# Wrong concept? Why?

1. Nov 12, 2004

### CartoonKid

I have a question which puzzles me since the day my tutor pointed out to me.

A uniform solid disk is set into rotation with an angular speed wi about an axis through its
center. While still rotating at this speed, the disk is placed into contact with a horizontal
surface and released as shown in the figure below.
(a) What is the angular speed of the disk once pure rolling takes place? (Hint: Consider
the torques about the centre of mass.) [Ans: wi/3 ]
(b) Find the fractional loss in kinetic energy from the time the disk is released until pure
rolling occurs. [Ans: −2/3 ]

I did the first part by using the method of conservation of angular momentum which involve the change in moment of inertia and I got the exact answer. However, my tutor said that my concept is wrong even though my answer is correct. After getting the first part, we can easily work out the second part. The thing is that I don't know why my concept is wrong in the first part. Can somebody please explain it to me.

2. Nov 12, 2004

### CartoonKid

Later I posted a question which has high similarity to this question to my tutor but he never reply me. The question is like this,
If I throw a rod in such a way that it rotates about the axis through its center of mass while flying in the air with an initial angular speed of, w0. Suddenly one of its end hooks onto a point that makes it rotates about the end of the rod at an angular speed, w1. How are you going to find w1 in term of w0? Neglect the air resistance.
For me, I will use the conservation of angular momentum which involve the change in the axis of rotation. Thus, (Ii)(Wi)=(If)(Wf) where Ii=moment of inertia about the axis at the center of rod while If=moment of inertia about the axis at the end of the rod.
Is this the right working? (i - initial , f - final)

3. Nov 12, 2004

### arildno

Angular momentum is conserved WITH RESPECT TO THE CONTACT POINT, but NOT with respect to C.M during the phase leading up to rolling!

It is completely wrong to change the position of the rotation axis during the phase.
Angular momentum is always computed relative to some point; stick to that point throughout the motion!

Here's how you SHOULD do it:
1) With respect to the moving contact point, there can be no net torques
2) Hence, the moment-of momentum equation with respect to the contact point reads:
$$\vec{0}=\frac{d\vec{L}_{C.P}}{dt}+M\vec{v}_{C.P}\times\vec{v}_{C.M}$$
3) Simplifications:
a) The second term is zero, since the velocities of the of the contact point and of the C.M are PARALLELL.
b) Hence, we gain conservation of angular momentum with respect to the contact point:
$$\frac{d\vec{L}_{C.P}}{dt}=\vec{0}$$
c) Simplified, between the times 0 and T, we have:
$$R\vec{k}\times{M}\vec{v}_{C.M.}(T)+\mathcal{I}_{C.M}{\omega}(T)\vec{j}=R\vec{k}\times{M}\vec{v}_{C.M.}(0)+\mathcal{I}_{C.M}{\omega}(0)\vec{j}$$
Here, $$\mathcal{I}_{C.M}$$ is the moment of inertia with respect to the C.M.
d) Rolling relation:
We have, at time T:
$$\vec{v}_{C.M}(T)=v_{C.M}(T)\vec{i}=R\omega(T)\vec{i}$$
e) at t=0, the velocity of C.M is clearly zero.
4) Hence, we get the scalar equation:
$$MR^{2}\omega(T)+\mathcal{I}_{C.M}\omega(T)=\mathcal{I}_{C.M}\omega(0)$$
Or:
$$\omega(T)=\frac{\mathcal{I}_{C.M}\omega(0)}{\mathcal{I}_{C.M}+MR^{2}}$$
By the parallell axis theorem, we have:
$$\mathcal{I}_{C.P}=\mathcal{I}_{C.M}+MR^{2}$$
That you could "derive" the relation in the manner you did is merely a curious fact, which should be regarded as such rather than some correct, alternative approach.
6) Note, in particular, that if you set down the disk such that the C.M has some non-zero initial horizontal velocity, then this will change the value of the final angular velocity.
It is, in fact, only because the initial horizontal velocity of the C.M. is zero that your erronous approach gave the right answer.

