The velocity of the center of the base of a rolling cone

[guv]

Homework Statement

A cone rolls without slipping on a table. The half-angle at the vertex is $\alpha$,
and the axis has height $h$, base radius $r$, and slant height $s$. If it takes $T$ seconds for the cone to complete one revolution with the vertex fixed. What is the velocity of the center of base $P$ in the lab frame?

Relevant Equations

Relative motion, spinning tangent velocity $$\vec v_t = \vec \omega \times \vec r$$

Let the vertex of the cone be $O$, the contact point on the cone all the way to the right be $D$ touching ground. Then $v_{\text{D relative to the table}} = v_{D/table} =0$ since it rolls without slipping.
Due to relative motion $$\vec v_{P/table} = \vec v_{P/D} + \vec v_{D/table} = \vec v_{P/D} + 0 = \vec v_{P/D} $$

One way to solve for the angular velocity of $D$ relative to $P$ is by thinking about the number of revolutions $D$ completes relative to $P$. If we take a bird view, it can be seen the number of revolutions can be found from
$$N = \frac{2 \pi s}{2 \pi r} = \frac{s}{r}$$
Therefore the angular velocity of the base spinning is $$\omega = \frac{N 2 \pi}{T} = \frac{2 \pi s}{T r}$$

Therefore the velocity of P relative to the table is
$$v_{P/table} = v_{D/P} = |\vec \omega \times \vec r| = \omega r = \frac{2 \pi s}{T}$$

This answer should be wrong since that's the velocity of a point along the bigger circle with circumference $2 \pi s$ but $P$'s trajectory projects to a smaller circle. However what's wrong with the arguments used in the calculation? Every argument seems deceivingly correct. Thanks,

 

[mfb]

If $\omega$ is based on your cone then you are missing the rotation of the cone axis itself. The full motion is not a simple rotation in any axis.

 

[guv]

Thanks, It’s understood angular velocity for the cone is more complicated but can we not use Spin omega for $v_{P/D}$?

 

[hutchphd]

Is point D on the cone or on the table? The equation is only good for an instant.
I suggest that you consider distances traveled instead (over, say, time T)

 

[guv]

$D$ is on the cone. No doubt I can obtain the correct answer through simple distance and time. What bothers me is where the fallacy in this argument presented is in the initial post?

 

[etotheipi]

The fallacy in the analysis was pointed out by @mfb in #2, in that the angular velocity of $P$ w.r.t the point on the ground is not $\frac{2\pi s}{Tr}$. That is instead the magnitude of the angular velocity in a rotating frame of reference in which the cone does not undergo translational movement, i.e. you ignored incorrectly the angular velocity of the axis of the cone in the lab frame.

Let $\hat{n}(t)$ be a unit vector in the direction $\overrightarrow{OP}$, and $\hat{y}$ be a unit vector that points vertically upward in the lab frame. In a frame of reference rotating with the cone about the $\hat{y}$ axis, the angular velocity of the cone is$$\vec{\omega} = \frac{2\pi s}{Tr} \hat{n}$$The rotating reference frame itself has angular velocity $$\vec{\omega}_{\mathcal{F}'} = -\frac{2\pi}{T} \hat{y}$$which means that the angular velocity of the cone in the lab frame is$$\vec{\Omega} = \vec{\omega}_{\mathcal{F}'} + \vec{\omega} = \frac{2\pi s}{Tr} \hat{n} - \frac{2\pi}{T} \hat{y}$$which derives from the law of addition of angular velocities. Now the velocity of a point $Q$ on the rigid body can be expressed as$$\vec{v}_Q = \vec{v}_P + \vec{\Omega} \times \overrightarrow{PQ} = \vec{v}_{P} + \left(\frac{2\pi s}{Tr} \hat{n} - \frac{2\pi}{T} \hat{y} \right) \times \overrightarrow{PQ}$$Now suppose we consider the point $Q = D$ on the edge of the disk, instantaneously in contact with the table, such that $\overrightarrow{PD} = r\hat{t}$ where $\hat{t}(t)$ is a unit vector parallel to $\overrightarrow{PD}$. Since $\vec{v}_D$ equals zero,$$\vec{v}_D = \vec{v}_P + \frac{2\pi s}{T} \hat{n} \times \hat{t} - \frac{2\pi r}{T} \hat{y} \times \hat{t} = \vec{0}$$ $$\vec{v}_P = - \frac{2\pi s}{T} \hat{n} \times \hat{t} + \frac{2\pi r}{T} \hat{y} \times \hat{t}$$Now $-\hat{n} \times \hat{t}$ is a unit vector in the direction of motion of $P$, which I'll denote $\hat{m}$, and we also have that$$\hat{y} \times \hat{t} = -\sin{\alpha} \hat{m}$$So putting this all together,$$\vec{v}_P = \frac{2\pi s}{T} \hat{m} - \frac{2\pi r \sin{\alpha}}{T} \hat{m} = \frac{2\pi}{T} \left(s - r\sin{\alpha} \right) \hat{m}$$Finally, since $s- r\sin{\alpha} = h\cos{\alpha}$, we notice that the magnitude of $P$'s velocity in the lab frame is$$v_P = \frac{2\pi h\cos{\alpha}}{T}$$And this is the result we obtain by looking at the simple circular motion calculation for $P$.

