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# hw9 - patel(kp9583 HW9 Mackie(20211 This print-out should...

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patel (kp9583) – HW9 – Mackie – (20211) 2 The mass on the right is 1 . 03 kg and on the left is 1 . 51 kg. 3 . 8 m 2 kg 24 . 9 cm ω 1 . 51 kg 1 . 03 kg What is the magnitude of the linear acceler- ation of the hanging masses? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 03612 m / s 2 . Explanation: Let : M = 2 kg , R = 24 . 9 cm , m 1 = 1 . 03 kg , m 2 = 1 . 51 kg , h = 3 . 8 m , and v = ω R . Consider the free body diagrams 1 . 51 kg 1 . 03 kg T 2 T 1 m 2 g m 1 g a a The net acceleration a = r α is in the direc- tion of the heavier mass m 2 . For the mass m 1 , F net = m 1 a = m 1 g - T 1 T 1 = m 1 g - m 1 a and for the mass m 2 , F net = m 2 a = T 2 - m 2 g T 2 = m 2 a + m 2 g . The pulley’s mass is concentrated on the rim, so I = M r 2 , and τ net = summationdisplay τ ccw - summationdisplay τ cw = I α T 1 r - T 2 r = ( m r 2 ) parenleftBig a r parenrightBig = m r a m a = T 1 - T 2 m a = ( m 1 - m 2 ) g - ( m 1 + m 2 ) a m a + ( m 1 + m 2 ) a = ( m 1 - m 2 ) g a = ( m 1 - m 2 ) g m + m 1 + m 2 = (1 . 03 kg - 1 . 51 kg) (9 . 8 m / s 2 ) 2 kg + 1 . 03 kg + 1 . 51 kg = 1 . 03612 m / s 2 .
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hw9 - patel(kp9583 HW9 Mackie(20211 This print-out should...

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