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 .