Question

L-M 7. A uniform flexible chain of length L, with mass per unit length A, passes over a small, frictionless, massless pulley. It is released from a rest position with a length of chain y hanging from one side and a length l-y from the other side. Find the acceleration a as a function of y a(y)-?

Answer 1

We first consider easier case of hanging chain of frictionless surface. In the above instance the frictional force is zero (0). and are two masses (m_1, m_2) connected by a string and a massless pulley. The equation of motion, can be written as 1) m_t middot a = T 2) m_h middot a = F - T \r\n On solving 1 & 2 equation 3. Rightarrow a = m_h middot g/m_t + m_h we can solve for the chain, if we notice that m_h = lambda middot y and m_t = lambda(L - y), where lambda = mass/unit length therefore a = d^2 y/dt^2 = g middot y/L For differential equation, y(t) = A middot e^(squareroot g/l) t + B e^-(squareroot g/l) t Initial condition of problem y(0) = y_0, where y_0 is the length of chain initially hanging over the table. V_0 = 0, chain released at rest. therefore y_0 = A + B: A & B are consonants determined by