Signs in torque - different analyses using different sign notations?

[mancity]

Homework Statement

Why are the sign notations different for the two derivations of MR(v_f-v_0)+I(w_f-w_0)=0?

Relevant Equations

MR(v_f-v_0)+I(w_f-w_0)=0.

In this derivation above, I have to account for the fact that v is translational and opposite to the sign of w, and similarly for v_0 and w_0, so the real equation should look something like this:

Now, what I don't understand is, in this second derivation using net change in linear and rotational angular momentum, we don't have to use the fact that v is opposite the sign of omega:

Is it somehow implied that, through the conservation of angular momentum, v would be opposite to w? I don't want to take this formula for granted, and a clarification of the "discrepancy" in these two approaches would greatly help me out. Thanks.

 

[kuruman]

Homework Statement: Why are the sign notations different for the two derivations of MR(v_f-v_0)+I(w_f-w_0)=0?
Relevant Equations: MR(v_f-v_0)+I(w_f-w_0)=0.

What is being derived here? What is the physical situation? Please explain the context.
It is certainly true that you can always express conservation of angular momentum as $$\Delta \vec{L}_{\text{linear}}+\Delta \vec{L}_{\text{translational}}=0.$$Note that this is a vector equation and that angular momentum must be expressed about a point that needs to be specified before you writing expressions in terms of the linear and angular velocities.

Is it somehow implied that, through the conservation of angular momentum, v would be opposite to w?

When angular momentum is conserved, the vector representing the translational velocity and the vector representing the rotational velocity cannot point in opposite directions. For rolling without slipping,$~\vec v=\vec{\omega}\times\vec r$. In this equation $\vec v$ is the linear velocity of a point on the rolling object at position $\vec r$ from the axis of rotation. This expression says that the linear velocity perpendicular to the angular velocity.

 

[mancity]

Thanks.

The context for my question is this problem, and its corresponding solution:

My logic was: here, the ball rotates in the counterclockwise direction, while the translational velocity of the ball is to the left (such that the point of contact of the ball with the ground rolls without slipping). But would that imply that instead of $\omega-\omega_0,$ we would have $-\omega-(-\omega_0)=\omega_0-\omega$, and that would be the "correct?" definition for the change in angular momentum for the first way of derivation.

My question is - why is it that for the second derivation, vec(delta L_trans)+vec(delta L_rot)=0, we can consider omega and omega 0 to be positive quantities, and not consider the fact that the rotational velocity is opposite that of the translational one?

 

[kuruman]

I am not sure whether the proof in the book is flawed, but it is certainly suspicious. The author states that "the initial angular momentum of the ball is $l_0=hmv_0=I\omega_0.$" This is clearly the angular momentum about the center of the ball O which is NOT conserved because friction exerts an external torque about point O. Then the author invokes the "results from E10-8" with which I am not familiar. Maybe it's some way to deal with the non-conservation of angular momentum but E10-8 is irrelevant if one applies angular momentum conservation correctly as shown below.

The key observation is that angular momentum is conserved about point P which is the point of contact between the ball and the surface when the ball is struck.
The linear momentum transfer is as stated in the solution $\Delta p = F\Delta t=mv_0.$
Now write the height above the center where the ball is struck as a fraction of the radius: $h=\alpha~R.$ The task is to show that $\alpha =\frac{4}{5}.$

The torque about point of contact P is $~\tau_{\text{p.c.}}=F(1+\alpha)R.$
The initial angular momentum of the ball about point of contact P is $$~L_{\text{p.c.}}=\tau_{\text{p.c.}}\Delta t=F\Delta t(1+\alpha~R)=mv_0(1+\alpha)R.\tag{1}$$ This angular momentum is conserved because no external torques act on the ball about point of contact P after the ball is struck and starts moving. Therefore, it is the same as the "no slip" angular momentum about the point of contact $L_{\text{n.s.}}$ when the ball starts rolling without slipping.

