Motion involving Translation & Rotation |Kleppner and Kolenkow

[warhammer]

My doubt is with Method 2 of the given example in KK.

I'm unable to understand why the torque around A (where we have chosen a coordinate system at A) becomes zero due to the R x F in z direction with a minus sign {Photo Attached}

I have tried to reason out that one way to formulate that term is R (perpendicular) which turns out to be 'b' essentially and is perpendicular to the line of action i.e. force. Employing the Right Hand Rule and curling the fingers from R perp to F we would get a torque in the upward/ +ve z direction so how come the value in the attached photo is having a minus sign?

Please help me out.

 

[BvU]

Right Hand Rule

Another reason I prefer the corkscrew rule.

I can't get my fingers to accommodate R and F with this picture (Wikipedia) so I have to get up and twist my whole upper body

so I will use $\vec R\times \vec F = - \vec F \times \vec R\$
In the picture $\vec F = a\ \ ,\ \ \vec R = b\$ so $\vec F \times \vec R\$ points up from the page to the viewer and $\vec R\times \vec F\$ points into the page, i.e. in the minus z direction. (*)

Corkscrew is easier: rotate the thing from $\vec R\$ to $\vec F \$ over the smallest angle and it goes into the page.(*) I have a hunch that is your only mistake ( z is towards you instead of away ) ?

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[kuruman]

In the torque equation $\vec \tau=\vec r \times \vec F$, position vector $\vec r$ is from the origin to the point of application of the force, not to the center of the wheel.

 

[BvU]

I agree. So what does K&K maen ?

 

[etotheipi]

To avoid making mistakes, it is often useful to denote by a subscript the point about which a moment, or an angular momentum, has been taken. For instance, relative to a coordinate frame with origin $\mathcal{O}$, the moment of a force $\mathbf{F}$ applied to a point $\mathcal{P}$ in the body is $\mathbf{M}_{\mathcal{O}} := \overrightarrow{\mathcal{OP}} \times \mathbf{F}$.

To give an example, in this problem you can construct two coordinate frames $(O; \boldsymbol{e}_x, \boldsymbol{e}_y, \boldsymbol{e}_z)$ and $(A; \boldsymbol{e}_x, \boldsymbol{e}_y, \boldsymbol{e}_z)$, where $O$ is at the centre of the wheel and $A$ is as in the figure. Denote the point where force is applied as $P$. Because $\overrightarrow{AO} + \overrightarrow{OP} = \overrightarrow{AP}$, you may write$$\mathbf{M}_{A} = \overrightarrow{AP} \times \mathbf{F} = (\overrightarrow{AO} + \overrightarrow{OP}) \times \mathbf{F} = \overrightarrow{AO} \times \mathbf{F} + \overrightarrow{OP} \times \mathbf{F}$$But since $\mathbf{M}_{O} = \overrightarrow{OP} \times \mathbf{F}$, you can re-write this as$$\mathbf{M}_{A} = \overrightarrow{AO} \times \mathbf{F} + \mathbf{M}_O$$As a final step, you can take the inner product of both sides with the basis vector $\boldsymbol{e}_z$ (the one pointing out of the page) to obtain an equation on the $z$-components.

 

[kuruman]

I agree. So what does K&K maen ?

If the torque is zero about A, K&K conclude that the angular momentum about point A is conserved. Then they proceed to find an expression for the angular momentum, namely $L_z=I_0 \omega-bMV.$ Approaches (a) and (b) to the solution of this problem work together to show that although the velocity of the CM increases and the cylinder spins faster under the action of the force, the difference between the "spin" angular momentum $I_0 \omega$ and the "orbital" angular momentum $bMV$ is fixed.

 

[A.T.]

Employing the Right Hand Rule and curling the fingers from R perp to F we would get a torque in the upward/ +ve z direction

Are you sure that you are using the the direction of R vector?

 

[warhammer]

Another reason I prefer the corkscrew rule.

I can't get my fingers to accommodate R and F with this picture (Wikipedia) so I have to get up and twist my whole upper body

View attachment 279871​

so I will use $\vec R\times \vec F = - \vec F \times \vec R\$
In the picture $\vec F = a\ \ ,\ \ \vec R = b\$ so $\vec F \times \vec R\$ points up from the page to the viewer and $\vec R\times \vec F\$ points into the page, i.e. in the minus z direction. (*)

Corkscrew is easier: rotate the thing from $\vec R\$ to $\vec F \$ over the smallest angle and it goes into the page.(*) I have a hunch that is your only mistake ( z is towards you instead of away ) ?

$\$

Sorry for such a late response. Thank you so much for helping me out. I was able to figure it out upon using your strategy. Yes, that was the only mistake I was committing, as correctly gauged by you.

I agree. So what does K&K maen ?

I used a shorthand to denote the famed text "An Introduction to Mechanics" by Kleppner & Kolenkow :)

 

[BvU]

I used a shorthand to denote the famed text "An Introduction to Mechanics" by Kleppner & Kolenkow :)

My question is ambiguous ! I meant: what do K&K try to tell us ? And @etotheipi (=$-$1, with a new Magritte avatar, probably from Wald's book, judging by the colors...) explained very well. And so did @kuruman.

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[kuruman]

My question is ambiguous ! I meant: what do K&K try to tell us ? And @etotheipi (=$-$1, with a new Magritte avatar, probably from Wald's book, judging by the colors...) explained very well. And so did @kuruman.

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It looks like @etotheipi got a promotion from specifically less than nothing (-1) to something more general.

 

[BvU]

I liked the previous avatar !

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