How Do You Calculate the Linear Speed of a Yo-Yo Using Energy Conservation?

[postfan]

Homework Statement

A yo-yo has mass 0.8 kg and rotational inertia 0.12 kg m2 measured about an axis perpendicular to the screen and passing through its center of mass. A light thin string is wrapped around the central part of the yo-yo that has radius 0.03 m. The string is attached to the ceiling as shown. The yo-yo is released from rest and begins to descend as the string unwraps.

Use conservation of energy to find the linear speed of the yo-yo by the time it descended a distance 0.8 m. Use g = 9.8 m/s2.

Homework Equations

The Attempt at a Solution

Used conservation of energy :mgh=.5mv^2+.5Iw. w=v^2/r^2. v=3.74.

What am I doing wrong?

 

[Delphi51]

Is the question to find the vertical velocity? At what distance fallen? You really should quote the entire question exactly as it is given to you. If you have, the answer cannot be found.
Which radius did you use in w=v/r ?

 

[Simon Bridge]

This is quite a common HS problem, it is reasonable to put I as about the center of rotation here:

Used conservation of energy :mgh=.5mv^2+.5Iw. w=v^2/r^2. v=3.74.

You need to explain your reasoning here and show the rest of your working for this to make any sense.
The attached diagram does not belong to this problem either - it's missing at least one force.
Then - there is no question in the problem statement.

The basic approach - that grav PE lost is stored as linear motion and in the rotation of the yoyo - is correct.

 

[postfan]

Well m=.8, g=9.8, h=.8 w=v/r. So I substituted these values in the above equation and got 3.74, which I know is wrong. So what am I doing wrong?

 

[postfan]

I resolved it and got the same answer 3.74, but it is wrong, can anyone please tell me why that's so?

 

[Doc Al]

Double check your arithmetic.

 

[postfan]

Is the answer 3.69?

 

[pgardn]

What is the radius of the entire yo-yo?

The whole thing has rotational K.

 

[postfan]

The radius is .03.

 

[pgardn]

That is the inner radius where the string is tied according to the diagram.

 

[postfan]

Can we say that the radius of the entire yo-yo is R?

 

[pgardn]

Can we say that the radius of the entire yo-yo is R?

You are given the rotational inertia. Since the problem does not state what type of extended mass the yo yo is we need the radius. Are you looking at the correct diagram?

 

[postfan]

The diagram is correct, and the problem doesn't give us the radius of the entire yo-yo.

 

[rcgldr]

The diagram is correct, and the problem doesn't give us the radius of the entire yo-yo.

The problem states the rotational inertia of the entire yo-yo so use that. From the first post in thread:

A yo-yo has mass 0.8 kg and rotational inertia 0.12 kg m^2

 

[Doc Al]

What is the radius of the entire yo-yo?

Why would you need that?

 

[postfan]

Wait , do I need to find the radius or not?

 

[pgardn]

Why would you need that?

I wast thinking v^2 / r^2 part?

But now I think I can find omega...

 

[postfan]

I'm really confused , do I need to find the radius or not?

 

[rcgldr]

I'm really confused , do I need to find the radius or not?

The problem states that the yo-yo has rotational inertia I = 0.12 kg m^2. There's no need to find the radius of the yo-yo, and that wouldn't help unless you knew how the mass in the yo-yo was distributed.

There's also a typo in your equation (you left out the ^2 for w^2), which should be:

m g h = .5 m v^2 + .5 I w^2

 

[postfan]

So do I just substitute and solve?

 

[rcgldr]

So do I just substitute and solve?

Yes. It seems you've already figured out how to relate w to v.

 

[postfan]

Is the answer 3.741657?

 

[Delphi51]

Yes, check the calc. I'm getting less than 1 for the velocity after falling 0.8 m.
You could show the details of your calculation if it still doesn't work out so we can compare with you.

 

[postfan]

Here's my calculation

mgh =.5mv^2+.5Iw^2
2gh = v^2 + .12*v^2
.62v^2=2*9.8*.8
v^2 = 2*9.8*.8/.62
v=sqrt(2*9.8*.8/.62)

What am I doing wrong?

 

[Delphi51]

mgh =.5mv^2+.5Iw^2
2gh = v^2 + .12*v^2

In this step you lost the I and forgot to divide the second term on the right by m. Also, you replaced w with v, when it should be w = v/r.

In the next step, I don't see how you got .62v².

Not a bad idea to hold the numbers until you have it solved for v.
After multiplying your first step by 2 and replacing w with v/r you have
mgh = v²(m + I/r²)
Dividing both sides by the brackets solves for v².

 

[postfan]

Ok, so is the answer 2.64575?

v^2=(mgh)/(m + I/r²)
V^2=(mgh)/(m+.12mr^2/r^2)
v^2=(mgh)/(1.12m)
v^2=(gh)/1.12
v=sqrt((gh)/1.12)= 2.64575

 

[rcgldr]

v^2=(mgh)/(m + I/r²)

you left out the .5's

v²=(mgh)/(.5 m + .5 (I/r²))

v²=(mgh)/(m+.12mr²/r²)

You multiplied the inertia term by r^2, the equation should still be:

v² =(mgh)/(.5 m + .5 (I/r²))

 

[postfan]

Ok so

v^2=mgh/(.5m+.5(I/r^2)
v^2=mgh/(.5m+.06(m*r^2/r^2)
v^2=mgh/(.5m+.06m)
v^2=gh/(.56)
v=sqrt(gh/.56)=3.74165

Doesn't the moment of inertia equal mr^2, and is this right?

 

[rcgldr]

Doesn't the moment of inertia equal mr^2?

The moment of inertia of the yo-yo I = .12 kg m^2 (where m is meters, not mass). The radius of the yo-yo is already taken into account for this stated moment of inertia and is not related to the radius of the central part that the string unwinds from. So just continue from here:

v² =(mgh)/(.5 m + .5 (I/r²))

 

[postfan]

v² =(mgh)/(.5 m + .5 (I/r²))
v^2=((.8)(9.8)(.8))/((.4)(.5*.12))
v=((.8)(9.8)(.8))/((.4)(.5*.12))^.5=16.1658.

How's that?

 

[Delphi51]

Yes, the moment of inertia will NOT be mr² because it is not a simple ring.

 

[Delphi51]

v² =(mgh)/(.5 m + .5 (I/r²))
v^2=((.8)(9.8)(.8))/((.4)(.5*.12))

What happened to the plus sign, lost between these two lines? Also I don't see dividing by .03 squared.

 

[postfan]

Ok so v=((.8)(9.8)(.8))/((.4)+(.5*.12/.03^2))^.5 = .3058. Is that right?

 

[rcgldr]

Ok so v=((.8)(9.8)(.8))/((.4)+(.5*.12/.03^2))^.5 = .3058. Is that right?

Yes, that seems correct.

 

[Delphi51]

Yep, I agree!