# Finding the *Time* it takes for yo-yo to travel 1 meter?

1. Feb 18, 2015

### Eugene Han

1. The problem statement, all variables and given/known data
This is for a lab report.

Research Question: Why does a yoyo fall slower when unraveled than a yoyo in free fall?

I need to find the time it takes for a yo-yo to travel 1 meter (falling) while unraveling/unwinding.

The height is 1 meter.

(I'm not looking for an actual answer, just the equations to solve it).

2. Relevant equations
h = 1/2 at^2 (free-fall of a yoyo)
I = 1/2 mr^2 (moment of inertia for a yoyo)

3. The attempt at a solution
h = 1/2 at^2
...
I do not know how to start this problem at all...

Edit:

mgh = 1/2 mv^2 + 1/2Iw^2
I = 1/2 mr^2
w = v/r

mgh = 1/2mv^2 + 1/4mv^2
mgh = 3/4mv^2
gh = 3/4v^2

wait ... why am I solving for velocity?!?!?!

I need to find time ... thanks!

2. Feb 18, 2015

### Nathanael

The same reason a block is slower when sliding down a frictional (inclined) plane than a frictionless one.

You could compare it with what the velocity that it would've had in free fall.

3. Feb 18, 2015

### lightgrav

is it constant acceleration? thhat is, are you treating the 1m string as being all wrapped at the inner radius?

4. Feb 18, 2015

### Eugene Han

The string is not necessarily 1 meter. A string is there, but I think it simply exists to slow it down. And yes, the string is all wrapped around the inner radius and we are dropping it while it unravels for a height of 1 meter.

5. Feb 18, 2015

### Eugene Han

I can't change the research question :( It's metaphorical, and we are using that question to solve for the time for comparison. I already solved for the time during free fall - I just need to solve the time during the falling of the yoyo while unraveling!

6. Feb 18, 2015

### Nathanael

I didn't mean to imply that you should change it. I was giving you a (half-way) answer the research question. The other half of the answer is 'because it takes energy to get all that matter moving!'

Sorry I thought you were trying to show it unravels slower. In post 1 you solved for the speed (while unraveling) after a given height, right? To find the time it falls 1 meter, just find the final velocity after 1 meter, and then find the time like a normal kinematics problem (the average velocity after falling 1 meter is half the final velocity, because acceleration is constant).

7. Feb 18, 2015

### Eugene Han

I'm confused on finding the final velocity.

What did I find? I got that v = 3.61478 m/s after solving that whole mess up there, using g = 9.8 m/s^2 and h = 1 meter.

Thanks.

8. Feb 18, 2015

### Eugene Han

HOLY MOLEY I might've just done something.

I got v = 3.61489.

Then I did vf = vi + at.

3.61458 = 0 + 9.8t

t = 0.368 seconds.

Wait ... the time it takes to fall while unraveling than the time it takes to fall in free fall (0.45 seconds) should be longer though ... gianeiogneoa I'm so lost! :(

9. Feb 18, 2015

### lightgrav

the acceleration is much less than caused by gravity alone ... the string pulls it upoward!
vavg happens to be = ½ vfinal , as Nathanael says. What is the definition for vavg?

10. Feb 18, 2015

### Eugene Han

I believe Vavg is change in position over change in time.

In my case, h / t, or 1 / t.

Hmm

11. Feb 18, 2015

### Eugene Han

I solved something out.

V avg = 1/2 Vfinal.

1/t = 1/2 (3.61m/s)

1/t = 1.805/1
cross multiple

1.805t = 1.

t = 0.5540166205 seconds ... Hmmm did I get it?! :D

12. Feb 18, 2015

### Nathanael

Yes.

13. Feb 18, 2015

### Eugene Han

One LAST question ...

My friends were telling me that I needed the inner radius of the yo-yo to solve this question.

Was this not true? I feel like I'm missing a vital component ...

If not, I guess I'm good...

14. Feb 18, 2015

### Nathanael

Your friends should've finished the problem! The angular velocity is dependent on the inner radius: ω=v/r and the rotational inertia is dependent on the the inner radius: I=0.5mr2 but these two dependencies cancel out.

They cancel out because the rotational energy is 0.5Iω2 but since I is proportional to r2 and ω2 is proportional to 1/r2, it cancels out.

Last edited: Feb 18, 2015
15. Feb 19, 2015

### haruspex

I disagree. The moment of inertia depends principally on the outer radius, while the relationship of linear speed to angular speed depends on the inner radius. A yo-yo with a large outer radius and small inner radius will descend very slowly.

16. Feb 19, 2015

### Nathanael

I was imagining it as an unknown mass distribution, which seemed to imply that the shape is irrelevant. I just took the moment of inertia to be measured as some arbitrary multiple of mr2. Is this wrong? I was essentially treating it as some cylinder of unknown mass distribution which happens to have moment of inertia of mr2/2

17. Feb 19, 2015

### haruspex

Are you saying you were taking the two equations in the OP, I=mr2/2 and v = wr, as correct? I suggest that since r is defined as the inner radius the expression for I is wrong.

18. Feb 19, 2015

### Nathanael

Yes I was taking it as correct. Realistically it wouldn't vary with r2. (Except perhaps for a set of yo-yo's with unusual mass distributions)

19. Feb 20, 2015

### dean barry

Differeny way of approaching this:
Imagine the yoyo body rolling without slipping (at a constant arbitrary speed, say 10 m/s) along a blade contacting on the inner radius.
Calculate the linear KE ( ½ * m * v ² ) and rotational KE ( ½ * I * ω ²) at this speed.
Calculate the effective mass ( em ) in kg from:
em = actual mass in kg * ( 1 + ( rotational KE / linear KE ) )
Then the effective mass is accelerated by the gravitational force.

20. Feb 20, 2015

### rcgldr

You need to know both the inner and outer radius, and also the mass distribution of the yo-yo. A simplified model of a yo-yo would treat it as a uniform (same density everywhere) solid cylinder. There are two approaches to solving for this.

One is to calculate the acceleration of the yo-yo, taking into account that the string exerts an upwards force at the axis of the yo-yo, and this upwards force times the inner radius is the torque exerted onto the yo-yo. The rate of angular acceleration of the yo-yo is the torque divided by the angular momentum of the yo-yo. The angular acceleration can be converted into downwards acceleration by multiplying by the innner radius. You also know that the downwards acceleration is equal to the net force divided by the mass of the yo-yo and that the net force is the downwards force due to gravity minus the upwards force exerted by the string . This gives you two equations where the downwards acceleration is a common factor, allowing you to solve the problem.

Another approach is to note the ratio of angular energy versus linear energy as the yo-yo descends, and note that the total increase in energy of the yo-yo equals m g height (the distance the yo-yo descends).

Last edited: Feb 20, 2015