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If we are writing torque equation about general point what points must be taken into account for getting the correct result. Also If that general point is accelerated then how do we deal with this.

Like the first one that should be taken into account is to have moment of inertia about that general point. Like in given below problem:

We have to calculate initial angular acceleration for above cases :

  1. if friction is absent.
  2. If friction is absent and pure rolling.
  3. If coefficient of friction is not enough to provide pure rolling.

For example how can we apply torque equation about point "P" to find initial angular acceleration. Figure given in image below:

If i understood this question then my concepts would be all clear.

I would really appreciate your help.

I'm adding original question and it's solution. I found it confusing as they used point C and didn't mentioned any pseudo force. Do you guys approve this.

Figure for question mentioned above .

Original question which is case I

its solution

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  • $\begingroup$ Can we write torque equation about any point as moment of inertia may not be same so there must be constraint? What is it !! $\endgroup$ – Amir RAZA Feb 11 at 16:46
  • $\begingroup$ the torque equality can only be used about an unaccelerated point if you don't want to involve a pseudo force in your calculations, and it can only be written about a point about which the body is performing fixed axis rotation, so you just cannot chose any arbitrary point in space and use the torque equality. CM of a body is always a safe point to work with as there is no torque due to pseudo force about this point (as the point of application of pseudo force is the CM itself). $\endgroup$ – Harsh Somani Feb 11 at 16:58
  • $\begingroup$ So, We can not calculate from the center of the full circle also. $\endgroup$ – Amir RAZA Feb 11 at 17:00
  • $\begingroup$ yes usually you cannot because that is not the CM of the body in your case. The CM is at a distance 4R/3π from the centre of the complete circle. But in the above problem, the line of action of pseudo force passes from that point (the centre of the complete body) and you might get the right result. $\endgroup$ – Harsh Somani Feb 11 at 17:02
  • $\begingroup$ Pseudo force passes through com in the direction opposite to that of point C as point C is accelrated in vertical direction how can this pass through Centre of full circle $\endgroup$ – Amir RAZA Feb 11 at 17:11
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$\let\om=\omega \def\rC{{\rm C}} \def\rP{{\rm P}} \def\rX{{\rm X}} \def\ns#1#2{#1_{\rm{#2}}} \def\Pd{\ns Pd} \def\Pg{\ns Pg} \def\dx{\dot x} \def\dy{\dot y} \def\ddx{\ddot x} \def\ddy{\ddot y} \def\bL{{\bf L}} \def\bM{{\bf M} \def\half{{\textstyle {1 \over 2}}}} \def\IPg{\ns I{P_g}} \def\D#1#2{{d#1 \over d#2}}$

Can you solve first case taking P as RP

So at last you let us know what questions were! I'll only study $\mu=0$, which isn't so easy. To understand the answer it's best not to consider $t=0$ alone, but a generic $t$ before.

To take P as RP requires qualification. Because of slipping P must be viewed as two different points: one P$_\rm g$ being the initial contact point as seen belonging to ground, the other P$\rm d$ belonging to disk. Only the former is fixed and qualifies as RP.

Let me introduce some coordinates: origin in P$_\rm g$, $x$-axis rightwards, $y$ axis upwards. Point C is constrained moving horizontally, so $\rC=(x,r)$ with $x(0)=0$, $\dx(0)=0$. Since no external horizontal forces are acting on half-disk, and X's initial velocity is null, its $x$ coordinate is fixed at $p=4r/(3\pi)$. Then $\rX=(p,y)$ with $y(0)=r$, $\dy(0)=0$.

In order to write $d\bL/dt=\bM$ with P$_\rm g$ as RP we only need to compute $\ns I{P_g}$: $$\ns I{P_g} = \ns IX + m\,\overline{\rm P_gX}^2 = \half m\,r^2 + m\,(p^2 + r^2).$$ Note that $\IPg$ is a function of $t$ because of $y$. But $$\D{}t \IPg = 2 m\,y\,\dy$$ which vanishes at $t=0$.

We're almost ready. But an important simplification may be made because the half-disk's motion is of a special kind: a plane motion. By this is meant that all velocities are parallel to one plane (the $(x,y)$ plane in our case) so that angular velocity and acceleration as well as angular momentum are orthogonal to that plane, i.e. are directed along $z$-axis. The same holds for force moments. Then there is no need to deal with vectors: $\om$, $M$, $L$ are simple scalars. Moreover only axes parallel to $z$ are of interest in computing moments of inertia, and the relationship between angular velocity and angular momentum is simply written $$L = I\,\om.$$

The only external force having not vanishing moment wrt P$_\rm g$ is weight, so that $$M = -m\,g\,x.$$ As to angular momentum $$L = \IPg\,\om$$ $$\D Lt = \D{}t \IPg\,\om + \IPg\,\D \om t.\tag1$$ At $t=0$ we have $x=p$, $y=r$ whereas first term in right-hand side of (1) vanishes. Then $$\D \om t = {m\,g\,p \over \IPg} = {2\,g\,p \over 3\,r^2 + 2\,p^2}.$$

Caution: I can't ensure no errors are present. Please check my calculations!

