What's the angular acceleration?

[NTesla]

Homework Statement

A heavy particle of mass m falls freely near the earths surface. What is the torque acting on this particle about a point 50 cm east to the line of motion ? Does this torque produce any angular acceleration in the particle ?

Relevant Equations

$$\tau = \overrightarrow{r} \times \overrightarrow{F}$$
$$\alpha = \frac{\tau}{I}$$

$\tau$ = 0.5*mg
$\alpha$= (0.5mg)/$mr^2$

This is what I'm finding difficult to understand, if a particle is in linear translation, how can it have angular acceleration. If by calculation, angular acceleration is non-zero, then shouldn't the body must move in a circular path with increasing $\omega$.
I think there is problem in either calculating torque or calculating moment of inertia. But we can calculate moment of inertia about any axis. Isn't it right. ?

 

[PeroK]

This is what I'm finding difficult to understand, if a particle is in linear translation, how can it have angular acceleration. If by calculation, angular acceleration is non-zero, then shouldn't the body must move in a circular path with increasing $\omega$.

This is common misconception. A particle traveling in a straight line has zero angular momentum about any point on that line; but, as can easily be seen from the definition, has non-zero angular momentum about any point not on that line.

For example, a particle traveling along the line $y = 1m$ has non-zero angular momentum about the origin. If the velocity of the particle is constant, then angular momentum (about the origin) is conserved - you should show that as a useful exercise. If, however, it is subject to a force and accelerates along that line, then angular momentum about the origin is not conserved. Hence the particle is subject to a torque about the origin.

 

[NTesla]

Ok. If calculating torque is not wrong, then there must be something wrong in calculation of moment of Inertia in my attempt. My understanding is that if a particle is at a distance of r from an point, then its moment of inertia is $mr^2$.
And if there's nothing wrong in calculation of torque, and in calculation of moment of Inertia, then there will definitely be an angular acceleration. If there is an angular acceleration, then the particle must be moving in a circular path, instead of an straight path. Something is wrong, as the particle does move only in a straight path.
I don't understand what is wrong and where am I going wrong.

 

[Gaussian97]

It is a common misconception to think that angular velocity and angular acceleration only occur in curved paths. Both quantities depend on your origin of coordinates (i.e. from where you measure this thing).
A good way of visualizing it is, try to imagine the movement of the body and imagine you follow the movement with your head (or with your eyes if it occurs in a small region). Then, if you have to move your head or your eyes to follow the movement, there's angular velocity (and probably angular acceleration).
So any object moving in a straight line, unless it comes directly to you, it will have angular velocity and angular acceleration.

 

[NTesla]

It is a common misconception to think that angular velocity and angular acceleration only occur in curved paths. Both quantities depend on your origin of coordinates (i.e. from where you measure this thing).
A good way of visualizing it is, try to imagine the movement of the body and imagine you follow the movement with your head (or with your eyes if it occurs in a small region). Then, if you have to move your head or your eyes to follow the movement, there's angular velocity (and probably angular acceleration).
So any object moving in a straight line, unless it comes directly to you, it will have angular velocity and angular acceleration.

If $\alpha$ is non-zero, then $\omega$ is non-zero, which means $\frac{d\theta }{dt}$ is non-zero. This means that $\Delta \theta$ is non-zero. Does that not mean that the body is moving in curved path.

 

[PeroK]

If $\alpha$ is non-zero, then $\omega$ is non-zero, which means $\frac{d\theta }{dt}$ is non-zero. This means that $\Delta \theta$ is non-zero. Does that not mean that the body is moving in curved path.

Look at a particle moving in a straight line. If you describe its path using polar cordinates, then $\theta$ is changing. $\theta$ is only constant for a radial path (i.e. straight line through the origin)

For example, a particle traveling along the line $y = 1m$ has non-zero angular momentum about the origin. If the velocity of the particle is constant, then angular momentum (about the origin) is conserved - you should show that as a useful exercise. If, however, it is subject to a force and accelerates along that line, then angular momentum about the origin is not conserved. Hence the particle is subject to a torque about the origin.

 

[NTesla]

Look at a particle moving in a straight line. If you describe its path using polar cordinates, then $\theta$ is changing. $\theta$ is only constant for a radial path (i.e. straight line through the origin)

Yes, that makes sense.
So what I understand is that even when a particle is in free fall due to gravity, It will have angular velocity and angular acceleration, and the particles moment of Inertia will be $mr^2$, if the origin of the coordinate system is away from its line of motion. Have I got it right ?

 

[PeroK]

Yes, that makes sense.
So what I understand is that even when a particle is in free fall due to gravity, It will have angular velocity and angular acceleration, and the particles moment of Inertia will be $mr^2$, if the origin of the coordinate system is away from its line of motion. Have I got it right ?

