How Do You Calculate the Maximum Speed in a Frictionless Pulley System?

[swannyboy]

Homework Statement

http://img524.imageshack.us/i/physics123.png/" The image is here. Sorry that I couldn't attach a file. The file I made was too big.

In the diagram given, the pulley is frictionless and the force of friction between the .5kg mass and the table is 4.8 N. The force equation for the spring is F = 50x. If the system starts at rest with the spring at its normal length, calculate the maximum speed the 1kg mass will reach.

P.S. I have the answer, I just don't know how to get there.

Homework Equations

Okay, so here are the equations I have during this unit:

Force of the Spring= kx (k is 50 in this question)
Kinetic Energy = 1/2 mv^2
Potential Energy = mgh
Potential Energy of the spring = 1/2kx^2
Total Energy = sum of all energy

Then from a previous unit, this equation might be needed:

Fnet = Fgravity -T where T is the tension in the string

The Attempt at a Solution

At Start:

KE = 1/2 mv^2
KE = 0.5kg (0m/s)^
KE = 0Joules

PE(spring) = 1/2 kx^2
PE(spring) = 1/2 (50) 0m^2
PE(spring) = 0 Joules

PE = mgh
PE = 1kg(9.8 N/kg)(0m)
PE = 0

TE= 0Joules

Fnet = Fg + T
1kg(a(1kg object)) = 9.8N/kg (1kg) -T T
A(1kg object) = (9.8N -T)/1kg

Fnet = Fg + T
0.5kg (a(0.5 kg object)) = 0N/kg (0.5kg) -T
a(0.5kg object) = -T/ 0.5kg

I don't know what to do from here, or what I did wrong. I missed a couple of days at school, so I'll admit that I'm a little bit lost.

 

[Galileo's Ghost]

You may want to reconsider your energy analysis at the start. You claim a PE = 0, but consider that the 1 kg mass will be falling. If it will fall, then it will experience a change in PE. Also consider that the 1 kg mass can only fall a distance equal to how far the spring is stretched. So there is a connection there to tap into.

 

[gneill]

Your "force equation for the spring", f = 50x, is that supposed to imply 50 N/m?

 

[swannyboy]

Your "force equation for the spring", f = 50x, is that supposed to imply 50 N/m?

Yes, that's right. x would be the number of meters.

 

[gneill]

Why don't you start by listing all the places that the energy can go or be stored, and an expression that is appropriate for each?

 

[swannyboy]

Okay, so here's what I've done since. I'm still not getting the answer, but I've set the two accelerations equal to each other.

a(1kg) = -a(.5kg)
(9.8N - T)/1kg = (-T)/.5kg
1.5T = 4.9N
T = 3.26 N

Fnet = fg - T - f(friction)
Fnet = 9.8N - 3.26N - 4.8N
Fnet = 1.74 N

F = ma
1.74N = 1kga
a = 1.74 N/kg

I guess it would stop when the force of the pulleys was equal to the force of the spring, right?

If that's the case, then:

F = 50x
50x = 1.74 N
x = 0.0348m

Am I doing any of this right so far? I may be overcomplicating things, but I missed these days at school, and I just can't get it.

 

[gneill]

The problem is going to be that the acceleration will not be constant since the spring stretching will alter it. So the "usual" approach that you're following is going to run into some difficulties (you'll probably end up with a second order differential equation to deal with).

I think that if you can just parcel out the energy appropriately for any given value of x, then you'll be able to isolate the kinetic energy and use that to determine the velocity. Note that the question isn't asking for the velocity with respect to time. Just the maximum velocity.

 

[swannyboy]

The problem is going to be that the acceleration will not be constant since the spring stretching will alter it. So the "usual" approach that you're following is going to run into some difficulties (you'll probably end up with a second order differential equation to deal with).

I think that if you can just parcel out the energy appropriately for any given value of x, then you'll be able to isolate the kinetic energy and use that to determine the velocity. Note that the question isn't asking for the velocity with respect to time. Just the maximum velocity.

I think part of it may be because I'm tired right now, and I'm freaking out a little bit because I have a test on Monday, and it's 15 minutes 'til Sunday, but I really don't see how to do that right now.

I mean I can try. If I do this at x= -0.01m when the start = 0, then:

PE(spring) = 1/2kx^2
PE(spring) = 25 (-.01)^2
PE(spring) = .0025

Here's where I have the problem though...

PE= mgh For g would I use 9.8N/kg, or would I use an approximation for what the acceleration would be?