We will now find an expression for $L_{\text{n.s.}}.$
$L_{\text{n.s.}}=I_{\text{p.c.}}\omega.$
Since there is no slipping, $\omega=\dfrac{v}{R}.$
The final speed of the ball is $v=\dfrac{9}{7}v_0$, therefore $\omega=\dfrac{9v_0}{7R}.$
The moment of inertia about the point of contact is given by the parallel axes theorem
$I_{\text{p.c.}}=\frac{2}{5}mR^2+mR^2=\frac{7}{5}mR^2$ so that the no-slip angular momentum about the point of contact is
$$L_{\text{n.s.}}=I_{\text{p.c.}}\omega=\frac{7}{5}mR^2\times\frac{9v_0}{7R}=\frac{9}{5}mv_0R. \tag{2}$$ Setting $L_{\text{p.c.}}=L_{\text{n.s.}}$ as demanded by angular momentum conservation, gives the expected result for $\alpha.$

I know I did not answer your question

My question is - why is it that for the second derivation, vec(delta L_trans)+vec(delta L_rot)=0, we can consider omega and omega 0 to be positive quantities, and not consider the fact that the rotational velocity is opposite that of the translational one?

mainly because it's about the solution in the book which I do not completely understand. However, I showed you how to solve the problem using angular momentum conservation correctly and transparently.

 

[haruspex]

The author states that "the initial angular momentum of the ball is $l_0=hmv_0=I\omega_0.$" This is clearly the angular momentum about the center of the ball O which is NOT conserved because friction exerts an external torque about point O. Then the author invokes the "results from E10-8" with which I am not familiar. Maybe it's some way to deal with the non-conservation of angular momentum but E10-8 is irrelevant if one applies angular momentum conservation correctly as shown below.

Since the author is taking velocity as positive to the left and angular velocity positive anticlockwise, the angular momentum about the point of contact is $mvR+I\omega$. Since this is conserved, $mvR+I\omega=mv_0R+I\omega_0$, whence the E10-8 result.

consider the fact that the rotational velocity is opposite that of the translational one?

I'm not exactly sure what your difficulty is. Because of the unusual sign convention adopted by the author, the angular velocity about the point of contact due to the linear motion is $Rv$ instead of $-Rv$. Does that help?

 

[kuruman]

Since the author is taking velocity as positive to the left and angular velocity positive anticlockwise, the angular momentum about the point of contact is $mvR+I\omega$. Since this is conserved, $mvR+I\omega=mv_0R+I\omega_0$, whence the E10-8 result.

I found the following statement by the author ambiguous and misleading.

Ambiguous because there is no mention anywhere about what point this angular momentum is calculated. It is clearly the angular momentum about the CM because the torque which gives rise to this angular momentum is about the CM. Misleading because this term is not, as claimed "the initial angular momentum of the ball $l_0=hmv_0=I\omega_0.$ It is only the initial spin angular momentum of the ball, it excludes the orbital part and is not conserved.

This is a nice example of spin angular momentum conversion to orbital angular momentum through friction without momentum loss. This solution has managed to muck it up, in my opinion.

 

[haruspex]

I found the following statement by the author ambiguous and misleading.
View attachment 354048
Ambiguous because there is no mention anywhere about what point this angular momentum is calculated. It is clearly the angular momentum about the CM because the torque which gives rise to this angular momentum is about the CM. Misleading because this term is not, as claimed "the initial angular momentum of the ball $l_0=hmv_0=I\omega_0.$ It is only the initial spin angular momentum of the ball, it excludes the orbital part and is not conserved.

This is a nice example of spin angular momentum conversion to orbital angular momentum through friction without momentum loss. This solution has managed to muck it up, in my opinion.

I still think you may be being unfair. The author may have consistently taken the CM as the axis for angular momentum. The "E10-8” result can be obtained by writing two equations involving friction then eliminating that term. If that is how it was done then at no point would the author be claiming or implying that spin angular momentum is conserved, and it would be true that the initial angular momentum is $hF\Delta t=hmv_0$.

