.0 \ m$,$t = 8 \ s$,$g = 10 \ m/s^2$. First,calculate the acceleration $a$ of the falling mass using the kinematic equation $s = ut + \frac{1}{2}at^2$. Since it starts from rest,$u = 0$. $2.0 = 0 + \frac{1}{2} \cdot a \cdot (8)^2 \Rightarrow 2.0 = 32a \Rightarrow a = \frac{2}{32} = \frac{1}{16} \ m/s^2$. Now,consider the forces on the mass: $mg - T = ma$,where $T$ is the tension in the string. $T = m(g - a) = 0.5 \cdot (10 - \frac{1}{16}) = 0.5 \cdot (\frac{160 - 1}{16}) = 0.5 \cdot \frac{159}{16} = \frac{159}{32} \ N$. The torque $	au$ on the wheel is $T \cdot r = I \cdot \alpha$,where $\alpha = \frac{a}{r}$. $T \cdot r = I \cdot \frac{a}{r} \Rightarrow I = \frac{T \cdot r^2}{a}$. Substituting the values: $I = \frac{(159/32) \cdot (0.2)^2}{1/16} = \frac{159}{32} \cdot 0.04 \cdot 16 = \frac{159 \cdot 0.04}{2} = 159 \cdot 0.02 = 3.18 \ kg \cdot m^2$.

As shown in the figure,a mass $m = 500 \ g$ hangs from the rim of a wheel of radius $r = 20 \ cm$. When released from rest,the mass falls $2.0 \ m$ in $8 \ s$. Then the moment of inertia of the wheel is .......... $kg \cdot m^2$. $(g = 10 \ m/s^2)$

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