Consider the Atwood machine which two masses, m1 and m2, are suspended by a string of constant length which passes over a massless pulley with frictionless bearings. Find the acceleration of the system.
Let x, x' and x'' be the position, velocity and acceleration of m1. Let y, y' and y'' be the position, velocity and acceleration of m2. Using Lagrangian mechanics, the acceleration of the Atwood machine is easy. The total kinetic energy of the system: T = (m1 * x' ^2)/2 + (m2*y'^2)/2 = (m1 + m2) * x' ^2 The total potential energy of the system: U = -(m1 - m2)*g*x The Lagrangian: L = T - U = (m1 + m2) * x' ^2 + (m1 - m2)*g*x The Euler Lagrange Equation: dL/dx = d/dt dL/dx' (m1 - m2) *g = (m1 + m2) x'' x'' = (m1 - m2) * g /(m1 + m2)
Which element has the larger atomic radius: Hydrogen or Helium.
Hydrogen has a larger atomic radius than helium. Both hydrogen and helium have valence electrons in the 1s orbital but helium has two protons in its nucleus whereas hydrogen only has one proton. The extra proton in helium's nucleus exerts a stronger force (pull) on its electrons thus reducing its atomic radius.
Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cc/s. How fast is the radius of the balloon increasing when the diameter is 50 cm?
Let V be the volume of the balloon and let r be its radius. In this problem, both volume and radius are functions of time (t). The rate of increase of the volume with respect to time is denoted dV/dt and the rate of increase of the radius with respect to time is dr/dt. We are given that dVdt = 100 cc/s and we want to know dr/dt when r = 25 cm (which is the same as when diameter equals 50 cm). Volume and radius of a sphere are related by the following equation: V = (4*pi*r^3 )/3 Differentiate with respect to time to get dV/dt = dV/dr * dr/dt = 4*pi*r^2 * dr/dt Solve for dr/dt when r = 25 cm dr/dt = 1/(25*pi) cm/s