# Some rotation without slipping problems.

1. Feb 5, 2009

### nns91

1. The problem statement, all variables and given/known data

1. A ball rolls without slipping down an incline of angle $$\theta$$. The coefficient of static friction is $$\mu$$s. Find the maximum angle of the incline for which the ball will roll without slipping.

2. Find the percentages of the total kinetic energy associated with rotation and translation, respectively, for an object that is rolling without slipping if the object is (a) a uniform sphere, (b) a uniform cylinder, or (c) a hoop.

2. Relevant equations

$$\tau$$=I*$$\omega$$

3. The attempt at a solution

1. I found out the acceleration and the frictional force. How do I relate these information to the maximum angle ?

I know that for the ball to roll without slipping the condition V= R*$$\omega$$ is hold.

2. I know that KE= (1/2)mv^2 + (1/2)I$$\omega$$^2.

How do I calculate the percentage from there ??

2. Feb 5, 2009

### Delphi51

If it was a block instead of a ball, you would just find when the component of gravitational force down the ramp is equal to the force of friction. Maybe that is all you need do in this case, too.

For (2) you must find the translation energy KE = .5mv^2 and the rotational energy separately and then what % each is of their sum. Of course the I varies for parts a, b, and c.

3. Feb 5, 2009

### nns91

Thanks.

For number 2.

I don't have specific number so how can calculate the percentage ? I know the formula for translation and rotation. I still don't get the idea of how I can get the percentage.

4. Feb 5, 2009

### Delphi51

Good question!
As the ball rolls down, the velocity = r*omega.
If you put that into the KE formula in your first post, you get a nice expression for the total energy that only has one unknown, omega. The m, r and I must be considered known quantities. Yes, looks good - you can find the ratio of rotational energy to total energy because the omegas cancel out. Interesting that the ratio stays the same as the rotational velocity and regular velocity increase as it rolls down.

Of course you will have m, r and I in your final answer. Oh, the m should cancel out when you put in the appropriate expression for moment of inertia. Maybe the r will, too.

I recall racing different objects down a ramp. There was quite a difference between a sphere and a cylinder and a big difference between a solid cylinder and a hollow one. No doubt due to these ratios being different.

5. Feb 5, 2009

### nns91

I got it nicely. Thank you delphi.

I still have one more problem.

A hollow cylinder and an uniform cylinder are roling horizontally without slipping. The speed of the hollow cylinder is v. The cylinders encounter an inclined plane that they climb without slipping. If the maximum height they reach is the same, find the initial speed v' of the uniform cylinder.

I know it's an energy problem so since they achieve the same height their potential energy has to equal so I can set their KEs equal to each other.

However, I of a hollow is (1/2)m*(r1^2+r2^2) so when I substitute for omega, will omega be v/r still ? or will it be something different than r ?

6. Feb 6, 2009

### nns91

I am still stucking there. Anyone know how to deal with the substitution for omega of the kinetic energy of the hollow cylinder since it has 2 radii.

7. Feb 7, 2009

### Staff: Mentor

The condition for rolling without slipping, v = ωr, always uses the outer radius. But just treat it as a thin hollow cylinder with a single radius. (Otherwise you don't have enough information.)

8. Feb 7, 2009

### nns91

thank you. That's what I did too. Thank you.

9. Dec 23, 2009

### GuitarJunkie

I just finished this problem for a webassign, anyone else who visits this page the question being asked is the percentage of Rotational Kinetic Energy to the total Kinetic Energy, and the same with translational.

So your equations will look like this for translation vs total

[(1/2)(M)(v)^2] / [(1/2)(M)(v)^2+(1/2)(I)(omega)^2]

v=R*omega

so your equations will look like this

[(1/2)(M)(R^2)(omega^2)] \ [(1/2)(M)(R^2)(omega^2)+(1/2)(Beta)(M)(R^2)(omega)^2]

Now remember, I, or the moment of Inertia, is some constant Beta * MR^2

The Beta values for a..

Sphere=(2/5)
Cylinder=(1/2)
Hoop=1

these values are determined by the integral r^2dm and are probably in your physics text book.

So when you do the algebra, and plug in the Beta values, you get the percentages

Sphere:
rotational=29
translational=71

Cylinder:
rotational=33
translational=67

Hoop:
rotational=50
translational=50