# Speed of sphere center

• January 2nd 2007, 04:17 AM
totalnewbie
Speed of sphere center
Sphere begins to wheel along incline.
The height of incline is 1.00 meters.
What is the speed of sphere center at the end of incline ?
• January 2nd 2007, 06:27 AM
CaptainBlack
Quote:

Originally Posted by totalnewbie
Sphere begins to wheel along incline.
The height of incline is 1.00 meters.
What is the speed of sphere center at the end of incline ?

Equate the loss of potential energy of the sphere to the gain in KE of
the sphere, which is the sum of the translational and rotational KE's.
(assume no slipping)

RonL
• January 2nd 2007, 06:57 AM
topsquark
Quote:

Originally Posted by totalnewbie
Sphere begins to wheel along incline.
The height of incline is 1.00 meters.
What is the speed of sphere center at the end of incline ?

I presume the sphere is rolling without slipping?

Conservation of energy says:
$\Delta E_{tot} = \Delta K + \Delta K_{rot} + \Delta U$
where $E_{tot}$ is the total energy of the sphere, K is the linear kinetic energy, and $K_{rot}$ is the rotational kinetic energy.

One thing about this kind of problem that can bother students is that in order for the sphere to roll without slipping there must be some friction between the sphere and the incline and there is no friction term (or equivalently no nonconservative work term) in the above energy equation. This term is accounted for as the change in rotational kinetic energy.

Since there is no work done other than the work done by friction in the rolling process we can state that
$\Delta E = 0$
so we have:
$0 = \left ( \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \right ) + \left ( \frac{1}{2}I \omega ^2 - \frac{1}{2}I \omega _0^2 \right ) + \left ( mgh - mgh_0 \right )$

where m is the mass of the sphere, v is the linear speed of the center of the sphere, I is the moment of inertia of the sphere, $\omega$ is the rotational speed of the sphere about the point of rotation, and h is the height above some defined 0 point.

One question: What point is the sphere rotating about as it rolls down the plane? The instantaneous point of rotation is the point of contact between the sphere and the plane, NOT the center of the sphere. (If you need help understanding why just ask and I'll explain that in a later post.) So the moment of inertia is taken as the point of contact between the sphere and the plane.

The moment of inertia of a (solid) sphere rotating about an axis through the center of the sphere is $I = \frac{2}{5}mR^2$ where m is the mass of the sphere and R is the radius. Using the parallel axis theorem we can move this axis of rotation to a point on the edge of the sphere. This gives us:
$I = I_{CM} + md^2 = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$.

I am going to define a zero level for the gravitational potential energy to be at the bottom of the ramp. So $h_0 = 1 \, m$ and $h = 0 \, m$. Additionally the sphere is stationary at the top of the incline, so $v_0 = 0 \, m/s$ and $\omega _0 = 0 \, rad/s$.

$0 = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{7}{5}mR^2 \omega ^2 - mgh_0$

We are almost done. What is the relationship between $v$ and $\omega$? Well, the sphere is rolling about the point of contact between the sphere and the plane, which is a distance R from the CM of the sphere. So we know that $v = \omega R$. We want the linear speed of the sphere at the bottom of the incline so we use $\omega = \frac{v}{R}$ in the energy equation:
$0 = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{7}{5}mR^2 \left ( \frac{v}{R} \right ) ^2 - mgh_0$

Simplifying:
$0 = \frac{1}{2}mv^2 + \frac{7}{10}mv^2 - mgh_0$

$0 = \frac{1}{2}v^2 + \frac{7}{10}v^2 - gh_0$ <-- That pesky m cancels.

$0 = \frac{12}{10}v^2 - gh_0$

$0 = \frac{6}{5}v^2 - gh_0$

$\frac{6}{5}v^2 = gh_0$

$v = \sqrt{\frac{5}{6}gh_0}$

(Note: This means that ANY sphere of any radius and any mass starting from rest at the top of the incline will have the same linear speed at the bottom of the incline.)

-Dan
• January 2nd 2007, 09:51 AM
totalnewbie
By the way.
Momentum of inertia should be $\frac{2}{5}$ because we want to find velocity in relation to center.

Then $v = \sqrt{\frac{10}{7}gh_0}$
• January 2nd 2007, 02:28 PM
topsquark
Quote:

Originally Posted by totalnewbie
By the way.
Momentum of inertia should be $\frac{2}{5}$ because we want to find velocity in relation to center.

Then $v = \sqrt{\frac{10}{7}gh_0}$

This relates to my comment of which point the sphere rotates about. The moment of inertia is calculated soley from the position and orientation of the axis of rotation, which is not the center of the sphere.

On any rotating body there is exactly one line about which it rotates. Now, look at the instantaneous translational speeds of all points of the sphere. There is only one point on the sphere that does not have a translational speed, and it is not the center of the sphere, it is the point of contact between the sphere and the plane. (Remember we have rolling without slipping!) This is the point of rotation. Now, this point DOES move so the axis of rotation moves along with the sphere, but that's a good thing since the axis of rotation will always have the same relationship to the CM of the sphere.

