Wheel rolling on a horizontal surface

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

Since the wheel is rolling , v_cm=ωr , a_cm = αr

The wheel is in pure rotation about the point O where it touches the surface .

The point P is a distance 4/3 from the bottom point .

Angular acceleration of the point = 3 rad/s²

v_P = 4m/s

Radial acceleration of the point P about bottom point O = v_P²/(4/3) = 12 m/s² .

Are the two components of acceleration correct ?

 

[kuruman]

One is to assume that you are looking for the components of the linear acceleration. Angular acceleration is not one of them.

 

[Jahnavi]

Is radial acceleration the net linear acceleration of point P ?

 

[Abhishek kumar]

Homework Statement

View attachment 214723

Homework Equations

The Attempt at a Solution

Since the wheel is rolling , v_cm=ωr , a_cm = αr

The wheel is in pure rotation about the point O where it touches the surface .

The point P is a distance 4/3 from the bottom point .

Angular acceleration of the point = 3 rad/s²

v_P = 4m/s

Radial acceleration of the point P about bottom point O = v_P²/(4/3) = 12 m/s² .

Are the two components of acceleration correct ?

From where u got 4/3

 

[Jahnavi]

From where u got 4/3

The point whose acceleration is to be calculated is at a distance (4/3)m from the bottom most point .

 

[haruspex]

Is radial acceleration the net linear acceleration of point P ?

The linear acceleration consists of the radial acceleration (which you found) and the tangential acceleration (which follows swiftly from the angular acceleration).

 

[Jahnavi]

The linear acceleration consists of the radial acceleration (which you found) and the tangential acceleration (which follows swiftly from the angular acceleration).

Tangential acceleration = 4m/s²

Radial acceleration = 12 m/s²

Net acceleration = √(12²+4²) = 4√10 m/s²

Is that correct ?

 

[haruspex]

Tangential acceleration = 4m/s²

Radial acceleration = 12 m/s²

Net acceleration = √(12²+4²) = 4√10 m/s²

Is that correct ?

Yes, except that what you called net acceleration is the magnitude only. Acceleration is a vector, so I would think you should give the components separately in the answer. Either that or magnitude and direction.

 

[Jahnavi]

Yes, except that what you called net acceleration is the magnitude only. Acceleration is a vector, so I would think you should give the components separately in the answer. Either that or magnitude and direction.

Ok.

I suppose the question is asking for the magnitude of linear acceleration of the point .

But unfortunately , this answer is wrong .

Do you see any mistake ?

 

[collinsmark]

The point P is a distance 4/3 from the bottom point .

Angular acceleration of the point = 3 rad/s²

v_P = 4m/s

Radial acceleration of the point P about bottom point O = v_P²/(4/3) = 12 m/s² .

I think there might be a problem with the way you are obtaining the radial acceleration. By the logic you used, the center of the wheel is 1 m above point O, so the center of the wheel has a radial acceleration of 9 m/s^2 (i.e., [3^2]/1). That's not right. The center of the wheel should have a radial acceleration of 0.

Now consider a point at the edge of the wheel. It will travel in a cycloid pattern:

As you can see, the radius of curvature when the point is on top is significantly greater than 2R. I think the flaw in your logic is the assumption that the entire wheel is rotating around point O. It doesn't really work that way. Even though things are also accelerating linearly (in addition to rotational acceleration), things are still "rotating" (strictly speaking) about the center of the wheel.

The point you are concerned about isn't at the center or the edge, but rather somewhere in between. So as a sanity check, I would expect the radial acceleration to be somewhere in between what you would get for those other two points (center or edge).

[Edit: Source of animated gif: https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif]

 

[collinsmark]

As another point of reference to ponder, a point at the very bottom of the rotating wheel (at point O) has a instantaneous velocity of 0 m/s. But is its instantaneous, radial acceleration also zero?

[Hint: The answer is "no."]

 

[Abhishek kumar]

The point whose acceleration is to be calculated is at a distance (4/3)m from the bottom most point .

Radius is 1m given

I think there might be a problem with the way you are obtaining the radial acceleration. By the logic you used, the center of the wheel is 1 m above point O, so the center of the wheel has a radial acceleration of 9 m/s^2 (i.e., [3^2]/1). That's not right. The center of the wheel should have a radial acceleration of 0.

Now consider a point at the edge of the wheel. It will travel in a cycloid pattern:
View attachment 214738
As you can see, the radius of curvature when the point is on top is significantly greater than 2R. I think the flaw in your logic is that the entire wheel is rotating around point O. It doesn't really work that way. Even though things are also accelerating linearly (in addition to rotational acceleration), things are still "rotating" (strictly speaking) about the center of the wheel.

The point you are concerned about isn't at the center or the edge, but rather somewhere in between. So as a sanity check, I would expect the radial acceleration to be somewhere in between what you would get for those other two points (center or edge).

