Wheel rolling on a horizontal surface

[collinsmark]

This is @Doc Al 's post

https://www.physicsforums.com/threads/instantaneous-axis-of-rotation.351097/#post-2421815

Unfortunately, Doc Al's particular examples in that thread are not good when acceleration is considered. They're fine for calculating instantaneous velocities, but fall apart if used to predict future or past events (even with infinitesimal time differences).

A better example would be a wheel slowly rolling off a sharp edge. During the segment of time where the wheel only touches the sharp edge, the point where the wheel touches the sharp edge can be considered the instantaneous axis of rotation. In this example, all points will instantaneously accelerate toward the bottom of the wheel. And, you guessed it, the wheel does have a component which actually accelerates downward (as it begins to fall off the cliff -- the center of the wheel does indeed accelerate with a downward component in this example).

 

[Jahnavi]

In summary: a rolling wheel is instantaneously a pure rotation about the lowest point in terms of position and velocity; but not in terms of acceleration and internal and external forces.

This seems to be an important information .

Please confirm one thing - If we take instantaneous axis through the bottom most point then , for this rotation , is the radius of curvature for point P 4/3 m ?

 

[PeroK]

This seems to be an important information .

Please confirm one thing - If we take instantaneous axis through the bottom most point then for this rotation, is the radius of curvature 4/3 m ?

What does "radius of curvature" mean in this case? Again, it must be best to take an inertial reference frame. Then, the curve traced out by point P is the cycloid, as posted earlier, and you can calculate the curvature from that. Although, in this case, P is acclerating along the cycloid.

Point P moves in a circle relative to point O. But, O's reference frame is non-inertial, so the curvature in that reference frame is not physically significant.

 

[Doc Al]

I've never heard the bottom point of a wheel on a flat surface called the "instantaneous axis of rotation." Can you provide a reference of it being called that?

It's also called the instantaneous center of rotation, a point where the velocity of a rotating body is instantaneously zero. Useful for describing the velocity of any point on the object and the object's kinetic energy, but (as PeroK points out) not particularly useful when considering acceleration.

PS In summary: a rolling wheel is instantaneously a pure rotation about the lowest point in terms of position and velocity; but not in terms of acceleration and internal and external forces.

Right!

 

[Jahnavi]

Useful for describing the velocity of any point on the object and the object's kinetic energy,

Yes , this was bothering me . So , even if instantaneous axis is at rest but accelerating , kinetic energy w.r.t this axis is equal to that in the lab frame ?

 

[PeroK]

Yes , this was bothering me . So , even if instantaneous axis is at rest but accelerating , kinetic energy w.r.t this axis is equal to that in the lab frame ?

KE depends on velocity and not on acceleration.

 

[Doc Al]

Yes , this was bothering me . So , even if instantaneous axis is at rest but accelerating , kinetic energy w.r.t this axis is equal to that in the lab frame ?

Right.

Please listen to PeroK!

KE depends on velocity and not on acceleration.

 

[Jahnavi]

Right.

But if point O is at rest (even though accelerating ) and wheel is rotating about it , shouldn't the radius of curvature of P about O be 4/3m ?

 

[Doc Al]

But if point O is at rest (even though accelerating ) and wheel is rotating about it , shouldn't the radius of curvature of P about O be 4/3m ?

If you mean: P is rotating about O (instantaneously) at a distance of 4/3 m, then sure. So?

 

[collinsmark]

But if point O is at rest (even though accelerating ) and wheel is rotating about it , shouldn't the radius of curvature of P about O be 4/3m ?

If the wheel was rotating about point O (at bottom edge of wheel), then the overall radius of curvature of the specified point would be 4.3 m.

But for this problem, the wheel is not rotating about point O [when also considering acceleration]. The overall radius of curvature is much larger.

---

Buy if you're looking for the radius of curvature at the top of the cycloid you're making this problem too complicated. There's a much easier way to solve this particular problem.

I hinted at this earlier, but find the following 4 things:

• x-component of the linear acceleration of the wheel's center.
• y-component of the linear acceleration of the wheel's center.
• Specified point's tangent component of the rotational acceleration about the wheel's center.
• Specified point's radial component of the rotational acceleration about the wheel's center.

Then, consider the angular relationship between the first two and the second two. Combine the vectors.

 

[kuruman]

Why so much concern about point O, the bottom-most point of the wheel and its acceleration? Just consider point Q, the point on the surface with which point O is in contact. Point Q is clearly on an inertial frame and at rest. Take a snapshot (i.e. draw the FBD) of the wheel when points O and Q are coincident. At that instant point O is at rest relative to Q and has zero velocity. Analyze the acceleration of point P relative to Q observing that (a) every point on the wheel has a velocity vector that is perpendicular to the position vector of that point with Q chosen as the origin and (b) that the center of the wheel has acceleration $a$ relative to point Q. It looks like this the direction OP was headed in the first post, except that OP did not consider the tangential component of the acceleration correctly.

The angular acceleration of the wheel about Q is $\alpha=a_{CM}/R$. What is the tangential acceleration of point P located at distance $r=4R/3$ from point Q?

