The cam is rotating with angular velocity omega and angular

The cam is rotating with angular velocity omega and angular acceleration alpha, as shown. Determine the velocity and acceleration of the rod R.

The acceleration of the rod is given by,

m double dot x + k*x = F cos (omega *t)

eliminate double dot x,

double dot x = F cos (omega*t) - k*x / m ----- eqn1

Since the rod is rolling over a cam of constant acceleration,

F cos (omega*t) = m * alpha, where alpha is the angular acceleration.

F cos (omega*t) = m * r^2 * omega, where omega could be represented as v/r.

So, the periodic input becomes simply,

F cos (omega*t) = m * v * r ------ the conservation of angular momentum

Replace the periodic input in the spring-mass system,

double dot x = m * v * r - k * x / m ------ eqn 2

double dot x (acceleration) = m * dot x * r - k * x / m where k/m is the nautal frequency, omega (n).

The acceleration, a is m * dot x * r - omega (n)^2 * x

Integrate eqn2 once to find the velocity,

dot x = m * x * r - k/m * x^2 / 2

dot x (velocity) = m * x * r - omega(n) * x^2 / 2 --- where omega (n) is the natural frequency