# Difference between Tangential and CoM acceleration

1. Apr 26, 2015

### prettydumbguy

1. The problem statement, all variables and given/known data
In rotational motion, the tangential velocity is defined as alpha multiplied the radius,
When an object is rolling without slipping, the acceleration of the center of the mass is defined as alpha multiplied by the radius. How, if at all, are these two alphas different?

2. Relevant equations
At = alpha * radius
Acm = alpha * radius

3. The attempt at a solution

2. Apr 26, 2015

### brainpushups

No. If, by alpha you mean the angular acceleration then the product of alpha and the radius of rotation is the tangential acceleration, not velocity.

3. Apr 26, 2015

### AlephNumbers

Maybe guy likes to use alpha to denote angular velocity instead of omega?

In any case, do you have any thoughts on the problem, guy?

4. Apr 26, 2015

### prettydumbguy

Indeed I misspoke, I did mean tangential acceleration as the product of alpha and the radius.

5. Apr 26, 2015

### prettydumbguy

I can't see how they would be different. When an object rolls without slipping, the linear distance it travels is equal to the angular distance multiplied by the radius: a ball with radius R that rotated by one radian will have translated a linear distance equal to R. So it makes sense to me that the magnitude of the tangential velocity is the same as the magnitude of the center of mass (the CoM rotates by theta degrees, a point on the edge of the ball rotates by theta degrees times the radius, the ball CoM has translated by the same amount). So I don't see how I can increase the speed of one without ruining that equation, so, I think that no, The tangential acceleration and the CoM linear acceleration should be equal, as long as we're not slipping.

6. Apr 26, 2015

### brainpushups

Sure. Suppose that a rigid object rolls down an incline so that it is accelerating. Relative to the center of the wheel, the speed of a point on the wheel's circumference is equal to the translational velocity of the CM. The acceleration of the CM and the tangential acceleration of the wheel are also equal.

7. Apr 26, 2015

### AlephNumbers

I politely disagree.

Consider a wheel that is rolling on the ground without slipping. Think of the wheel as rotating about the point of contact with the ground. I think we can all agree that v = ωr. But the distance from a point on the wheel in contact with the ground to the point of rotation is zero. So we come to the conclusion that the point in contact with the ground has a tangential velocity of zero. This can be extrapolated to account for tangential acceleration, I think.

So is the question asking how the tangential accelerations differ, or how the angular accelerations differ?

8. Apr 26, 2015

### brainpushups

I do not disagree with you. Notice that I was careful to say 'relative to the center of the wheel' (though I should have said tangential 'speed,' not velocity). Treating the axis of rotation to be the instantaneous point of contact with the ground is different. The speed of the center of the wheel is half that of the top of the wheel from this point of view.

9. Apr 26, 2015

### haruspex

Sure, but you are talking about two different set-ups, right? Motion of centre of rolling wheel versus tangential motion of wheel rotating about fixed centre?

10. Apr 26, 2015

### prettydumbguy

Is the tangential acceleration zero though, or is it just the net acceleration that is zero due to the opposite acceleration cause by friction?

It's actually part of a larger question that stumped me, where my tangential acceleration at the point of contact was not equal to the translational acceleration of a cylinder and it confused the heck out of me. The tangential acceleration was used to find the torque caused by friction (IIRC) and it was different than the translational acceleration. I'd have to look the problem back up.

11. Apr 26, 2015

### AlephNumbers

As it is, the problem statement is a little unclear. Why don't you look up that old problem and make a new thread for it. I'm sure someone could help you work through it, and maybe it will help you to understand this problem too.

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