Question

7. 30 pointy) The pulley situation shuown bekom hem051 kg, and m-0.35 kg. The puley has mass m0.1 kg and radius Rp0.05 m,so that it's mement of inertia is 1-1/2 m, Rp: Ignore friction. a. (20) Caleulate the tensions, Tnd T LN m2 mj b. (10) Use conservation of mechanical enerzy to calculate the speed of both masses after m, falls a distance of 0.5m and m, rises by the same distance. Don't forget the kinetic energy of the pulley. m/s

please solve parts a and b. please please show all work.

Answer 1

For m_1 m_1 g - T_1 = m_1 a thatis (i) for m_2 T_2 - m_2 g = m_2 a thatis (ii) for pulley T_1 R_p - T_2 R_p = I a/R_p (gamma = I alpha) T_1 - T_2 = 1/2 M_p R^2_p middot a/R^2_p T_1 - T_2 = 1/2 M_p a thatis (iii) using (i) & (ii) in eq^n (iii) (m_1 g - m_1 a) - (m_2 g + m_2 a) = 1/2 m_p a m_1 g - m_1 a - m_2 g - m_2 a = 1/2 m_p a implies a = (m_1 - m_2) g/m_1 + m_2 + m_p /2 = (0.51 - 0.35) times 9.8/0.51 + 0.35 + 0.1/2 = 1.72 m/s^2 from (i) T_1 = m_1 g - m_1 a = 0.51 (9.8 - 1.72) = 4.1208 N from (ii) T_2 = m_2 g + m_2 a = 0.35 (9.8 + 1.72) = 4.032 N b) increase in P.E of m_1 = m_1 g h increase in P.E of m_2 = m_2 gh therefore net decrease in P.E