Question

6. (BONUS) The two masses are released from rest. the pulley is my Ass Idiak = 1/2M R. Find thi other variables (10 pt) Wo masses (m, and ma) in the following Adwood rest. If the height of me from the floor is h, the mass of Assume the pulley is uniform disk with radius T and . Find the equation of the muulley massm in terms of the in the following Adwood's machine iak 1/2 F=ma T-mg=m, a mag-T= mza ir wing wema

Answer 1

From Newton's law of motion, Fy = Fri-mg = ma Fri = mg+ma----(1) Fy2 = m,g-F12 = m a Fn = m,g-ma----(2) The torque acting on the system is, { t = Frr- Frjr = Ia=1 From equation 1 and 2 (m;g-m,a)r -(m,g+ma)r=1 (m,g–ma)–(mg+ma)=1C) (m.g–ma)– (mg+ma)=(mx) ) mo m. -21°((m,g-m,a)-(m_g+m,a)) 2(m,g-m,a)-(m,g+m,a)) 2(m2-m,)g-2(m+ m2 )a 20m, -m.)8-2(m, + m2)