# Why do we assume this?

1. Aug 4, 2007

### pardesi

why do we assume this??

most problems on rotational mechanics which we do we are not provided with the rotational axis ...like in case of a sphere which when given some initial translational speed and left finally starts rolling we assume that it rotates about the axis passing through the c.o.m and perpendicular to it's plane...is this just an intutive conclusion based on say experiments or is that a 'theoritical' proof for it

2. Aug 4, 2007

### Staff: Mentor

Since friction opposes slipping, the friction force will be opposite to the velocity of the sphere. That friction creates a torque which rotates the sphere. (Not sure if that's what you're looking for.)

3. Aug 4, 2007

### pardesi

yes that's ok but there is torque about many points but why do we take the rotaional axis as the axis through centre

4. Aug 4, 2007

### Staff: Mentor

You can take the torque about any point you like, it's just easier to describe the motion as the sum of a rotation about the center of mass and the translation of the center of mass.

5. Aug 4, 2007

### pardesi

yes that's right but how do we 'know' the body sotates about an axis passing throughthe c.o.m yes it is highly intutive but is intution enough or do we have a 'proof'

6. Aug 6, 2007

### Meir Achuz

There is a theorem (easily proven) that shows that the angular momentum about any axis of a moving and rotating body equals the angular momentum of the center of mass of the object about the chosen axis plus the angular momentum (I\omega) of the rotating object about its center of mass.

7. Aug 7, 2007

### ice109

the parallel axis theorem?

8. Aug 7, 2007

### Meir Achuz

Not quite the parallel axis theorem, which gives I about an axis other than one through the cm.
This one says: J=L+S, where J is the total angular momentum,
L=MRXV, and S=I.omega. R and V refer to the center of mass and
I is the tensor of inertia.

9. Aug 7, 2007

### uiulic

For a sphere, we want to know the evolution of the configuration of each point attached to the sphere body (the contact may be more of concern). We then want to know the position of the center first and then check how the other points (especially contact points for a rigid sphere) move relative to this center.

regards

Last edited: Aug 7, 2007
10. Aug 8, 2007

### matematikawan

Is it Chasle's theorem? A displacement can be considered as a translation plus a rotation about its center of mass.

11. Aug 8, 2007

### rcgldr

The axis of a rolling sphere goes through the center of the sphere. If the sphere has uniform density, then the center of the sphere is also it's center of mass.

Continuing with the assumption of a uniform sphere, then the torque force is the result of the equal and opposing, but non-alligned friction force and the reaction force of deceleration of the sphere. The reaction force effectively occurs at the sphere's center of mass. These two opposing forces are seperated by the distance of the radius of the sphere. The center of these forces is 1/2 the radius of the sphere above the surface the sphere is sliding on. The torque is equal to the sum of each force times 1/2 the radius, which translates into (reaction force + friction force) x 1/2 radius = friction force x radius (since reaction force == friction force).

In the case of simple objects, like a sphere, I'm pretty sure that it can be proven with calculus that the center of rotation is the center of mass.

Last edited: Aug 8, 2007
12. Aug 9, 2007

### BobG

If you just push a sphere without giving it any spin, then the angular velocity vector and angular momentum vector all have to line up with the torque vector simply because you only have one component to your motion.

Does that mean that's the only way a sphere can rotate? No.

A bowling ball thrown with a mean hook would be quite a challenge for an early physics problem. So would a billiard ball struck with a little 'English'. So would a spinning satellite undergoing nutation because of fluid slosh (or whatever).