Question

4. Blocks of masses 11.0 kg and 25.0 kg hang on opposite sides of a pulley (Atwood's machine) with a massless string that rotates the pulley without slippage. The mass of the pulley is 1.5 kg and it has a radius of 15.0 cm. The heavier mass is 2.5 m above the ground and the lighter mass is 1.2 m above the ground. Using energy consideration, determine the velocity of the two masses when the heavier mass hits the ground.

Answer 1

Because There is no slipping on pully, Then velocity of point A = V_A = V i.e pure rolling present V = WR = V_A V = WR equation (1) Hence rotational K.E of pully [REP] = 1/2 Iw^2 = 1/2 mR^2 middot V^2/R^2 = 1/2 mv^2 equation (2) Apply Energy conservation (K.E + P.E)_i = (K.E + P.E)_f (K.E)_m_1 + (K.E)_m_2 + (P.E)_m_1 + (P.E)_m_2 + rotational K.E oF pully]_i = (K.E)_m_2 + (K.E)_m_1 + (P.E)_m_1 + (P.E)_m_2 + REP]_f [0 + 0 + 11 times g times 1.2 + 2.5 times g times 2.5 + 0] = [1/2 11 times v^2 + 1/2 times 25 times v^2 + 11 times g times 3.7 + 0 + 1/2 times 1.5 v^2] 75.7g = 1/2 v^2 (11 + 25 + 1.5) + 40.7g 70g = v^2 times 37.5 v = squareroot 70 times 10/37.5 = 4.32m/sec