I'd either use:

PE= mgh
PE= 1kg(1.74N/kg)(-.01m)
PE = -.0174 Joules

OR

PE = mgh
PE = 1kg(9.8N/kg)(-0.01m)
PE = -0.098 Joules

Now if I was estimating, would I assume that the value of the velocity here approached zero?

Hypothetically, if I do this, I get:

0 = TE + KE + PE + PE(spring)
0 = TE + 0 + -.0174joules + .0025 Joules
TE = 0.0149 Joules

OR

0 = TE + KE + PE + PE(spring)
0 = TE + 0 + -.098joules + .0025 Joules
TE = 0.0955 Joules

Now what? I seriously don't know.

Edit: So I think I've got it. I have to continue from my second post:

I had the A(1kg)= (9.8N -T)/1kg
but I needed to put the Ffriction into the A(0.5kg)

That leaves me with:

A(0.5kg) = (4.8N-T)/0.5kg

Now setting them equal to each other I end up with 1.5T = 9.7N
So T=6.46(the six repeating) N

Fnet = Fg - T
Fnet =9.8N - 6.46
Fnet = 3 1/3 N

I was right that if you set F= 50x equal to the net force of the pulleys you get the total distance before it stops.
That total distance = 2/30 m

Then I do:

PE = mgh
PE= .222

PE(spring) = 1/2kx^2
PE(spring) = 0.111

Now since it's stopped at 2/30 m, and the velocity-time graph for this setup would be concave down, and symmetrical, you know that the local max (or the maximum velocity) would be in the center.

Now if you plug into KE = 1/2mv^2

KE = 1/2mv^2
1/3Joules = 1kgv^2
V^2 = 1/3 m^2/s^2
V = 0.58 m/s

 

[gneill]

I think part of it may be because I'm tired right now, and I'm freaking out a little bit because I have a test on Monday, and it's 15 minutes 'til Sunday, but I really don't see how to do that right now.

I mean I can try. If I do this at x= -0.01m when the start = 0, then:

In my experience, plugging in numerical values before you have all the formulae and overall plan of solution tends to confuse things, as details get lost in, well, more details.

PE(spring) = 1/2kx^2

Okay, I’ll stop you right there. You’ve identified a place where energy can hide. In this case it’s potential energy in the spring.

{snip}

PE= mgh For g would I use 9.8N/kg, or would I use an approximation for what the acceleration would be?

Gravitational potential energy is strictly a function of relative location in the Earth’s gravitational field. g is g. Accept no substitutes!

So, you’ve identified a second place for energy to hide: the gravitational potential energy associated with the suspended mass.

{snip}

Now if I was estimating, would I assume that the value of the velocity here approached zero?

Hypothetically, if I do this, I get:

0 = TE + KE + PE + PE(spring)
0 = TE + 0 + -.0174joules + .0025 Joules
TE = 0.0149 Joules

OR

0 = TE + KE + PE + PE(spring)
0 = TE + 0 + -.098joules + .0025 Joules
TE = 0.0955 Joules

Now what? I seriously don't know.

What does TE represent? The Total Energy? Again, I think that the numerical speculations at this stage are not entirely profitable.

Edit: So I think I've got it. I have to continue from my second post:

I had the A(1kg)= (9.8N -T)/1kg
but I needed to put the Ffriction into the A(0.5kg)

That leaves me with:

A(0.5kg) = (4.8N-T)/0.5kg

Now setting them equal to each other I end up with 1.5T = 9.7N
So T=6.46(the six repeating) N

Fnet = Fg - T
Fnet =9.8N - 6.46
Fnet = 3 1/3 N

I was right that if you set F= 50x equal to the net force of the pulleys you get the total distance before it stops.
That total distance = 2/30 m

What makes you think that the system will stop if those forces balance? If you look carefully at the setup you might notice that it’s got all the elements of a classic spring-mass oscillator, including damping (the friction). The inertial mass is the sum of the two block masses, while the mass producing the driving force is just that of the suspended block.

Then I do:

PE = mgh
PE= .222

PE(spring) = 1/2kx^2
PE(spring) = 0.111

Now since it's stopped at 2/30 m, and the velocity-time graph for this setup would be concave down, and symmetrical, you know that the local max (or the maximum velocity) would be in the center.

Now if you plug into KE = 1/2mv^2

KE = 1/2mv^2
1/3Joules = 1kgv^2
V^2 = 1/3 m^2/s^2
V = 0.58 m/s

Well, you’ve identified another place for energy to go in the system: the kinetic energy of its moving parts. Identify all the moving parts that can hold kinetic energy.