 

[kuruman]

I still think you may be being unfair. The author may have consistently taken the CM as the axis for angular momentum. The "E10-8” result can be obtained by writing two equations involving friction then eliminating that term. If that is how it was done then at no point would the author be claiming or implying that spin angular momentum is conserved, and it would be true that the initial angular momentum is $hF\Delta t=hmv_0$.

Perhaps I am being unfair. I note that in post #5 you very appropriately specified that angular momentum about the point of contact is conserved hence the equation $mvR+I\omega=mv_0R+I\omega_0$ which is E10-8 with terms moved around to bring it to the form $~\Delta \vec{L}_{\text{linear}}+\Delta \vec{L}_{\text{translational}}=0.$ If E10-8 also mentions that "angular momentum about the point of contact is conserved", then I will concede that I have been unfair in that I did not fully consider the facts.

 

[mancity]

E10-8 is attached.

I think where the "discrepancy" lied for my first approach was that I considered delta L_lin=Rm(v-v_0), but I thought that v was the translational velocity (incorrect), not the tangential one. Delta L_rot=I(w-w_0) is a rotational/tangential velocity. When writing I(w_f-w_0)+m(v_f-v_0), I need to be careful about which velocity (either rotational/tangential or translational velocity, which are opposite to each other due to rolling without. slipping) I am referring to.

I understand now that the real equation would be this one:

$$\Delta \vec{L}_{\text{linear}}+\Delta \vec{L}_{\text{translational}}=0.$$Note that this is a vector equation and that angular momentum must be expressed about a point that needs to be specified before you writing expressions in terms of the linear and angular velocities.

 

[haruspex]

I thought that v was the translational velocity (incorrect), not the tangential one.

I don’t understand the distinction you are making. $v$ is the linear velocity of the centre of the ball. How are you defining tangential velocity here?

 

[kuruman]

E10-8 is attached.

Thank you providing it. It is informative. Note that applies to a ball that is constrained to rotate about a fixed axis o=passing through a diameter. The results from E10-8 cannot be applied to the present case without justification and expansion to the more general case of a ball whose center mass is free to move.

That said, I think that you are understandably confused and I believe that your confusion stems from a lack of understanding of the direction of the vector quantities in the scalar equation $$L=mv_0R+I\omega_0.$$ A derivation of this equation appeared as post #282 here. Because this thread has a total of 336 posts, I have pasted the derivation with minor editorial changes to spare you the trouble of sifting through everything that was said then.

Consider an irregular rigid object of mass $M$ simultaneously rotating and translating in space in the absence of external forces and torques (see figure on the right). This means that both its linear and angular momentum are conserved. We will calculate its angular momentum about an arbitrary point A (see diagram below) given that the velocity of the center of mass is $\vec V_{cm}$ and its rotational angular velocity is $\vec \omega.$

First we point out that the axis of rotation must pass through the CM. That's because linear momentum is conserved which means that $\vec V_{cm}= \rm{const.}$ If the axis of rotation did not pass through the CM, the CM would rotate about that hypothetical axis and its velocity would change direction which means that it cannot be constant. Thus, the linear velocity relative to the CM of an arbitrary point P at position $\vec {r}'$ from the CM is $\vec {v}'=\vec{\omega}\times \vec {r}'.$

Now consider mass element $dm$ at point P the position vector of which relative to A is $\vec r$. Let $\vec {r}'$ be the position vector of $dm$ relative to the CM and $\vec R$ be the position of the CM relative to A. Finally, let $\vec v_P$ be the instantaneous velocity of point P relative to point A. Addition of velocities requires that the velocity of point P relative to A is $\vec v_P=\vec V_{cm}+\vec{\omega}\times \vec {r}'.$

The angular momentum contribution of $dm$ about point A is $$d\vec L=(dm)\vec r\times \vec v_P=(dm)(\vec R+\vec {r}')\times(\vec V_{cm}+\vec{\omega}\times \vec {r}')$$ We multiply out the cross product on the RHS and consider each of the four resulting terms separately.

