A rod of mass m slides within a fixed vertical cylinder of the same length L , s

Question

A rod of mass m slides within a fixed vertical cylinder of the same length L , subject to the force of gravity, and to a frictional force proportional both to the velocity v of the rod's motion and to the area of the contact surface between the rod and the cylinder, so that the drag force has the form F_d = -cv(L-x) , where x is the length of the rod protruding from the end of the cylinder and c is a positive drag constant. The rod starts at rest, flush with the ends of the cylinder. Assume that, at any moment during its descent, the rod is travelling at the "terminal velocity" determined by the instantaneous equilibrium between gravity and drag forces. (This velocity keeps changing as the rod slides further down.) Find the time it takes the rod to drop out of the cylinder.

Explanation / Answer

Since at every instant, the rod is travelling at the terminal velocity, the net force on the rod at every instant will be zero.
Fnet = mg - cv(L - x) =0
v = dx/dt
mg/c = dx/dt(L - x)
dt = c/mg (L - x)dx
Integrating,
t = c/mg [Lx - x2/2], x from 0 to L
t = c/mg [L2 - L2/2] = cL2/2mg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.