You need to be careful: moment of inertia applies to rotation about a given axis. That is specific to rotational motion.

 

[NTesla]

You need to be careful: moment of inertia applies to rotation about a given axis. That is specific to rotational motion.

In the present case, the particle is in straight line motion. Since we have established that the particle will have torque and angular acceleration, and since $\tau = I\alpha$, the particle must have a moment of inertia. Isn't it ?
which will be equal to $mr^2$. How does your statement apply here ?

You've stated that moment of Inertia is specific to rotational motion, and here no rotational motion is happening, then according to your statement, moment of Inertia in present case is zero ?

 

[PeroK]

In the present case, the particle is in straight line motion. Since we have established that the particle will have torque and angular acceleration, and since $\tau = I\alpha$, the particle must have a moment of inertia. Isn't it ?
which will be equal to $mr^2$. How does your statement apply here ?

You've stated that moment of Inertia is specific to rotational motion, and here no rotational motion is happening, then according to your statement, moment of Inertia in present case is zero ?

It's simply not defined. The moment of inertia relates to rotation about a fixed axis. That is not the case here.

If you analyse the straight line motion, there is no torque, but there is angular acceleration.

 

[haruspex]

the particle will have torque

No, a force has torque. In this case the force is mg.

moment of Inertia is specific to rotational motion, and here no rotational motion is happening

From the perspective of an axis not in the line of motion, the particle has rotation.
Suppose at some instant it is distance r from the axis and at angle theta from the horizontal above it. Its linear acceleration here is g, and the tangential component is g cos(θ). That constitutes an angular acceleration $\frac gr\cos(\theta)$.
The torque mg has about the axis is $m gr\cos(\theta)$, and the particle's moment of inertia is mr²:
$m gr\cos(\theta)=mr^2\frac gr\cos(\theta)$.

 

[NTesla]

No, a force has torque. In this case the force is mg.

From the perspective of an axis not in the line of motion, the particle has rotation.
Suppose at some instant it is distance r from the axis and at angle theta from the horizontal above it. Its linear acceleration here is g, and the tangential component is g cos(θ). That constitutes an angular acceleration $\frac gr\cos(\theta)$.
The torque mg has about the axis is $m gr\cos(\theta)$, and the particle's moment of inertia is mr²:
$m gr\cos(\theta)=mr^2\frac gr\cos(\theta)$.

@haruspex , I understand your calculation. This makes sense. But, even from the point of view of an observer who is on an axis separate from line of motion, the particles motion will seem to be a straight line. The particle will have angular velocity, and angular acceleration, but the trajectory of the particle will be straight line. How will he see it any different.

In post#8, @PeroK has mentioned that moment of inertia applies to rotation about a given axis. That is specific to rotational motion. And since in the present case, the body is not in rotational motion, the concept of moment of inertia is not applicable, therefore in post#10, he has mentioned that in the present case, moment of inertia is non-existent.

Now, you've mentioned that the moment of inertia in present case is $mr^2$. How are these 2 different viewpoints of moment of inertia co-existent.?

 

[haruspex]

from the point of view of an observer who is on an axis separate from line of motion, the particles motion will seem to be a straight line.

The observer is free to interpret the motion either way, and both are valid according to the respective laws. It would be a concern if the laws governing the angular aspect of motion suddenly stopped being true just because the path is a straight line. What about a path which is neither straight nor an arc of constant radius about the chosen axis?

 

[PeroK]

In post#8, @PeroK has mentioned that moment of inertia applies to rotation about a given axis. That is specific to rotational motion. And since in the present case, the body is not in rotational motion, the concept of moment of inertia is not applicable, therefore in post#10, he has mentioned that in the present case, moment of inertia is non-existent.

In general, in polar coordinates $\alpha = \ddot \theta$ may be non-zero for inertial motion. So, if you try to use $$\tau = \alpha I$$ you'll get non-zero torque in cases of inertial motion.

PS when $r$ is a function of $t$, then $I = mr^2$ would be a function of time. Whereas, the above formula must assume that $I$ is constant with time. Otherwise, you'll end up with time derivatives of your moment of inertia in these equations. I've never seen $I$ vary with time.

 

[NTesla]

The observer is free to interpret the motion either way, and both are valid according to the respective laws. It would be a concern if the laws governing the angular aspect of motion suddenly stopped being true just because the path is a straight line. What about a path which is neither straight nor an arc of constant radius about the chosen axis?

@haruspex
My understanding is that, In deep space, where there's no gravity, if a particle is traveling in straight path as seen from one inertial frame, then from any other inertial frame also, the trajectory of the particle will be a straight line. Likewise, in the present case, when seen from line of motion of the particle, if the particle is traveling in straight line, then from any other axis which is about 0.5m away from its line of motion, the path will seen to be a straight line.

