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
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
3 answers
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 :)
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
0 answers