# Homework Help: Yo Yo and Inertia!

1. Mar 25, 2005

### BlasterV

Yo-Yo's Inertia A yo-yo has a rotational inertia of 810 g · cm2 and a mass of 190 g. Its axle radius is 3.6 mm, and its string is 70 cm long. The yo-yo rolls from rest down to the end of the string.

I've tried 4 or 5 ways to get linear accel, but none work. b, c, d, and e are straight forward once you get a, but I haven't been able to successfully get it. Can anyone show me how to go about getting linear accel? Thanks

(a) What is its linear acceleration?
cm/s2
(b) How long does it take to reach the end of the string?
s
(c) As it reaches the end of the string, what is its linear speed?
cm/s
(d) As it reaches the end of the string, what is itstranslational kinetic energy?
J
(e) As it reaches the end of the string, what is its rotational kinetic energy?
J
(f) As it reaches the end of the string, what is its angular speed?

2. Mar 25, 2005

### marlon

you will need to use $$a = \frac {+}{-} \alpha R$$

where r is the radius of the yo-yo. You will need to find oud whether to take the + or the minus sign. Just figure out whether a (tangential acceleration) and alpha (angular acceleration) are in the same direction or opposite to each other.

You will also need both versions of Netwon's second law

F = ma
$$I \alpha = \tau$$

marlon

3. Mar 25, 2005

### marlon

As an addendum : realize that there are two forces acting on th yo-yo : the tension T and gravity Mg. Suppose that the y-axis is pointing upwards. Then also realiz that the rope is fixed with the yo-yo on the right side so that if the yo-yo is falling down, you have a rotation counterclockwise. This yields that alpha is positive and the torque is also positive.

The acceleration a is negative because the movement is downwards.

Ma = T-Mg

What will the second versio of Newton's second law yield ?

The solution for a must be $$a = \frac {-g}{1+ \frac{I}{MR^2}}$$

Where I is the rotational inertia of the yo-yo and R it's radius

marlon

4. Mar 25, 2005

Ive done this proof befoure and I believe that the linear acceleration is $$\frac{2}{3} g$$

Regards,

5. Mar 26, 2005

### marlon

How do you come to this conclusion ???

marlon

6. Mar 26, 2005

### Sirus

And how does this help the OP in understanding the problem? BlasterV, why don't you try the problem using what marlon has posted.

7. Mar 26, 2005

I am simply giving the answer to him so he can make sure his work is correct. I am not going to post how I got it since marlon alwready posted the method. If you would like to know, feel free to pm me and I will write it all out for you so you can see the solution.

Regards,

8. Mar 26, 2005

If you would like to know, pm me and I will write it out for you. The answer you wrote can be generalized for all kinds of yo-yo's.

Regards,

9. Mar 27, 2005

### Staff: Mentor

I think you are confusing this (yo-yo) problem with another: A falling spool (cylinder) unwinding as it falls. In that special case, $I = 1/2 MR^2$ where R is the radius of the cylinder (the cylinder is the axle). Plug that into marlon's general solution and you'll get $a = 2/3 g$.

But marlon gave the correct solution for the given yo-yo problem.

10. Mar 27, 2005

My mistake. Sorry about that marlon.

Regards,