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The torque equation $$ \text{net torque = change of net angular momentum per unit time} $$

can be used with any fixed point of space as reference point, in inertial reference frame.

In case 1, where there is no friction, the center of mass will move along straight line downwards, while the lowest mass point of the body moves to the left. Thus the center of rotation is initially at C. We can choose this point of space as our fixed point of reference. At $t=0$, there is only single force that has torque around this point, the gravity force.

In case 2, the body is rolling, so at $t=0$, the lowest point of the body in contact with the ground must not move with respect to the ground. Thus the center of rotation is this contact point and it is easiest to use this point of space as our fixed point of reference.

In case 3, there is some friction force due to the ground, which points to the right horizontally. Actual center of rotation depends on how strong this force is, so we cannot point beforehand where this point is. Then we have to choose a different point of reference. The most advantageous point seems to be the point that initially coincides with the center of mass: at $t=0$, torque of gravity is zero there.

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  • $\begingroup$ Sir, what is the logic behind implying centre of rotation to be C in first case just by the movement of lowest point to the left. Is it because perpendicular vectors to the direction of movement of all particles passes through instantaneous axis of rotation? $\endgroup$ – Amir RAZA Feb 13 at 1:32
  • $\begingroup$ I got it all now i understood the importance of predicting instantaneous axis of rotation. $\endgroup$ – Amir RAZA Feb 13 at 1:36
  • $\begingroup$ Yes, C is the intersection of the lines perpendicular to velocities of the two points: the center of mass of the body, and its lowest point. $\endgroup$ – Ján Lalinský Feb 13 at 2:50
  • $\begingroup$ Can you tell me how to apply the same via point P on the ground. $\endgroup$ – Amir RAZA Feb 13 at 2:51
  • $\begingroup$ Not sure what you're asking - how to use that same method to find that point P is the center of rotation in case 2? $\endgroup$ – Ján Lalinský Feb 13 at 2:53
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enter image description here

We first write the EOM's for the center mass $C_m$

Translation

$$m\,\vec{a}=\vec{F}\tag 1$$

and

Rotation

$$I_{cm}\,\vec{\alpha}_{cm}+\vec{\omega}_{cm} \times (I_{cm}\,\vec{\omega}_{cm})=\vec{\tau}\tag 2$$

Where :

$\vec{a}$ translation acceleration

$\vec{\alpha}_{cm}$ angular acceleration

$\vec{\omega}_{cm}$ angular velocity

$\vec{F}$ sum of all apply forces

$\vec{\tau}$ sum of all apply torque

$I_{cm}$ Inertia Tensor in center of mass coordinate system

But we can write equation (2) for point $P$

$$I_{p}\,\vec{\alpha}_{p}+\vec{\omega}_p \times (I_{p}\,\vec{\omega}_p)=\vec{r}\times \vec{F}+\vec{\tau}\tag 3$$

with : $I_{p}$ Inertia Tensor in $p$ coordinate system

$I_p=I_{cm}+m\,\tilde{\vec r}^2$

and

$\tilde{\vec r}=\left[ \begin {array}{ccc} 0&-r_z&r_y\\ r_z&0&-r_x \\ -r_y&r_x&0\end {array} \right] $

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  • $\begingroup$ Sir will the angular accleration about point P would not br same as alpha CM. $\endgroup$ – Amir RAZA Feb 12 at 16:53
  • $\begingroup$ Here are you implying for the outside point or inside. As far i know angular acceleration about any point on the body remains same and equals to angular accleration of CM. If I'm wrong please do correct me i would appreciate your help. $\endgroup$ – Amir RAZA Feb 12 at 16:55
  • $\begingroup$ Isn't alpha about every point on body same as alpha about com. If point is outaide then it would be different i think so. Bring some light to me!! $\endgroup$ – Amir RAZA Feb 12 at 17:08
  • $\begingroup$ Look at the equations, equation (2) give you the angular momentum in coordinate system of the com, equation (3) in system parallel to the com system that fixed in point p. If the equations are different, the angular momentum is also different? $\endgroup$ – Eli Feb 12 at 17:33
  • $\begingroup$ Got it sir. And one last question i have to ask if someone gives me very simple case with initial angular velocity equal to zero and friction is absent. But there is no comment about pute rolling. Is it possible to find the initial angular accleration or we need more data to get it. $\endgroup$ – Amir RAZA Feb 12 at 17:37
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$\def\rA{{\rm A}} \def\rB{{\rm B}} \def\rP{{\rm P}} \def\br{{\bf r}} \def\bv{{\bf v}} \def\bF{{\bf F}} \def\bL{{\bf L}} \def\bM{{\bf M}} \def\D#1#2{{d#1 \over d#2}}$ I've been in SE for less than six months but I've seen that questions about rotatory motions, torques, moments of inertia and so on are among the most frequent (perhaps only beaten by relativity questions). I understand the matter is not easy but I have also some doubts about how it's taught. I'm afraid there are some delicate points aren't often correctly discussed.

