Solving Rope on Table Homework: Time for Length of 1m

In summary, the conversation involved finding the time at which the end of a flexible rope, initially hanging over the edge of a frictionless table, will reach the end of the table. The solution involved using conservation of energy to calculate the potential and kinetic energy of the rope, and solving for the time using a linear equation with constant coefficients. The final solution was t=\sqrt{\frac L g} \cosh^{-1} \left(\frac x {x_0}\right).
  • #1
msd213
25
0

Homework Statement



A flexible rope of length 1.0 m slides from a frictionless table top. The rope is initially released from rest with 30 cm hanging over the edge of the table. Find the time at which the end of the rope left on the table will reach the end of the table (basically when the length of the rope hanging of the table will be the whole 1 m. This is problem 9-21 in Thornton/Marion's Classical Dynamics).


Homework Equations



Let L be the total length of the rope and x be the portion hanging off the table.


The Attempt at a Solution



I started off with [tex]dp/dt = mg x/L[/tex]

because only the part hanging off the table is accelerated due to gravity.

I then rewrote that as [tex]dp=mgx/L \:dt[/tex]. Since the initial momentum is zero [tex]dp=p(t)-p(0)=p(t)=mdx/dt \:dt[/tex] where t is just some later time.

So I'm stuck at [tex]dx/dt=gx/L \:dt[/tex]. I can integrate over position but I'll have this (dt)^2 term that I don't know what to do with.
 
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  • #2
we know that the rope will gain mass as it falls so we will have to do this in terms of mass density which will be represented with [tex]\sigma[/tex]
force of gravity=mg=( [tex]\sigma[/tex] x) [tex]\cdot[/tex] x [note a dot equals the time derivative]
momentum= [tex]\sigma[/tex] L [tex]\cdot[/tex] x
F=ma=( [tex]\sigma[/tex] x)g= [tex]\sigma[/tex] L [tex]\cdot[/tex] [tex]\cdot[/tex] x
now integrate and you will be able to get your length of rope as a function of time.
 
  • #3
hi there
I have a solution for your problem but I am not sure it is right. Anyway, here it is.
I started with the relation W = (integral)F dx
Work done equals change in kinetic energy and initial KE is zero. Plugging in W = 0.5mv(squared) and F = (x/L)mg and integrating it, I got Lv(squared) = gx(squared)
Using v = v(initial)+at and equating x and 0.7L because the question is about when the rope travels 0.7 m that is on the table, I finally got: [tex]\ t = 0.7 \sqrt{\frac{L}{g}}[/tex].
Frankly, I was shocked by this equation's similarity with the simple harmonic motion. =D
Thanks
 
  • #4
I am really sorry, I did a big mistake. I assumed the mass to be constant. Mine is flawed
 
  • #5
sleventh said:
we know that the rope will gain mass as it falls so we will have to do this in terms of mass density which will be represented with [tex]\sigma[/tex]
force of gravity=mg=( [tex]\sigma[/tex] x) [tex]\cdot[/tex] x [note a dot equals the time derivative]
momentum= [tex]\sigma[/tex] L [tex]\cdot[/tex] x
F=ma=( [tex]\sigma[/tex] x)g= [tex]\sigma[/tex] L [tex]\cdot[/tex] [tex]\cdot[/tex] x
now integrate and you will be able to get your length of rope as a function of time.

That's going to involve solving a second order linear ODE. I was thinking that you can go through this problem with only separation of variables.
 
  • #6
How about using the conservation of energy?

[tex]\begin{array}{rcl}U&=&\int\limits_0^x -\mu g h{\rm d}h \\ &=&-\frac{\mu g} 2 x^2\end{array}[/tex]

and

[tex]K=\frac 1 2 \mu L v^2^[/tex]

and we know that:

[tex]\Delta (U+K)=0[/tex]

hence

[tex]v=\sqrt{\frac g L} \sqrt{x^2-x_0^2^}} \Leftrightarrow \frac{{\rm d} x}{\sqrt{x^2-x_0^2}}=\sqrt{\frac g L} {\rm d}t[/tex]

[tex]\fbox{t=\sqrt{\frac L g} \cosh^{-1} \left(\frac x {x_0}\right)}[/tex]([tex]\mu=\frac m L[/tex], [tex]L[/tex]=1m, [tex]x_0[/tex]=30 cm)
 
Last edited:
  • #7
Donaldos said:
How about using the conservation of energy?

