The illustration for this problem is at ds055uzetaobb<dot>cloudfront<dot>net/image_optimizer/142fdba0bc53bb9ae12ecea6de058f57fef01274.png
A rope of length 90cm 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 in m/s of the rope (to 2 decimal places) at the instant it loses contact with the table?
Details: g=-9.81 m/s²
let the mass of the rope be M
The initial potential energy of the rope is gives up is Mass(.9/2) *9.8 since the rope falls half it's length.
the final KE at that point is 1/2 M V^2
setting them equal
1/2 MV^2=M*9.8*.45
V=sqrt(9.8*.9) m/s
rho = mass /meter, call it 1 kg/meter ( it will cancel)
amount down hole = x
so mass pulling down = 1 x
force pulling down = 1 g x
total mass = 1 * .9 = 0.90 kg
g x = F = m a = .9 * a
g x = .9 d^2x/dt^2
.9 d^2x/dt^2 - g x = 0
let x = c e^kt
d^2x/dt^2 = ck^2 e^kt = k^2 x
.9 k^2 x - g x = 0
k^2 = g/.9
k = 3.3
so
x = c e^3.3 t
well to start it off, we need a tiny bit over the lip
let x = 0.0001 meter at t = 0
.0001 = c * 1
so
x = .0001 e^3.3 t
when x = 0.90
0.90/.0001 = e^3.3 t
9.1 = 3.3 t
t = 2.76 seconds
now speed
dx/dt = .00033 e^3.3 t
at t = 2.76
dx/dt = 2.98 m/s
now try that with a different starting x, like 0.00001 :)
LOL, my way is more fun
To find the speed of the rope at the instant it loses contact with the table, we need to use the principle of conservation of energy. At the beginning, the rope has potential energy (due to its height) and no kinetic energy. At the end, when it loses contact with the table, it has no potential energy but gains kinetic energy.
To solve the problem, we can use the equation for conservation of energy:
Potential energy (PE) at the start = Kinetic energy (KE) at the end
The potential energy at the start is given by the formula:
PE = mgh
Where:
m = mass of the rope
g = acceleration due to gravity
h = height of the rope above the table
Since the rope is hanging vertically, the height h is equal to the length of the rope. So, h = 90 cm = 0.9 m.
Let's assume the mass of the rope is 'm'. Therefore:
PE = mgh
The kinetic energy of the rope at the end is given by the formula:
KE = (1/2)mv^2
Where:
v = velocity of the rope when it loses contact with the table
Setting up the equation for conservation of energy:
mgh = (1/2)mv^2
The mass 'm' cancels out from both sides of the equation, giving:
gh = (1/2)v^2
Now, we can solve for the velocity 'v':
v^2 = 2gh
v = √(2gh)
Substituting the given values:
g = -9.81 m/s² (negative because the rope is falling)
h = 0.9 m
v = √(2 * (-9.81) * 0.9)
Now we can calculate the value of v using a calculator or software:
v ≈ 4.21 m/s
Therefore, the speed of the rope at the instant it loses contact with the table is approximately 4.21 m/s.