Last edited: Nov 12, 2004
4. Nov 12, 2004

### CartoonKid

$$\vec{0}=\frac{d\vec{L}_{C.P}}{dt}+M\vec{v}_{C.P}\times\vec{v}_{C.M}$$
I don't know how to come to this equation, can you elaborate somemore. Thanks. Btw, what is the general law you are applying here? Thanks again.

5. Nov 12, 2004

### arildno

Sure enough:
1.Consider your rigid body as composed of tiny particles with individual masses "dm"
2. Let the position vector from the contact point to particle "p" be $$\vec{r}_{p}$$
3) Clearly, we have:
$$\vec{r}_{p}=\vec{r}_{C.M}+\vec{r}'_{p}$$
where $$\vec{r}_{C.M}$$ is the position vector from the contact point to the C.M, whereas $$\vec{r}'_{p}$$ is the position vector from the C.M to "p".
4. The moment-of momentum equation for the individual particle "p" with respect to the contact point is simply Newton's 2.law for that particle, cross-multiplied with the position vector from the contact point:
$$\vec{r}_{p}\times\vec{F}_{p,ext}+\vec{r}_{p}\times\vec{F}_{p,int}=\vec{r}_{p}\times{dm}\vec{a}_{p}$$
Here, $$\vec{F}_{p,ext}$$denotes those forces acting on particle "p" which places its reaction force OUTSIDE the whole system (i.e, exterior forces), whereas $$\vec{F}_{p,int}$$ denotes those forces acting on particle "p" which places its reaction force INSIDE the system (interior forces).
5)When summing (integrating) over all particles in the rigid body, only eventual torques from external forces with respect to the contact point will appear.
Hence, we have at the start for our problem:
$$\vec{0}=\int_{V}\vec{r}_{p}\times{dm}\vec{a}_{p}$$
(The integration sign means summing over all particles in the rigid body, occupying region V)
Now:
$$\vec{r}_{p}\times\vec{a}_{p}=\frac{d}{dt}(\vec{r}_{p}\times\vec{v}_{p})-\frac{d\vec{r}_{p}}{dt}\times\vec{v}_{p}$$

Clearly, the derivative of the position vector is the relative velocity between "p" and the contact point "C.P":
$$\frac{d\vec{r}_{p}}{dt}=\vec{v}_{p}-\vec{v}_{C.P}$$
Where $$\vec{v}_{C.P}$$ is the velocity of the contact point.
Or, we gain:
$$\int_{V}\vec{r}_{p}\times{dm}\vec{a}_{p}=\frac{d}{dt}\int_{V}\vec{r_{p}}\times{dm}\vec{v}_{p}+\int_{V}\vec{v}_{C.P}\times{dm}\vec{v}_{p}$$
(we've eliminated the cross-product $$\vec{v}_{p}\times\vec{v}_{p}\equiv\vec{0}$$)
6) Since $$\vec{v}_{C.P}$$ is constant with respect to all particles, the last integral simplifies to:
$$\int_{V}\vec{v}_{C.P}\times{dm}\vec{v}_{p}=\vec{v}_{C.P}\times\int_{V}\vec{v}_{p}dm=\vec{v}_{C.P}\times{M}\vec{v}_{C.M}$$
by definition of the center of mass.
The angular momentum with respect to the contact point is defined as:
$$\vec{L}_{C.P}=\int_{V}\vec{r}_{p}\times{dm}\vec{v}_{p}$$
Hence, we gain our formula:
$$\vec{0}=\frac{d\vec{L}_{C.P}}{dt}+\vec{v}_{C.P}\times{M}\vec{v}_{C.M}$$

6. Nov 12, 2004

### CartoonKid

Oh I think I got the idea. Thank you very much. It take me so long time to digest. Basically we just apply the following formula for this question, right?
mv(0)r+Iw(0)=mv(t)r+Iw(t)
As long as the procedure only takes place in the vertical through the contact point, this equation will eventually resort to my wrong concept, right?

7. Nov 12, 2004

### arildno

You've got it; as long as the initial horizontal velocity of C.M (your v(0)) is zero, your wrong approach will yield the same answer, oddly enough (regard that as an oddity, don't put more weight on it than that).