The discussion revolves around the concepts of tangential acceleration and the acceleration of the center of mass (CoM) in the context of rotational motion. Participants are exploring how these two forms of acceleration relate to each other when an object rolls without slipping.
The discussion is active, with multiple viewpoints being expressed. Some participants are exploring different setups and interpretations of the problem, while others are questioning the assumptions about the relationship between tangential and CoM accelerations. There is no explicit consensus, but productive dialogue is ongoing.
Participants note that the problem statement may be unclear and suggest revisiting related problems for further understanding. There is mention of specific scenarios, such as rolling without slipping and the effects of friction, which may influence the discussion.
prettydumbguy said:the tangential velocity is defined as alpha multiplied the radius,
brainpushups said: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.
Indeed I misspoke, I did mean tangential acceleration as the product of alpha and the radius.brainpushups said: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.
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.AlephNumbers said: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?
brainpushups said: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.
prettydumbguy said:The tangential acceleration and the CoM linear acceleration should be equal, as long as we're not slipping.
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?prettydumbguy said: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.
Is the tangential acceleration zero though, or is it just the net acceleration that is zero due to the opposite acceleration cause by friction?AlephNumbers said: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.
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.So is the question asking how the tangential accelerations differ, or how the angular accelerations differ?
haruspex said: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?
prettydumbguy said:I'd have to look the problem back up.