Two masses 400 g and 350 g are suspended fr | vedclass

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(A) Given: $m_1 = 0.4 \ kg$,$m_2 = 0.35 \ kg$,$R = 0.02 \ m$,$s = 0.81 \ m$,$t = 9 \ s$,$g = 9.8 \ m/s^2$.Using the kinematic equation $s = ut + \frac

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$9.5 	imes 10^{-3}$

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(A) Given: $m_1 = 0.4 \ kg$,$m_2 = 0.35 \ kg$,$R = 0.02 \ m$,$s = 0.81 \ m$,$t = 9 \ s$,$g = 9.8 \ m/s^2$.Using the kinematic equation $s = ut + \frac{1}{2}at^2$,where $u = 0$:$0.81 = 0 + \frac{1}{2} \cdot a \cdot (9)^2$$0.81 = \frac{81}{2} \cdot a \implies a = 0.02 \ m/s^2$.For the masses,the equations of motion are:$m_1g - T_1 = m_1a$$T_2 - m_2g = m_2a$For the pulley,the torque equation is:$(T_1 - T_2)R = I \alpha = I \cdot \frac{a}{R} \implies T_1 - T_2 = \frac{Ia}{R^2}$.Adding the equations for the masses:$(m_1 - m_2)g - (T_1 - T_2) = (m_1 + m_2)a$$(m_1 - m_2)g - \frac{Ia}{R^2} = (m_1 + m_2)a$$I = \frac{R^2}{a} [(m_1 - m_2)g - (m_1 + m_2)a]$$I = \frac{(0.02)^2}{0.02} [(0.4 - 0.35)(9.8) - (0.4 + 0.35)(0.02)]$$I = 0.02 [0.05 	imes 9.8 - 0.75 	imes 0.02]$$I = 0.02 [0.49 - 0.015] = 0.02 	imes 0.475 = 0.0095 \ kg \cdot m^2 = 9.5 	imes 10^{-3} \ kg \cdot m^2$.

Two masses $400 \ g$ and $350 \ g$ are suspended from the ends of a light string passing over a heavy pulley of radius $2 \ cm$. When released from rest,the heavier mass is observed to fall $81 \ cm$ in $9 \ s$. The rotational inertia of the pulley is . . . . . . $kg \cdot m^2$. $(g = 9.8 \ m/s^2)$

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