My first post on this subject may be found [here] (For a solid sphere rolling (pure roll) up a slope (with friction) does friction play a role in slowing it down?). In this answer I'll not give proofs or an accurate discussion. I prefer to focus on the most relevant points and give what I think is the right approach.


First principles

1) I'm not going to use the term "torque" - it may be cause of confusion. I prefer to speak of moment of a force.

2) In most problems there's no use shifting to a different frame. I'll always remain in one and the same (inertial) frame. So no pseudo-forces are present.

3) A moment (of forces, of momentum, or else) always requires a reference point, not to be confused with a reference frame. In my native language (italian) there is a term ("polo") to name that point but I can't use its english translation (pole) as it has other meanings too. I'll use the shorthand RP.

4) If A is the RP a force $\bF$ applied in point P has a moment $$\bM_\rA = (\br_\rP - \br_\rA) \times \bF$$ It's independent of possible A's motion, velocity, acceleration, if not because a change in A's position will change $\bM_\rA$.

5) If B is another RP, $\bM_\rB$ is defined analogously and will generally differ from $\bM_\rA$. That's all.

6) The same also holds for angular momentum (i.e. moment of momentum). For a mass point P it's defined as $$\bL_\rA = (\br_\rP - \br_\rA) \times (m\,\bv_\rP).$$ For a different RP, say B, you have only to change $\br_\rA$ into $\br_\rB$. It hasn't the least relevance if A and B have different motions: their velocities don't appear in definitions. Of course if A (or B) are moving $\bL_\rA$ will change in time not only because of P's motion but also of A's. This is automatically taken into account in the definition.

7) If you aren't given a single point but a system both $\bM$ and $\bL$ are to be summed on all points. In computing $\bM$ only external forces need to be taken into account.


The main equation (torque equation)

And now the most important point. When does equation $$\D\bL t = \bM \tag1$$ hold true? Is it true for any RP?

The answer is no. It holds

  • if RP is the com of system, whatever its motion

  • if RP velocity is parallel to the one of com: in particular if it's the same or if it's null (still RP).

Don't worry about an acceleration of RP, but remember: you must stick to one (inertial) frame, and all velocities must be computed in that frame, even if the RP is moving.


The half-disk

Let's apply all this to your half-disk. You may use eq. (1) taking as RP

  • the point P

  • the com of half-disk (is it X? you didn't say).

It would be wrong to use C as it's moving with a velocity not parallel to com's.

As to P, there's no problem for its being - so to say - split in two after the half-disk rotates. You must keep clear that in computing moments only one instant of time matters, not what happens afterwards.

Note: I'm assuming you know that if $\bL$ is computed by $I\omega$ then the right $I$ must be taken ($\omega$ stays the same irrespective of your choice of RP).


Comments

Of course I'm worried about what you write:

I got the correct answer via applying from centre of the circle of full disc but I'm not getting via lower point.

This is in contradiction with what I wrote above. Why do you say so? Did you compare your solution with another, maybe your teacher's?

Unfortunately you didn't tell us what question you were requested to answer. I may only guess: "If the half-disk is left alone at rest in the shown position, it will begin to fall. What coefficient of friction is required in order that it doesn't slip on the ground?"

My answer is $\mu\ge16/(15\pi)$. What was yours?

Edit

There was an error: $\mu\ge8/(9\pi)$.

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  • $\begingroup$ We had to calculate intial angular acceleration see i put the req information. $\endgroup$ – Amir RAZA Feb 12 at 16:51
  • $\begingroup$ How crystal class are your concepts. Could i know what books do you read? Sir, i have edited my question please review again. $\endgroup$ – Amir RAZA Feb 12 at 17:05
  • $\begingroup$ Can you solve first case taking P as RP. $\endgroup$ – Amir RAZA Feb 13 at 2:11

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