A rope of length L lies in a straight line on a frictionless table, except for a very small piece at one end which hangs down through a hole in the table. This piece is released, and the rope slides down through the hole. What is the speed of the rope at the instant it loses contact with the table?
Let σ be the mass density of the rope. From conservation of energy, we know that the rope’s final kinetic energy, which is (σL)v^2/2, equals the loss in potential energy. This loss equals (σL)(L/2)g, because the center of mass falls a distance L/2. Therefore, v = (gL)^1/2 This is the same as the speed obtained by an object that falls a distance L/2. Note that if the initial piece hanging down through the hole is arbitrarily short, then the rope will take an arbitrarily long time to fall down. But the final speed will be always be (arbitrarily close to) (gL)^1/2
let x= tan^-1(4/3), so tan x = 4/3 . since cot^2 x + 1 = csc^2 x = 1 / sin^2 x (3/4)^2 + 1 = 25/16 = 1/ sin^2 x , sin^2 x = 16/25 sin x= 4/5 so sin(tan^-1(4/3))= 4/5
Determine the length of the parametric curve given by the set of parametric equations. For this problem you may assume that the curve traces out exactly once for the given range of t’s. x=8t^3/2 y=3+(8-t)^3/2 0 <= t <= 4
We’ll need the ds for this problem. ds= ([12t^1/2]^2+[-3/2(8-t)^1/2]^2)^1/2 dt = (144t+9/4(8-t))^1/2 dt = (567/4t+18)^1/2 dt The integral for the arc length is then, L= ſ ds= ſ(0,4) (567/4t+18)^1/2 dt This is a simple integral to compute with a quick substitution. Here is the integral work, ſ(0,4) (567/4t+18)^1/2 dt = (4/567)(2/3)((567/4)t+8)^3/2](0'4) = 66.1865