Question

5. An Atwood machine consists of two masses mi and m2 (with mi > m2) attached to the ends of a light string that passes over a light, frictionless pulley Problem Setup1 FBD ty m2 mi 1-png When the masses are released, the mass mi is easily shown to accelerate down with an acceleration mi m2 mi +m2 Suppose that and m 2 are measured as mn = 100 ± 1 and mg = 50 ± 1, both in grams. Derive a formula of the uncertainty in the expected acceleration in terms of the masses and their uncertainties, and then calculate ca for the given numbers

Answer 1

Delta a = squareroot ((partial differential alpha/partial differential m) Delta m)^2 + ((partial differential a/partial differential m) Delta m)^2 rightarrow (1) a = g(m - m_2/m_1 + m_2) partial differential a/partial differential m_1 = 2mg/(m_1 + m_2)^2 rightarrow (2) Therefore partial differential a/partial differential m_2 = -2 mg/(m_1 + m_2)^2 rightarrow (3) Put (2) & (3) in (1), we get, Delta alpha = 2g/(m_1 + m_2)^2 squareroot m_1^2 Delta m_2^2 + m_2^2 Delta m_1^2 Now, acceleration expected = a plusminus Delta a = g (m_1 - m_2)/(m_1 + m_2) plusminus Delta a. = 9.8 (100 - 50)/(100 + 50) plusminus Delta a. = 9.8 times 1/3 plusminus Delta a. = 3.267 plusminus Delta a. Delta_a = 2 times 9.8/(100 + 50) squareroot 100^2 times 1^2 + 50^2 times 1^2 = 0.0971 Therefore acceleration = (3.267 plusminus 0.097) m/s^2