Thus the moment of inertia needs to be calculated for an axis tangent to the surface of the sphere. Which gives $I = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$ via the parallel axis theorem as I claimed before.

-Dan
• January 2nd 2007, 08:48 PM
CaptainBlack
Quote:

Originally Posted by topsquark
This relates to my comment of which point the sphere rotates about. The moment of inertia is calculated soley from the position and orientation of the axis of rotation, which is not the center of the sphere.

On any rotating body there is exactly one line about which it rotates. Now, look at the instantaneous translational speeds of all points of the sphere. There is only one point on the sphere that does not have a translational speed, and it is not the center of the sphere, it is the point of contact between the sphere and the plane. (Remember we have rolling without slipping!) This is the point of rotation. Now, this point DOES move so the axis of rotation moves along with the sphere, but that's a good thing since the axis of rotation will always have the same relationship to the CM of the sphere.

Thus the moment of inertia needs to be calculated for an axis tangent to the surface of the sphere. Which gives $I = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$ via the parallel axis theorem as I claimed before.

-Dan

I don't think so. Consider the situation if you remove the plane but leave the
sphere moving with the same translational and angular velocities through
space. The rotational and translational KE's are unchanged, but it is clear
that if you were to increase the spin the sphere while its translational
velocity remains unchanged the point about which we would calculate the
new angular velocity for energy considerations is the centre of mass.

Alternatively consider a swarm of particles, then the KE of the swarm can
be written as the translational KE of the swarm (that of a virtual particle
of the same total mass situated at the centre of mass of the swarm)
and the rotational energy of the swarm about the centre of mass.

RonL
• January 3rd 2007, 05:30 AM
topsquark
Quote:

Originally Posted by CaptainBlack
I don't think so. Consider the situation if you remove the plane but leave the
sphere moving with the same translational and angular velocities through
space. The rotational and translational KE's are unchanged, but it is clear
that if you were to increase the spin the sphere while its translational
velocity remains unchanged the point about which we would calculate the
new angular velocity for energy considerations is the centre of mass.

However these are not the same physical situations. Recall that in the case where the plane is present there is a frictional force acting on the sphere. This frictional force is naturally absent for the case where the plane is no longer there, so we should expect a difference in the equations. And recall that the disk accelerates down the plane, it does not move with a constant speed.

Consider not a sphere, but a thin disk rolling without slipping along a horizontal plane. (To keep things simple.) Pick two points on the disk, the point of contact (A), and the CM of the disk (B). Assuming the disk is moving with a constant linear speed v, what are the translational speeds at points A and B?

By definintion the translational speed at B is the same as the linear speed v, since B is the CM. Since A is the point of contact we know the speed is 0 because we are rolling without slipping. (The point at the top of the disk in line with A and B is moving (instantaneously) with a translational speed of 2v.)

So where is the axis of rotation? It MUST be the point A, since this is the only point on the disk that is not moving. So instantaneously the disk is rotating about point A, not point B. The trouble most people have with this concept is that the axis of rotation moves along with the disk, so it would appear that the axis of rotation must be at the CM of the disk.

As a further verification of this argument, consider that there is a frictional force on the disk that keeps it rolling without slipping. How can we possibly argue that the disk will continue to roll with the same linear velocity indefinitely if there is a friction force on it (which should cause the disk to decelerate)? The frictional force will decelerate the system only if it is doing (rotational) work on the disk. This will be true if the axis of rotation is at the CM because the work done would be -fR where f is the friction force and R is the radius of the disk. So if the disk has an axis of rotation about the CM we predict it will have an angular deceleration and thus slow down, contrary to observation (assuming no other resistive forces on the disk). But if the axis of rotation is at the point of contact between the disk and the plane then friction can do no work since the distance between the friction force and the axis of rotation is 0. So we predict that the disk will continue to rotate with the same angular velocity. Since the angular velocity is constant, so too must the linear velocity be constant, in accord with observation.

(If you think about it carefully the above argument implies there is a centripetal acceleration on the CM. This acceleration will be constant since the CM is always the same distance from the axis of rotation and is always in the same orientation with respect to the axis of rotation (it is always directly above it) and thus gives the CM a constant linear speed, which again says that the disk should not slow down.)

In reality the only thing to do is an experiment to see if the concept comes up with the correct numbers. I did the labwork in my Undergrad days and I can tell you that it does indeed give the correct results.

-Dan
• January 3rd 2007, 05:43 AM
CaptainBlack
Quote:

Originally Posted by topsquark
However these are not the same physical situations. Recall that in the case where the plane is present there is a frictional force acting on the sphere. This frictional force is naturally absent for the case where the plane is no longer there, so we should expect a difference in the equations. And recall that the disk accelerates down the plane, it does not move with a constant speed.

All irrelevant to an energy argument.

The no slippage condition establishes the relation between angular velocity and
translational velocity, but the energy equations are oblivious to the presence of
the surface.