1. $~(dm) \vec R \times \vec V_{cm}$
The contribution from the first term to the total angular momentum is $$\vec L_1=\int (dm) \vec R \times \vec V_{cm}=\left( \int dm \right) \vec R \times \vec V_{cm}=M\vec R \times \vec V_{cm}.$$ I have seen this called the angular momentum of the center of mass, the translational angular momentum or the orbital angular momentum.

2. $~(dm)\vec R\times (\vec{\omega}\times \vec {r}')$
Using the triple-product rule,
$$\vec R\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec R\cdot \vec {r}') -\vec {r}'(\vec R \cdot \vec{\omega}).$$The second term on the RHS is zero because $\vec R$ is perpendicular to $\vec {\omega}$. To find the contribution of the first term, we integrate $$\vec L_2=\int (dm)\vec{\omega}(\vec R\cdot \vec {r}')=\vec{\omega}\left(\vec R\cdot \int (dm)\vec {r}'\right)=0$$ Note that by definition of the CM coordinates, $\int (dm)\vec {r}'=0.$

3. $~(dm)\vec {r}'\times \vec V_{cm}$
This term also vanishes upon integration for the same reason when we take constant $\vec R$ out of the integral. $$\vec L_3=\int (dm)\vec {r}'\times \vec V_{cm}=\left(\int (dm)\vec {r}'\right)\times \vec V_{cm}=0.$$ 4. $~(dm)\vec {r}'\times (\vec{\omega}\times \vec {r}')$
Triple-product rule once more
$$\vec {r}'\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec {r}'\cdot \vec {r}')-\vec {r}'(\vec{\omega}\cdot \vec {r}').$$The second term on the RHS vanishes because the vectors are orthogonal. Integrating the first term, $$\vec L_4=\int \vec{\omega}(\vec {r}'\cdot \vec {r}')=\vec{\omega}\int (dm){r'}^2=I_{cm}~\vec{\omega}.$$I have seen this called the angular momentum about the center of mass or the rotational angular momentum or the spin angular momentum.

Thus, the total angular momentum about point A of the translating and rotating rigid body is the sum of four terms two of which are zero, so that $$\vec L=M\vec R \times \vec V_{cm}+I_{cm}~\vec{\omega}.\tag{1}$$ The first term in this expression, angular momentum of the center of mass, depends on the choice of reference point A. The second term, angular momentum about the center of mass, does not.

The bottom line is that one can always write the angular momentum of a simultaneously translating and rotating rigid body in the form of Equation (1).

In this equation the first term is $\vec L_{\text{translatonal}}=M\vec R \times \vec V_{cm}$ is the translational angular moment of the center of mass about point A, i.e. the angular momentum that any moving point mass has relative to a fixed point. Its direction is perpendicular to the plane defined by the position vector $\vec r$ from point A and the velocity vector $\vec v$ (right hand rule.)

The second term is the angular momentum about the center of mass, i.e. the additional angular momentum if it so happens that the mass is rotating about a fixed axis through its CM. It is given by $\vec L_{\text{rotational}}=I_{cm}~\vec{\omega}.$ Here, $I_{cm}$ is the moment of inertia about the fixed axis and $\vec{\omega}$ points along the axis.

Note that, in general, the translational angular momentum and the rotational angular momentum need not point in the same direction. For example, as the Earth orbits the Sun it has $\vec L_{\text{translatonal}}$ directed perpendicular to the plane of its orbit (the ecliptic). However, the Earth's $L_{\text{rotational}$ is directed at an angle of 23.4° away from the perpendicular to the ecliptic.

I hope this helps.

 

[kuruman]

I still think you may be being unfair. The author may have consistently taken the CM as the axis for angular momentum. The "E10-8” result can be obtained by writing two equations involving friction then eliminating that term. If that is how it was done then at no point would the author be claiming or implying that spin angular momentum is conserved, and it would be true that the initial angular momentum is $hF\Delta t=hmv_0$.

As seen in post #9, E10-8 considers the rotation of a ball about a fixed axis. Invoking E10-8 in the book's solution to go from $L=I\omega$ to $L=mvR+I\omega$ without justification would require a leap of faith from the reader.