There’s one more place that energy can go. You’ve mentioned it before but didn’t write an expression for the energy that goes there. Can you think of what it is?

 

[swannyboy]

There’s one more place that energy can go. You’ve mentioned it before but didn’t write an expression for the energy that goes there. Can you think of what it is?

I mentioned another place energy can go?

I guess I got these:
kinetic energy of the moving parts
gravitational potential energy
potential energy in the spring

What else is there?
Energy of friction?
rotational energy in the pulley?

I've looked through my posts, and I don't see where I've mentioned another for energy to hide.

 

[gneill]

I mentioned another place energy can go?

I guess I got these:
kinetic energy of the moving parts
gravitational potential energy
potential energy in the spring

What else is there?
Energy of friction?
rotational energy in the pulley?

I've looked through my posts, and I don't see where I've mentioned another for energy to hide.

Friction. Energy is being lost to friction. Write the expression.

It seems that the pulley is massless and frictionless, so nothing to see there! Keep moving...

 

[swannyboy]

So here are all the equations for where energy can be found:

PE(s)=1/2kx^2

KE = 1/2mv^2

PE(g) = mgh

So I guess the expression for friction would be 4.8N x distance in meters.
It's energy lost to friction, so it would be negative, right?

 

[gneill]

So here are all the equations for where energy can be found:

PE(s)=1/2kx^2

KE = 1/2mv^2

PE(g) = mgh

So I guess the expression for friction would be 4.8N x distance in meters.
It's energy lost to friction, so it would be negative, right?

You're definitely on the right track. Now, write a single expression relating all the energies involved for a given value of x.

 

[swannyboy]

You're definitely on the right track. Now, write a single expression relating all the energies involved for a given value of x.

mgh + 1/2mv^2 + 1/2kx^2 + F(friction)x = constant

 

[gneill]

You need to specify which masses are which. You've got two masses involved, one dangling and one on the table. Also, since the masses are joined, any motion x of the mass on the table is going to be the same as the change in height h of the dangling mass. So do away with extraneous symbols.

If you assume that the system begins at rest, what's the value of your constant?

Check to make sure that all the signs make sense. You can think of it this way: Starting from rest, the energy for the system is being delivered by the PE (mgx). This energy is going to go into the KE of the masses, PE of the spring, and Frictional loss. So you'd expect:

mgx = 1/2(m?)v2 + 1/2kx2 + Ffx

 

[swannyboy]

You need to specify which masses are which. You've got two masses involved, one dangling and one on the table. Also, since the masses are joined, any motion x of the mass on the table is going to be the same as the change in height h of the dangling mass. So do away with extraneous symbols.

If you assume that the system begins at rest, what's the value of your constant?

Check to make sure that all the signs make sense. You can think of it this way: Starting from rest, the energy for the system is being delivered by the PE (mgx). This energy is going to go into the KE of the masses, PE of the spring, and Frictional loss. So you'd expect:

mgx = 1/2(m?)v2 + 1/2kx2 + Ffx

If it starts at rest, the value of the constant is zero.

Also, with the 1/2(m?)V^2, I think this mass would be the 0.5kg mass. I'm just guessing because the mass used for mgx must be the one that has the gravitational force acting on it, and the other one must be used some time, right?

 

[gneill]

If it starts at rest, the value of the constant is zero.

Also, with the 1/2(m?)V^2, I think this mass would be the 0.5kg mass. I'm just guessing because the mass used for mgx must be the one that has the gravitational force acting on it, and the other one must be used some time, right?

That's an interesting use of logic! Unfortunately, it's not an equal-opportunity equation!

Hint: Everything with mass that's moving should have a KE term.

 

[swannyboy]

So if m1 = 1kg
m2 = .5kg

m1gx = 1/2 m1 v^2 + 1/2 m2 V^2 + 1/2kx^2 + Ff xor maybe

m1gx = 1/2 m1m2 v^2 + 1/2kx^2 + Ffx

Or

m1gx = m1/2m2 v^2 + 1/2kx^2 + Ffx

 

[gneill]

Kinetic energies ADD. Both masses have the same velocity at all times. So the total kinetic energy is...

 

[swannyboy]

1/2 (m1 + m2) v^2

 

[gneill]

Okay, you're coming into the home stretch. Carry on.