 

[Jahnavi]

kuruman ,

I would like to clarify one thing .

Are points O and Q two separate points ? When we refer to the instantaneous axis , does it pass through a point on the surface or does it pass through the bottom most point on the wheel ?

 

[PeroK]

Why so much concern about point O, the bottom-most point of the wheel and its acceleration? Just consider point Q, the point on the surface with which point O is in contact. Point Q is clearly on an inertial frame and at rest. Take a snapshot (i.e. draw the FBD) of the wheel when points O and Q are coincident. At that instant point O is at rest relative to Q and has zero velocity. Analyze the acceleration of point P relative to Q observing that (a) every point on the wheel has a velocity vector that is perpendicular to the position vector of that point with Q chosen as the origin and (b) that the center of the wheel has acceleration $a$ relative to point Q. It looks like this the direction OP was headed in the first post, except that OP did not consider the tangential component of the acceleration correctly.

The angular acceleration of the wheel about Q is $\alpha=a_{CM}/R$. What is the tangential acceleration of point P located at distance $r=4R/3$ from point Q?

Using that method, what do you get for the acceleration of the centre of the wheel?

 

[Abhishek kumar]

If

If the wheel was rotating about point O (at bottom edge of wheel), then the overall radius of curvature of the specified point would be 4.3 m.

But for this problem, the wheel is not rotating about point O [when also considering acceleration]. The overall radius of curvature is much larger.

---

Buy if you're looking for the radius of curvature at the top of the cycloid you're making this problem too complicated. There's a much easier way to solve this particular problem.

I hinted at this earlier, but find the following 4 things:

• x-component of the linear acceleration of the wheel's center.
• y-component of the linear acceleration of the wheel's center.
• Specified point's tangent component of the rotational acceleration about the wheel's center.
• Specified point's radial component of the rotational acceleration about the wheel's center.

Then, consider the angular relationship between the first two and the second two. Combine the vectors.

If we work from centrr of mass frame then point looks circulating about centre but the question is circular motion is uniform or nonuniform. If motion is uniform then we can esily calculate the radial accn and addin noninertial accn.but the point is if motion of that point is nonuniform then how to calculate tangential accn?

 

[Jahnavi]

@Abhishek kumar , if you have any query , please make a new thread . Do not divert the discussion .

Allow me to resolve my doubts .

 

[Abhishek kumar]

@Abhishek kumar , if you have any query , please make a new thread . Do not divert the discussion .

I am diverting?
This is concerned to the problem

 

[PeroK]

If the wheel was rotating about point O (at bottom edge of wheel), then the overall radius of curvature of the specified point would be 4.3 m.

But for this problem, the wheel is not rotating about point O [when also considering acceleration]. The overall radius of curvature is much larger.

In fact, to emphasise this consider a particle moving horizontally above the x-axis. When it reaches the y-axis, its motion is the same as a particle in a circular orbit around the origin with the same speed. In terms of position and velocity they are indistinguishable, but one is accelerating towards the origin and the other is not.

The same issue would result from any motion that is instantaneously horizontal at the y-axis. The acceleration cannot be calculated from the instantaneous position and velocity alone.

 

[Jahnavi]

It looks like this the direction OP was headed in the first post, except that OP did not consider the tangential component of the acceleration correctly.

Please see post#7 .

The angular acceleration of the wheel about Q is $\alpha=a_{CM}/R$. What is the tangential acceleration of point P located at distance $r=4R/3$ from point Q?

4m/s²

 

[Jahnavi]

P is rotating about O (instantaneously) at a distance of 4/3 m, then sure.

But for this problem, the wheel is not rotating about point O [when also considering acceleration].

The experts do not seem to agree .

Is wheel rotating instantaneously about O ?

 

[kuruman]

Are points O and Q two separate points ? When we refer to the instantaneous axis , does it pass through a point on the surface or does it pass through the bottom most point on the wheel ?

They are two separate points, separated by an infinitesimally small distance. Point O is on the wheel and point Q is on the surface on which the wheel is rolling without slipping. You may call point Q the instantaneous axis of rotation, if you wish, because for all intents and purposes it behaves as such.

 

[PeroK]

The experts do not seem to agree .

Is wheel rotating instantaneously about O ?

I'll give you my answer. Everything in physics is only true in a certain context; under certain assumptions. For example, Newton's second law:

$F = ma$

Well, strictly speaking, that is only true in Inertial Reference Frames. We all need to understand that, but it's not necessary to emphasise that every time.

In your case, you've discovered today that the "rotation" of a rolling wheel about the point of contact with the ground is only true up to a point and there are subleties involved:

a) The wheel is definitely rotating about any physical point on its rim. But, a physical point on the rim is non-inertial.

b) The wheel is not, strictly speaking, rotating about the fixed point on the ground where the rim instantaneously touches. It's rotating about this as far as velocity is concerned, but not as far as acceleration and forces are concerned.

What you do with this deeper understanding is up to you. You're not going to change all the textbooks, but you do now have a deeper insight into how far you can use instantaneous configurations. The same issue will arise in other areas where you will have to work out whether and how you can apply a well-chosen instantaneous frame of reference to a problem.

