What is the Vertical Velocity of a Falling Rope Sliding Over a Peg?

[Matt951]

Homework Statement

A limp rope with a mass of 2.4 kg and a length of 1.3 m is hung, initially at rest, on a frictionless peg that has a negligible radius, as shown in the Figure. y1 is equal to 0.44 m. What is the vertical velocity of the rope just as the end slides off the peg?

Homework Equations

ΔU+ΔKE= Uο+KEο
U=mgh
Ke= .5mV²

The Attempt at a Solution

I've gotten Vf=√2gh after having Vο equaling 0 m/s
mass cancels out throughout the formula
Shouldn't Δh just be y1?
but I'm still incorrect what i am I forgetting or doing wrong?

 

[Matt951]

 

[berkeman]

Homework Statement

A limp rope with a mass of 2.4 kg and a length of 1.3 m is hung, initially at rest, on a frictionless peg that has a negligible radius, as shown in the Figure. y1 is equal to 0.44 m. What is the vertical velocity of the rope just as the end slides off the peg?

Homework Equations

ΔU+ΔKE= Uο+KEο
U=mgh
Ke= .5mV²

The Attempt at a Solution

I've gotten Vf=√2gh after having Vο equaling 0 m/s
mass cancels out throughout the formula
Shouldn't Δh just be y1?
but I'm still incorrect what i am I forgetting or doing wrong?

Welcome to the PF.

Can you show your detailed calculations? Also, I would have started with the centers of mass of the two pieces for the initial PE calculation, and compated that to the PE of the COM of the full rope just as it finishes slipping over the peg...

 

[Matt951]

U + Ke = Uο + Keο
mgΔh + .5mVf² = mg(Δh = 0) + .5mV²(Vi should equal 0)
cancel all the Ms since there in each individual formula
gΔh = -.5Vf²
multiply -2 on each side to get rid of the .5 on the right and then square root
Vf = √-2(-9.81)*Δh
so you're saying that my Δh should begin in the center of mass of the rope?

 

[berkeman]

so you're saying that my Δh should begin in the center of mass of the rope?

I'm just saying that my approach to this problem would be to find the initial PE of both pieces of rope (from the lengths and the mass that is hanging on each side). I would use the COM for each of the 2 pieces of rope. That gives the two initial PEs.

Then I would calculate the final PE with the whole rope hanging on one side, and use the full rope's mass at the COM for that calculation. The delta PE should give the KE gained (as you know)...

Can you try sketching the initial and final COMs (with lengths down from the peg and the masses in each case)?

 

[Matt951]

So you're saying I should do this:
Uο=ΔU
mgh_y1+mgh_y2=ΔU
and since ΔU= -ΔKE
mgh_y1+mgh_y2=-.5mv²
get rid of Ms and take out gravity on the left
g(h_y1+h_y2= -.5v²
multiply -2 on both sides with the negatives of -2 and gravity canceling
2g(h_y1+h_y2)= v²
then square root
v=√2g(h_y1+h_y2)
or should I keep ΔU and not change it to KE (on line 4)

 

[berkeman]

Uο=ΔU

I don't know what that means. How can the initial PE equal some change in PE?

What is the initial mass of the left rope piece? What is the distance down from the peg to the COM of the left rope piece?

Same questions for the right rope piece...

And for the final position, what is the distance down from the peg to the COM (easy)? And the mass of the rope?

So what do those calculations give you for initial PE and final PE?

 

[gneill]

A lot of your math can be dealt with by visual inspection of the sketch you made (you made the sketch, right?).

Here's my version of the sketch. Centers of mass for the rope segments are indicated by the black dots:

 

[Matt951]

Ok so the left side's mass is (.44/1.3)*2.4 = .81 kg and its change in height is zero since the whole rope goes down by the length of the left height
and the right side's mass is (.86/1.3)*2.4 = 1.59 kg and its change in height is just the length of the left side
When you ask for final distance of the peg to the rope's COM do you mean the whole rope of the COM from just the right side?
And its mass is still 2.4 as a whole
So i should use ΔU+ΔKE= Uο+KEο with KEο = 0 J
and so ΔKE = Uο-ΔU ?

 

[gneill]

For the system as a whole you can write (thanks to energy conservation): ΔPE + ΔKE = 0.

In this case the net change in PE is due to only the mass "y2" falling by Δh = y₁, confirmed by visual inspection of the sketch. So calculate its change in PE.

The change in PE translates to KE via energy conservation. The KE applies to the entire rope (entire mass), so find the velocity accordingly.

 

[Matt951]

Okay so here are my calculations
ΔPE = -ΔKE
(m_y2*g*h_y2)-(m*g*Δh) = -.5mv² (initial KE is 0)
than multiply over -2 from KE and divide by m
(-2(m_y2*g*h_y2)-(m*g*Δh))/m = v²

 

[gneill]

You seem to be making your life difficult by blending changes in quantities with initial values. Really all that matters is the changes. Sometimes when things are complicated you need to detail initial and final positions of several objects in order to determine the changes, but here it's really straightforward: There's only one mass' change in position that matters.

You can pick out the relevant change from the diagram. You have one mass, $m_{y2}$, which changes in height by an amount $\Delta h = -y_1$. So $\Delta PE$ is $-m_{y2}~g~ y_1$. Calculate its value.

Then change the sign and call it $\Delta KE$ (thanks to conservation of energy). After that you can write the usual formula for KE and solve for v.

 

[Matt951]

Thank you! I finally got the answer. I just keep on over complicating my formulas which makes everything so much harder than it needs to be

 

[gneill]

Thank you! I finally got the answer. I just keep on over complicating my formulas which makes everything so much harder than it needs to be

A diagram can do a lot of the math for you if you can pick out the relevant bits visually.