 

[swannyboy]

so mgx =

mgx = 1/2(m1+m2) V^2 + 1/2kx^2 + Ffx

I don't see why I can't figure it out from here.

mgx = KE + 1/2kx^2 + Ffx
-KE = 1/2kx^2 + Ffx -mgx
-KE = 25Nmx^2 + 4.8Nx -9.8Nx
KE = 5.0Nx -25x^2
KE =5x(1-5x) Yes, I know I currently don't have units.

 

[gneill]

so mgx =

mgx = 1/2(m1+m2) V^2 + 1/2kx^2 + Ffx

I don't see why I can't figure it out from here.

mgx = KE + 1/2kx^2 + Ffx
-KE = 25Nmx^2 + 4.8Nx -9.8Nx
KE = 5.0Nx -25x^2
KE =5x(5x-1) Yes, I know I currently don't have units.

I'm not sure where you're taking this. You seem to be un-deriving what you've spent the time deriving!

You want to find the speed as a function of x, right? Don't you already have the speed (v) in the energy equation above? Stir and serve!

Once you've got the velocity vs displacement you should be able to find a maximum (a touch of calculus). Keep in mind that v^2 is maximum when v is maximum -- this could save you dealing with square roots when you differentiate.

 

[swannyboy]

I'm not sure where you're taking this. You seem to be un-deriving what you've spent the time deriving!

You want to find the speed as a function of x, right? Don't you already have the speed (v) in the energy equation above? Stir and serve!

Once you've got the velocity vs displacement you should be able to find a maximum (a touch of calculus). Keep in mind that v^2 is maximum when v is maximum -- this could save you dealing with square roots when you differentiate.

Okay, if you want to find speed as a function of x, then you want to rewrite the equation in the form of v^2 =...

So when I do that, I get

V^2 = (50Nmx^2 -10Nx)/1.5kg

When I did this before, I forgot the 1.5 kilograms, so I went an entirely different direction.
Now when I do this,

I let V^2 = y.

Now again, I'll rewrite it

Y = (50x^2 -10x)/1.5
Y' = 0 when x =.0999, and the value of the function is equal to 1/3.

So Y = 1/3

I'll plug v^2 back in for y.

V^2 = 1/3.
V = about .58

I honestly don't know what I was thinking before with this.

 

[gneill]

Okay, if you want to find speed as a function of x, then you want to rewrite the equation in the form of v^2 =...

So when I do that, I get

V^2 = (50Nmx^2 -10Nx)/1.5kg

I get something similar, so you might want to check:

v² = (1/M)*(-k*x² + 2*(m1*g - Ff)*x
where: M = m1 + m2

Which, when values are plugged in, boils down to

v² = -33.33(N/kg/m)*x² + 0.1380(N/kg)*x

 

[swannyboy]

I get something similar, so you might want to check:

v² = (1/M)*(-k*x² + 2*(m1*g - Ff)*x
where: M = m1 + m2

Which, when values are plugged in, boils down to

v² = -33.33(N/kg/m)*x² + 0.1380(N/kg)*x

I think it must be an equivalent form.

I got that it was the square root of 1/3. I've had this confirmed by 2 people now.
Edit: Nope, you're right. Using exact values, it still works out to be the square root of 1/3.

Well... I tried it using the form you got. I got slightly more than the square root of 1/3. However, I just saw the same question done using de Moivre's theorem, and the answer was the square root 1/3.

 

[gneill]

I think it must be an equivalent form.

I got that it was the square root of 1/3. I've had this confirmed by 2 people now.
Edit: Nope, you're right. Using exact values, it still works out to be the square root of 1/3.

Well... I tried it using the form you got. I got slightly more than the square root of 1/3. However, I just saw the same question done using de Moivre's theorem, and the answer was the square root 1/3.

The reason I have my doubts is that your formula for V^2 is negative near x = 0. I would expect V^2 to be positive and increasing.

So, in all its glory,

$$m1\;g\;x = \frac{1}{2}(m1 + m2)v^2 + \frac{1}{2}k\;x^2 + F_f\;x$$

I don't see how you could arrive at an exact answer of sqrt(1/3), with the constant g mixed up in the result...

$$v^2 = \frac{1}{m1 + m2}(-k\;x^2 + 2\;(m1\;g - F_f)\;x)$$

differentiating, setting to zero, and solving for x yields

$$x = \frac{m1\;g - F_f}{k}$$

$$x = 2.066 x 10^{-3}\;m$$

It also yields (symbolically)

$$v^2_{max} = \frac{(m1\;g - F_f)^2}{k(m1 + m2)}$$

$$v = \frac{m1\;g - F_f}{\sqrt{k(m1 + m2)}}$$

$$v = 11.93 mm/sec$$