(It's a consequence of integrating the KE of the mass elements constituting the
sphere over the sphere that the KE is the sum of the KE corresponding to the
translational motion of the centre of mass, and the KE of the rotational motion
RonL
• January 3rd 2007, 09:44 AM
topsquark
Quote:

Originally Posted by CaptainBlack
All irrelevant to an energy argument.

The no slippage condition establishes the relation between angular velocity and
translational velocity, but the energy equations are oblivious to the presence of
the surface.

(It's a consequence of integrating the KE of the mass elements constituting the
sphere over the sphere that the KE is the sum of the KE corresponding to the
translational motion of the centre of mass, and the KE of the rotational motion
RonL

However the energy argument is NOT immune to the presence of the friction if the axis of rotation is through the CM of the sphere. The friction would create a net torque on the sphere, which would cause it to lose energy over the distance it moves. As this does not happen, the friction force must be located at the rotation axis.

-Dan
• January 3rd 2007, 10:21 AM
CaptainBlack
Quote:

Originally Posted by topsquark
However the energy argument is NOT immune to the presence of the friction if the axis of rotation is through the CM of the sphere. The friction would create a net torque on the sphere, which would cause it to lose energy over the distance it moves. As this does not happen, the friction force must be located at the rotation axis.

-Dan

The friction does no work as the point of application is stationary in the
no slip condition.

Anyway I am not continuing this discussion

RonL
• January 3rd 2007, 10:35 AM
totalnewbie
Anyway my physic lector told me the correct answer was: $v = \sqrt{\frac{10}{7}gh_0}$
• January 4th 2007, 04:36 AM
topsquark
(sigh) I hate it when this happens. My discomfort isn't so much that I got the wrong answer, but why I was wrong. (That and the fact I argued so strenously for it.)

All of my previous arguments were correct, but there is one slight problem with them. I have been mixing two problem solving methods. Let's go back to my disk rolling on a horizontal plane, which is slightly simpler. What I was trying to do was to look at the problem from the point of view that the wheel is (instantaneously) in pure rotational motion. The axis of rotation is indeed the point where the disk makes contact with the plane.

However if this is the case then I canNOT include a translational kinetic energy term in the conservation of energy equation because from my method I am assuming no translational motion.

What happens is this: We need to use the parallel axis theorem to figure the moment of inertia about the actual axis of rotation. This means that
$I = I_{CM} + mR^2$
The ultimate effect of the second term is to give the translational kinetic energy term.

So using the axis of rotation I was using I should have used the moment of inertia that I used, but not included a translational kinetic energy term. Alternately since I included a translational kinetic energy term in the conservation of energy equation I should have used the moment of inertia about an axis through the CM. Both methods give the same answer, the one that totalnewbie just posted.

Many apologies for the confusion!

-Dan
• January 4th 2007, 05:32 AM
CaptainBlack
Quote:

Originally Posted by topsquark
(sigh) I hate it when this happens. My discomfort isn't so much that I got the wrong answer, but why I was wrong. (That and the fact I argued so strenously for it.)

All of my previous arguments were correct, but there is one slight problem with them. I have been mixing two problem solving methods. Let's go back to my disk rolling on a horizontal plane, which is slightly simpler. What I was trying to do was to look at the problem from the point of view that the wheel is (instantaneously) in pure rotational motion. The axis of rotation is indeed the point where the disk makes contact with the plane.

However if this is the case then I canNOT include a translational kinetic energy term in the conservation of energy equation because from my method I am assuming no translational motion.

What happens is this: We need to use the parallel axis theorem to figure the moment of inertia about the actual axis of rotation. This means that
$I = I_{CM} + mR^2$
The ultimate effect of the second term is to give the translational kinetic energy term.

So using the axis of rotation I was using I should have used the moment of inertia that I used, but not included a translational kinetic energy term. Alternately since I included a translational kinetic energy term in the conservation of energy equation I should have used the moment of inertia about an axis through the CM. Both methods give the same answer, the one that totalnewbie just posted.

Many apologies for the confusion!

-Dan

Damn, I had just got the experimental apparatus together to time a
sphere rolling down an inclined track, and solved the equations of
motion so I could calculate the time using the two candidate models.

The experiment was going to be on for this evening:(

RonL
• January 5th 2007, 09:49 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Damn, I had just got the experimental apparatus together to time a
sphere rolling down an inclined track, and solved the equations of
motion so I could calculate the time using the two candidate models.

The experiment was going to be on for this evening:(

RonL

I did the experiment anyway. The effective track length was 1.47m.

The times measured, and calculated from the CaptainBlack and topsquark
models are plotted against height difference for the ends of the track are
shown in the attached plot.

These results are consistent with the CB model, but not the ts model as
we expected:) .

(the experimental times should always be longer than no-slip condition
predicts as there are other losses in the system that are not modelled,
but never less than the prediction)

RonL
• January 5th 2007, 10:02 AM
Jameson
Wow! Never knew MHF debates could bring about experiments like this. Very cool. :)