[Edit: Source of animated gif: https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif]

I think there might be a problem with the way you are obtaining the radial acceleration. By the logic you used, the center of the wheel is 1 m above point O, so the center of the wheel has a radial acceleration of 9 m/s^2 (i.e., [3^2]/1). That's not right. The center of the wheel should have a radial acceleration of 0.

Now consider a point at the edge of the wheel. It will travel in a cycloid pattern:
View attachment 214738
As you can see, the radius of curvature when the point is on top is significantly greater than 2R. I think the flaw in your logic is that the entire wheel is rotating around point O. It doesn't really work that way. Even though things are also accelerating linearly (in addition to rotational acceleration), things are still "rotating" (strictly speaking) about the center of the wheel.

The point you are concerned about isn't at the center or the edge, but rather somewhere in between. So as a sanity check, I would expect the radial acceleration to be somewhere in between what you would get for those other two points (center or edge).

[Edit: Source of animated gif: https://en.wikipedia.org/wiki/Cycloid#/media/File:Cycloid_f.gif]

Pattern you have shown in figure is when we work from frame other than wheel frame

 

[Jahnavi]

I think the flaw in your logic is the assumption that the entire wheel is rotating around point O.

Isn't the wheel rotating around an axis passing through the bottom most point O ?

I think there might be a problem with the way you are obtaining the radial acceleration.

What exactly is the problem with my approach ? I have done few problems considering the bottom most point as the instantaneous axis and obtained correct answers .

So why exactly is it failing this time ?

 

[collinsmark]

Pattern you have shown in figure is when we work from frame other than wheel frame

Yes, the pattern in the animated gif shows a frame corresponding to the flat, horizontal surface. I'm guessing that the values in the problem statement (the 3 m/s velocity of the center and the 3 m/s^2 horizontal acceleration of the wheel's center) are all relative to the flat, horizontal surface.

But for the moment, it might help to use a non-rotating, accelerating frame that corresponds to the center of the wheel, by subtracting out the 3 m/s^2 in the horizontal direction. This way, the center of the wheel isn't accelerating in the new frame. (You can always add it back into the x-component later, before your final answer). Now everything is rotating around the center. What's the instantaneous, radial acceleration of the specified point now? What direction is it in?

So when we change back to the frame of reference of the flat surface, is the acceleration that we add back in perpendicular to the point's instantaneous, radial acceleration?

 

[PeroK]

Isn't the wheel rotating around an axis passing through the bottom most point O ?

What exactly is the problem with my approach ? I have done few problems considering the bottom most point as the instantaneous axis and obtained correct answers .

So why exactly is it failing this time ?

The bottom of the wheel is accelerating relative to the centre of the wheel. So, its acceleration is not the same as the centre.

Acceleration must be the same in all inertial frames. What happens if you consider a frame moving at $3m/s$?

 

[haruspex]

Ok.

I suppose the question is asking for the magnitude of linear acceleration of the point .

But unfortunately , this answer is wrong .

Do you see any mistake ?

Yes, I do, belatedly.

 

[PeroK]

Tangential acceleration = 4m/s²

Radial acceleration = 12 m/s²

Net acceleration = √(12²+4²) = 4√10 m/s²

Is that correct ?

That is acceleration relative to point O, whose acceleration you would then need to calculate.

 

[Jahnavi]

Yes, I do, belatedly.

Please explain the mistake ?

 

[Jahnavi]

That is acceleration relative to point O, whose acceleration you would then need to calculate.

Isn't wheel undergoing pure rotation about O ?

 

[PeroK]

Isn't wheel undergoing pure rotation about O ?

Yes, but point O is accelerating. The bottom of the wheel is instananeously at rest, but it is accelerating!

 

[Jahnavi]

Yes, but point O is accelerating. The bottom of the wheel is instananeously at rest, but it is accelerating!

What does it mean when it is said that the bottommost point is the instantaneous axis ?

 

[PeroK]

What does it mean when it is said that the bottommost point is the instantaneous axis ?

You can calculate the instantaneous acceleration of this axis of rotation. It is not zero.

If you consider a simpler problem where the wheel is not (linearly) accelerating, then every point on the rim of the wheel is accelerating. In fact, every point except the centre is accelerating towards the centre.

You cannot ignore this acceleration.

 

[PeroK]

PS In fact, a good test of your method would be to calculate the acceleration of the centre of the wheel.

 

[Jahnavi]

I would like to understand what does it mean when a rolling object is said to be undergoing pure rotation about instantaneous axis at the bottom most point ?

If it's pure rotation then why is radial acceleration of point P w.r.t O wrongly calculated ?

 

[PeroK]

I would like to understand what does it mean when a rolling object is said to be undergoing pure rotation about instantaneous axis at the bottom most point ?

If it's pure rotation then why is radial acceleration of point P w.r.t O wrongly calculated ?

It's not. You correctly calculated the acceleration of point P relative to point O. Your problem is that point O is accelerating.

