In Example 4.3 we found the equation of motion of a block on an inclined plane, using the generalized coordinate \(X\), the distance of the block from the bottom of the incline. Solve the equation for \(X(t)\) in terms of an arbitrary initial position \(X(0)\) and velocity \(\dot{X}(0)\).

Data from Example 4.3

Transcribed Image Text:

A block of mass m slides down a frictionless plane tilted at angle a to the horizontal, as shown in Figure 4.4. There is only a single degree of freedom here, say the distance X of the block from the bottom, measured along the incline. The gravitational potential energy is mgh = mgX sin a. Using X as the generalized coordinate, the velocity is X, and the Lagrangian of the block is 1 1 L=T-U = m - U= mx - mgX sin a, (4.13) which depends explicitly upon the single coordinate (X) and the single velocity (X). The Lagrange equation is L d L - dt X d -mg sin amX = 0, or - mg sin = mx, dt which is indeed the correct F = m a equation for the block along the tilted plane. (4.14) Block sliding down an inclined plane.