[tex]\begin{array}{rcl}U&=&\int\limits_0^x -\mu g h{\rm d}h \\ &=&-\frac{\mu g} 2 x^2\end{array}[/tex]

and

[tex]K=\frac 1 2 \mu L v^2^[/tex]

and we know that:

[tex]\Delta (U+K)=0[/tex]

hence

[tex]v=\sqrt{\frac g L} \sqrt{x^2-x_0^2^}} \Leftrightarrow \frac{{\rm d} x}{\sqrt{x^2-x_0^2}}=\sqrt{\frac g L} {\rm d}t[/tex]

[tex]\fbox{t=\sqrt{\frac L g} \cosh^{-1} \left(\frac x {x_0}\right)}[/tex]


([tex]\mu=\frac m L[/tex], [tex]L[/tex]=1m, [tex]x_0[/tex]=30 cm)

That looks correct but I'm just wondering what you're motivation was for integrating the potential energy?
 
  • #8
Potential energy varies with height. You need to calculate the potential energy of a small segment at a given height first and then integrate that expression over the length of the rope.
 
  • #9
Donaldos said:
Potential energy varies with height. You need to calculate the potential energy of a small segment at a given height first and then integrate that expression over the length of the rope.

So you've chosen the tabletop as your zero reference point?
 
  • #10
msd213 said:
So you've chosen the tabletop as your zero reference point?

Yes.
 
  • #11
Donaldos said:
[tex]\fbox{t=\sqrt{\frac L g} \cosh^{-1} \left(\frac x {x_0}\right)}[/tex]

I get the same thing by just using F = m.a

the weight of the piece that's over the edge is [tex] \frac {m g x} {L} [/tex], but
the entire rope has to move so the mass wil be just m
(x = the length of the rope over the edge)

this getst you [tex] \frac {m g x} {L} = m \frac {d^2x} {dt^2} [/tex] or

[tex] \frac {d^2x} {dt^2} = \frac {g} {L} x [/tex]

this is a very easy linear equation with constant coefficients.

the solution of is [tex] C_1 e^{\sqrt {\frac{g}{L}} t} + C_2 e^{-\sqrt \frac{g}{L} t} [/tex]

using what we now about x(0) you get [tex] C_1 + C_2 = x(0) [/tex]

and [tex] x'(0) = 0 = C_1 - C_2 [/tex] so [tex] C_1 = C_2 = x(0)/2 [/tex]

the solution now becomes [tex] x = x(0) \frac { e^{\sqrt {\frac{g}{L}} t} + e^{-\sqrt \frac{g}{L} t} } {2} = x(0) cosh(\sqrt {\frac{g}{L}} t) [/tex]

and solving that gets you

[tex] t=\sqrt{\frac L g} \cosh^{-1} \left(\frac x {x_0}\right) [/tex]
 

1. How do you solve a rope on table homework?

To solve a rope on table homework, you will need to use the formula Time = Length/Speed. First, measure the length of the rope in meters. Then, determine the speed at which the rope moves on the table. Finally, plug these values into the formula to calculate the time it takes for the rope to move 1 meter on the table.

2. What is the formula for solving rope on table homework?

The formula for solving rope on table homework is Time = Length/Speed. This formula calculates the time it takes for a rope to move a certain distance on a table, based on the length of the rope and the speed at which it is moving.

3. What units should be used for the length and speed in the formula?

The length should be measured in meters and the speed should be measured in meters per second. This will ensure that the time is calculated in seconds, which is the standard unit for time.

4. Can this formula be used for any type of rope and table?

Yes, this formula can be used for any type of rope and table, as long as the length is measured in meters and the speed is measured in meters per second. However, the accuracy of the formula may vary depending on the surface of the table and the material of the rope.

5. How can I check if my answer is correct?

You can check your answer by plugging the values back into the formula and solving for time. If your calculated time matches the given time or is very close, then your answer is likely correct. You can also double check your measurements and calculations to ensure accuracy.

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