 

[collinsmark]

Isn't wheel undergoing pure rotation about O ?

You could, hypothetically, choose an accelerating frame where point O is the center (where O is on the bottom of the wheel) but it's not very useful in solving this problem.

When using the frame of the flat surface, then no, the wheel is not rotating around point O, not even instantaneously.

Instantaneously, every point on the wheel is rotating around the center of the wheel. (The center of the wheel is accelerating linearly too, meaning that that bit of acceleration carries over to all points on the wheel.)

Let me put it another way. For every point on the wheel, you can break up its acceleration vector in the following ways:
The linear acceleration corresponding to the acceleration of the wheel's center, plus
its rotational acceleration about the wheel's center.

You can further break up the point's linear acceleration (which matches the acceleration of the wheel's center, for all points on the wheel) into its
x-component and
y-component.

You can further break up the point's rotational acceleration (depends on the point's distance from the wheel's center and the wheel's angular acceleration) into
tangential component and
radial component.

Once you know all 4, then you can add things (vector sum) back together at the end, once you know how the angles line up (how tangential&radial correspond to x&y).

 

[Jahnavi]

It's not.

Sorry . Isn't wheel undergoing pure rotation about the bottom most point O ?

 

[Jahnavi]

When using the frame of the flat surface, then no, the wheel is not rotating around point O, not even instantaneously.

Then why is an axis through the bottom most point called as the instantaneous axis of rotation ?

 

[PeroK]

Sorry . Isn't wheel undergoing pure rotation about the bottom most point O ?

I'm not going to discuss this any more until you have calculated, using your method, the acceleration of the centre of the wheel.

In fact, I'll do it for you:

Radial acceleration of centre of wheel relative to point O is $9m/s^2$ (vertically downward).

Is the centre of the wheel accelerating downward? Or, is the bottom of the wheel accelerating vertically upward at $9m/s^2$? In the inertial reference frame of the surface?

 

[Jahnavi]

Is the centre of the wheel accelerating downward? Or, is the bottom of the wheel accelerating vertically upward at $9m/s^2$? In the inertial reference frame of the surface?

The bottom is accelerating vertically upward w.r.t the center . Since center is accelerating horizontally , in the frame of the surface , net acceleration is the resultant of the two .

 

[PeroK]

The bottom is accelerating vertically upward w.r.t the center . Since center is accelerating horizontally , in the frame of the surface , net acceleration is the resultant of the two .

Exactly. The best approach is to analyse the problem relative to the centre of the wheel, as its acceleration is known.

Also, note that the instantaneous point of contact with the surface is accelerating upward absolutely (not just relative to the centre). The centre has no vertical acceleration.

 

[Jahnavi]

I pretty much agree with you and @collinsmark .

I am confused about something else .

My doubt is - Is wheel undergoing pure rotation about an axis passing through the bottom most point ?

If not , then why is it called instantaneous axis of rotation ?

Please address this .

 

[collinsmark]

I pretty much agree with you and @collinsmark .

I am confused about something else .

My doubt is - Is wheel undergoing pure rotation about an axis passing through the bottom most point ?

If not , then why is it called instantaneous axis of rotation ?

Please address this .

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?

(I could imagine it being called that if a rigid wheel was moving over a sharp rock, or maybe a wheel slowly rolling off of a sudden cliff. But I don't think it's an accurate description for a wheel on a flat surface.)

 

[Jahnavi]

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?

This is @Doc Al 's post

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

 

[PeroK]

I pretty much agree with you and @collinsmark .

I am confused about something else .

My doubt is - Is wheel undergoing pure rotation about an axis passing through the bottom most point ?

If not , then why is it called instantaneous axis of rotation ?

Please address this .

It all depends what you mean by "pure" rotation. In this case, the centre of the rotation is actually an accelerating reference frame. In fact, in the reference frame of any point on the wheel, the motion is a "pure" rotation about that point. You can always decompose the motion of a rigid body as the motion of any single point and a rotation of the body about that point. The most useful case is often to consider the motion of the COM, because the motion of the COM is determined by the external forces.

In the case of a rolling wheel, you could pick any point on the rim and decompose the motion into the motion of that point and a rotation of the wheel about that point. Then, when that point hits the ground it is instantaneously at rest and the motion becomes "pure" rotation. But, that is still relative to an accelerating reference frame.

Look at it another way. Take the two cases. a) The rolling wheel. b) The wheel is (really) rotating about a fixed point on the rim.

If you take a snapshot of the motion, then in both cases the position and velocity of very point on the wheel could be the same - assuming you catch the rotating wheel just at the right moment But, the motions are different because of the acceleration. So, a) and b) are not equivalent, even instantaneously, as the accelerations are different. Each point may have the same position and the same velocity, but not the same acceleration.

You have a simpler case with a ball reaching its highest point under gravity, compared with a ball at rest with no net force acting on it. Both are stationary but they are not in the same state of rest as one is accelerating.

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.

 

[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 .