 

[hutchphd]

where the fallacy in this argument presented is in the initial post?

The fallacy is that P does not rotate with a moment of length s but rather of length $h cos(\alpha)$ as @etotheipi has pointed out

 

[etotheipi]

The fallacy is that P does not rotate with a moment of length s but rather of length $h cos(\alpha)$. I don't think it is more complicated than that or is that still not the question at hand?

The intention of the problem setter is indeed to find the velocity by dividing the circumference traced out by $P$, $2\pi h\cos{\alpha}$, by $T$. However @guv asks for the flaw in his approach using angular velocities, and the resolution is that the angular velocity of the cone was calculated incorrectly. I hope I have explained how to do so in #6 to the satisfaction of the OP!

 

[guv]

@etotheipi that's all very good, I have no dispute against what you wrote.
@hutchphd no dispute with what you wrote either. I have in fact worked out the problems in many different ways.

I guess what bothers me is that I have not seen a direct correction and explanation which step was done incorrectly for the steps shown in the initial post. Alternative methods are good but they do not seem to explain what went wrong. Does this make more sense now? Thanks,

 

[etotheipi]

I guess what bothers me is that I have not seen a direct correction and explanation which step was done incorrectly for the steps shown in the initial post. Alternative methods are good but they do not seem to explain what went wrong. Does this make more sense now? Thanks,

You assumed incorrectly that the angular velocity of the point $P$ w.r.t. the point of contact with the ground is $\frac{2\pi s}{Tr}$. It is not!

 

[guv]

To be clear I deduced $\omega$ at which $D$ spins around $P$ is $\frac{2 \pi s}{Tr}$.

 

[etotheipi]

To be clear I deduced $\omega$ at which $D$ spins around $P$ is $\frac{2 \pi s}{Tr}$.

If $D$ is a point on the edge of the base of the cone, then $D$ revolves around $P$ with an angular velocity of magnitude $\frac{2 \pi s}{Tr}$ in the frame rotating with the cone about the $\hat{y}$ axis. But when you transform back into the lab frame of reference, as I explained, you will find that the angular velocity at which $D$ revolves around $P$ is $\frac{2\pi s}{Tr} \hat{n} - \frac{2\pi}{T} \hat{y}$, using my unit vector definitions. Your error is forgetting to transform the angular velocity of the cone between frames in relative motion!

 

[guv]

Excellent!

 

[etotheipi]

By the way, I've found it's really easy to go wrong with the angular velocity vector, since it's quite a peculiar object. I really like D. Gregory's Classical Mechanics textbook, especially for its discussion of angular velocities and rotating frames. I think you'd probably like it too, if anything it's a great reference!

 

[jbriggs444]

I guess what bothers me is that I have not seen a direct correction and explanation which step was done incorrectly

If the steps are individually fine but lead to an incorrect conclusion, check for the fallacy of equivocation. Do the definitions for the quantities calculated in each step match the definitions of the quantities used in the next?

Since few, if any, of the quantities are carefully defined, it is hard to know. It is oh, so easy, to assume a definition that is convenient for the current step rather than to use a consistent definition that is valid for every step.

 

[hutchphd]

Homework Statement:: A cone rolls without slipping on a table. The half-angle at the vertex is $\alpha$,
and the axis has height $h$, base radius $r$, and slant height $s$. If it takes $T$ seconds for the cone to complete one revolution with the vertex fixed. What is the velocity of the center of base $P$ in the lab frame?
Relevant Equations:: Relative motion, spinning tangent velocity $$\vec v_t = \vec \omega \times \vec r$$

View attachment 268340
Let the vertex of the cone be $O$, the contact point on the cone all the way to the right be $D$ touching ground. Then $v_{\text{D relative to the table}} = v_{D/table} =0$ since it rolls without slipping.
Due to relative motion $$\vec v_{P/table} = \vec v_{P/D} + \vec v_{D/table} = \vec v_{P/D} + 0 = \vec v_{P/D} $$

One way to solve for the angular velocity of $D$ relative to $P$ is by thinking about the number of revolutions $D$ completes relative to $P$. If we take a bird view, it can be seen the number of revolutions can be found from
$$N = \frac{2 \pi s}{2 \pi r} = \frac{s}{r}$$
Therefore the angular velocity of the base spinning is $$\omega = \frac{N 2 \pi}{T} = \frac{2 \pi s}{T r}$$

Therefore the velocity of P relative to the table is
$$v_{P/table} = v_{D/P} = |\vec \omega \times \vec r| = \omega r = \frac{2 \pi s}{T}$$

This answer should be wrong since that's the velocity of a point along the bigger circle with circumference $2 \pi s$ but $P$'s trajectory projects to a smaller circle. However what's wrong with the arguments used in the calculation? Every argument seems deceivingly correct. Thanks,