 

[Jahnavi]

you're making this problem too complicated. There's a much easier way to solve this particular problem.

Yes .

But what has been bothering me is why the use of instantaneous axis is failing in this particular problem when it has been giving correct results up till now ?

What if in a time bound exam I had gone ahead with this without knowing the limitations of the use of instantaneous axis ?

My textbook by a local author encourages the use of instantaneous axis . In fact almost every rotational problem is solved using the instantaneous axis .

I hope you understand my concern and do not simply think it as my stubbornness

 

[collinsmark]

The experts do not seem to agree .

Is wheel rotating instantaneously about O ?

If you choose your reference frame at point O, (where O is part of the wheel, on the edge of the wheel, and is not part of the flat surface) it makes the problem more complicated than it needs to be. I'm not saying you can't do it, but it's not the approach I would take. You'd have to determine the acceleration of point O in addition to how other accelerations in reference to the acceleration of point O. There's a simpler way.

If you are going to choose an accelerating frame of reference, choose the center of the disk. Yes, the center if the wheel is accelerating. But that choice allows you to separate out the rotational acceleration of all points about the wheel's center of mass. It allows you to separate out the angular accelerations from the linear acceleration of the wheel's COM. And that's a lot easier to work with.

 

[PeroK]

They are two separate points, separated by an infinitesimally small distance. Point O is on the wheel and point Q is on the surface on which the wheel is rolling without slipping. You may call point Q the instantaneous axis of rotation, if you wish, because for all intents and purposes it behaves as such.

The point on the ground does not behave as the instantaneous point of rotation for all intents and purposes. As the OP has discovered, it behaves as such in terms of velocity but not in terms of acceleration.

 

[collinsmark]

Yes .

But what has been bothering me is why the use of instantaneous axis is failing in this particular problem when it has been giving correct results up till now ?

What if in a time bound exam I had gone ahead with this without knowing the limitations of the use of instantaneous axis ?

My textbook by a local author encourages the use of instantaneous axis . In fact almost every rotational problem is solved using the instantaneous axis .

I hope you understand my concern and do not simply think it as my stubbornness

Yes, you could still choose point O as the center of rotation. But in your final answer you would have to add the acceleration of point O itself because point O itself is accelerating. It's possible to do, but it adds an additional step.

Edit: The reason (well, at least one reason) I suggested picking the center of the wheel, is because you already know what the acceleration there is. You don't need to calculate it. If you choose point O you have more to calculate.

 

[Jahnavi]

@kuruman , I have replied your question in post#53 . Would like to read your views :)

 

[kuruman]

The point on the ground does not behave as the instantaneous point of rotation for all intents and purposes. As the OP has discovered, it behaves as such in terms of velocity but not in terms of acceleration.

Yes, of course. Bad choice of words.

The suggestion by @collinsmark in post #50 is simple and works well. One can also write an equation giving the position vector of point P on the wheel as a function of time, take the double derivative and evaluate the acceleration at a time when the position vector is vertical. The results are the same.

 

[Jahnavi]

Yes, you could still choose point O as the center of rotation. But in your final answer you would have to add the acceleration of point O itself because point O itself is accelerating.

@PeroK ,

Tangential acceleration of point P w.r.t O is 4m/s² .

Radial acceleration of P w.r.t O is 12m/s² (vertically down ) .

Acceleration of O in the lab frame is 9 m/s² ( vertically up ).

Net acceleration of point P in the lab frame would be vector sum of the above three components .

Is that correct ?

 

[PeroK]

@PeroK ,

The tangential acceleration of point P w.r.t O is 4m/s² .

Radial acceleration of P w.r.t O is 12m/s² (vertically down ) .

Acceleration of O in the lab frame is 9 m/s² ( vertically up ).

Net acceleration of point P in the frame of O would be vector sum of the above three components .

Is that correct ?

My suggestion for this problem is to test any solution against the known acceleration of the centre of the wheel.

As long as you are happy with how you got the acceleration of O, then yes that's correct.

The final answer is in the lab frame, of course.

 

[Jahnavi]

As long as you are happy with how you got the acceleration of O, then yes that's correct.

OK.

So all I had to do was to add the acceleration of O vectorially to the accelerations in post#7 so as to get acceleration of P in lab frame .

The answer also matches .

 

[collinsmark]

OK.

So all I had to do was to add the acceleration of O vectorially to the accelerations in post#7 so as to get acceleration of P in lab frame .

The answer also matches .

Yes, that's essentially correct. In post #7 you neglected to convert back to an inertial frame. Once you account for the acceleration of frame O, you'll get the correct answer.

If you haven't done so yet, you owe it to yourself to try the problem again but this time choose the wheel's center as your accelerating frame of reference. The process is the same, i.e., you'll need to add (vector wise) the acceleration of your accelerating frame (the wheel's center in this case) to your preliminary result before you are finished, in the process of converting back to the inertial frame.

So either way gives you the right answer. But now, since you've done it both ways, you can compare which method is the more straightforward of the two.