Solving the Yoyo Problem: Velocity After Fall of h

[KiNGGeexD]

Yoyo Problem:(

Question:

A cylindrical yoyo of mass m and radius R falls a height h from rest. What is its velocity after this fall?

My attempt:

I drew a diagram and at rest determined that

T-mg=ma

Where T is tension and acceleration is centripetal acceleration v^2/r

So

v= sqrt of T-gR

I also know the moment of inertia is

1/2mR^2

I'm not sure how to find the velocity after fall h?
I think it is right in front of me but I can't seem to see it :(Thanks for your help in advanced:):):)

 

[voko]

Conservation of energy?

 

[KiNGGeexD]

So kinetic energy =0 at the top and kinetic energy after height h is =1/2mv^2?

So I can get potential energy with my velocity I calculated and the that would equal the kinetic energy after height h?

 

[KiNGGeexD]

Ok I done

mgh=1/2Iω^2

Worked through to isolate for v

Since ω=v*r

And I got

v=sqrt of 4*g*h

Not sure if this is ok or notAlso can I just ignore v initial?

 

[TSny]

Ok I done

mgh=1/2Iω^2

Using I = (1/2)MR² in the expression (1/2)Iω² gives the rotational kinetic energy about the center of the yo-yo. But there is some additional kinetic energy due to the fact that the yo-yo is "falling".

 

[KiNGGeexD]

Ah so there is angular kinetic energy and translational kinetic energy which when subtracted from the potential energy is zero?

 

[TSny]

Yes. Both forms of KE must be included.

It is usually best to apply the conservation of energy principle in the form E_f = E_i. The way you have phrased it as taking the total KE and subtracting the PE to get zero seems to me a little dangerous even though I think you will probably get the correct answer that way.

 

[KiNGGeexD]

Ok thanks! I think I'm getting the hang of the physics malarkey!

Cheers friend:)

 

[KiNGGeexD]

Would E final just be kinetic energy and E initial would be potential!
This would be the same as saying subtracting the two would be zero wouldn't it?

Potential energy is at a max when it is stationary and kinetic is at a maximum when it is moving after a time, assuming it doesn't reach the bottom of its path

 

[TSny]

Would E final just be kinetic energy and E initial would be potential!
This would be the same as saying subtracting the two would be zero wouldn't it?

Yes, this is ok as long as you realize that you have made a choice of taking the potential energy to be zero at the final height of the yoyo. Whenever you use the formula mgh for the potential energy, you just want to be clear on where you are choosing h to be zero. If I chose h to be zero on the ground instead of at the lowest point of travel of the yoyo, then E final of the yoyo would not be all kinetic energy.

Maybe I'm belaboring a point that is already clear to you!

 

[KiNGGeexD]

Ok I shall express the conservation of energy in the way you quoted!

 

[KiNGGeexD]

I got the final answer v= square root of 2gh?Looks nice and beat but I could have made a mistake in my maths:)
Wouldn't be the first time

 

[TSny]

v= square root of 2gh?

That's not the correct answer. Were you able to combine the rotational and kinetic energies into one expression?

 

[KiNGGeexD]

Yea

mgh= 1/2mv^2 + 1/2 I ω^2

Then I solved for v?

 

[TSny]

That's the right setup. Devil is in the details.

Can you show the steps that lead from this equation to your final answer?

 

[KiNGGeexD]

Yea sure...

mgh= mv^2+Iω^2

The halfs add to be 1

So

mgh/Iω^2= mv^2

v^2= mgh/m*Iω^2

I= 1/2MR^2

Have I made a mistake yet?

 

[TSny]

Yea sure...

mgh= mv^2+Iω^2

The halfs add to be 1

No. Note that (1/2)(4) + (1/2)(6) = 2 + 3 = 5

But, if you said the halfs add to 1 you would get

(1/2)(4) + (1/2)(6) = 4 + 6 = 10 (wrong)

So

mgh/Iω^2= mv^2

No, you can't do that (legitimately).
If you have a = b + c, it is not true in general that a/c = b. If you wanted to get b by itself, you could subtract c from both sides. But, you don't want to do that anyway.

Going back to mgh = (1/2)mv² + (1/2)Iω², substitute for ω in terms of v and try to simplify the right hand side.

 

[KiNGGeexD]

Ok I will give it a bash!

 

[KiNGGeexD]

Also I for some reason thought the two terms were multiplied and meant to put a subtract sign not a divide I'm am not that dumb!

 

[KiNGGeexD]

Would it then be

v^2= (4gh)/3

Then take the square root to get v! I thought it would be neater to do it this way!

 

[KiNGGeexD]

I ended up with the RHS as

1/2nv^2 + 1/4 mv^2

Once I substituted in omega and moment of inertia

 

[KiNGGeexD]

Although the dimensions don't add up

 

[TSny]

Would it then be

v^2= (4gh)/3

Then take the square root to get v! I thought it would be neater to do it this way!

Yes, that looks right!

 

[TSny]

I ended up with the RHS as

1/2nv^2 + 1/4 mv^2

Once I substituted in omega and moment of inertia

Right: 1/2 mv² + 1/4 mv² = 3/4 mv²

 

[TSny]

Although the dimensions don't add up

I'm not following you here.

 

[KiNGGeexD]

Yea so when I solved that I got v^2= 4/3 gh

 

[KiNGGeexD]

I just meant in terms of dimensional analysis

v^2 = (m/s)^2

And

g*h = (m/s^2) * m

So both are equal:)

 

[TSny]

Good work!