Question

A long, uniform rod of mass m and length L is connected to a frictionless pivot. It's held in place horizontally and allowed to swing down under the influence of gravity. 3. Sketch the: net torque on the rod as a function of time, and as a function of angular position (for the interval of horizontal to vertical ). (the fatび ) Derive an expression for the angular velocity at the bottom of the rod's swing. a. b.

Answer 1

when the rod makes an angle of theta with the horizontal, it has descended by a distance of x: which can be calculated as sin theta = x/L/2 material implication x = L/2 sin theta Now, using the principle of conservation of energy, we have 1/2 Ia^2 = mgL/2 sin theta - (1) downarrow gain in kinetic energy downarrow decrease in potential energy I, moment of inertia of the rod about the frictionless pivot is I = mL^2/3 which on substituting in equation gives 1/2(mL^2/3)w^2 = mgL/2 si theta w^2 = 3gsin theta/L w = squareroot 3gsin theta/L or d theta/dt = squareroot 3gsin theta/L - (2) Now, torque tau = I alpha = mL^2/3 ties dw/dt = mL^2/3 times squareroot 3g/L cos theta/2 squareroot sin theta or tau = squareroot m^2gL^3/3 sin theta times cos theta Time dependence of torque can be obtained by solving the differential equation (2), which is a difficult task.