However, the 2nd issue which I think is more important to discuss is that, whether moment of Inertia is applicable in present case or not. You've mentioned that it does, and have shown it by calculation, @PeroK has mentioned that it doesn't and has shown it by calculation. What is right. Can someone clarify.

 

[NTesla]

In general, in polar coordinates $\alpha = \ddot \theta$ may be non-zero for inertial motion. So, if you try to use $$\tau = \alpha I$$ you'll get non-zero torque in cases of inertial motion.

PS when $r$ is a function of $t$, then $I = mr^2$ would be a function of time. Whereas, the above formula must assume that $I$ is constant with time. Otherwise, you'll end up with time derivatives of your moment of inertia in these equations. I've never seen $I$ vary with time.

@PeroK , You have mentioned that the formula $\tau = \alpha I$, must have I constant with time. I couldn't understand why is that restriction imposed on this formula.

 

[haruspex]

if a particle is traveling in straight path as seen from one inertial frame, then from any other inertial frame also, the trajectory of the particle will be a straight line

Quite so, but that does not contradict the fact that the particle may have a nonzero angular velocity about a chosen axis.
If a boat floats straight N, passing you at a distance of 10m to the E, the angle you perceive the boat's position makes to N (the bearing) changes over time. That constitutes an angular velocity. If the boat is moving at constant linear velocity then that angular velocity will increase until the boat is at closest approach, then decrease; so we also have angular acceleration.

whether moment of Inertia is applicable in present case

Imagine now that the boat is only moving because you are holding a long stick which you press against the stern. (Say the boat is constrained to a straight channel.) The force on the boat is via the torque you exert on the stick, so you can either think of it as the force on the stern creating linear acceleration via F=ma, or as the torque you exert creating angular acceleration via τ=Iα.
Just as mass is a body's resistance to linear acceleration, moment of inertia is its resistance to angular acceleration.

 

[NTesla]

@haruspex, I already agree to your first part of argument.

For the 2nd part, concerning moment of Inertia: In the present case, it seems to me that what you are saying is opposite of what @PeroK is saying. How would you counter his argument regarding moment of Inertia that the concept of moment of Inertia doesn't even apply in the present case.

 

[haruspex]

the above formula must assume that I is constant with time.

I don't see why. If a cart leaking water is subject to a force F(t) then F(t)=m(t)a(t) still applies.

 

[PeroK]

I don't see why. If a cart leaking water is subject to a force F(t) then F(t)=m(t)a(t) still applies.

What I meant is this. Consider a particle moving along a trajectory $x = vt, y = d$. And we measure $\theta$ clockwise from the y-axis: $$\ddot \theta = -\frac{2v^2}{d^2}\sin \theta \cos^3 \theta$$ And using $\tau = \frac{\ddot \theta}{I}$ we would need a non-zero torque about the origin to maintain linear inertial motion.

More generally, we have the acceleration in the angular direction: $$a_{\theta} = r\ddot \theta + 2\dot r \dot \theta$$ and when $\dot r \ne 0$ we have $$\tau = mra_{\theta} = mr^2(\ddot \theta + \frac{2\dot r \dot \theta}{r}) $$
Where we have an extra term involving $\dot r$.

PS This might be my mistake in thinking we were looking for $\alpha = \ddot \theta$, but more generally we have: $\alpha = \ddot \theta + \frac{2\dot r \dot \theta}{r}$

What @haruspex says all makes sense now.

 

[NTesla]

What I meant is this. Consider a particle moving along a trajectory $x = vt, y = d$. And we measure $\theta$ clockwise from the y-axis: $$\ddot \theta = -\frac{2v^2}{d^2}\sin \theta \cos \theta$$ And using $\tau = \frac{\ddot \theta}{I}$ we would need a non-zero torque about the origin to maintain linear inertial motion.

More generally, we have the acceleration in the angular direction: $$a_{\theta} = r\ddot \theta + 2\dot r \dot \theta$$ and when $\dot r \ne 0$ we have $$\tau = ma_{\theta} = mr^2(\ddot \theta + \frac{2\dot r \dot \theta}{r}) $$
Where we have an extra term involving $\dot r$.

PS This might be my mistake in thinking we were looking for $\alpha = \ddot \theta$, but more generally we have: $\alpha = \ddot \theta + \frac{2\dot r \dot \theta}{r}$

What @haruspex says all makes sense now.

@PeroK :
I tried calculating $\ddot \theta$, this is what I'm getting:
$$\ddot \theta=\frac{-2v^{2}}{r^{2}}Sin\theta Cos\theta$$
In the denominator, you've written $d^2$, where'as in my calculation I'm getting denominator as $r^2$.
Here's my calculation :

I couldn't understand, why you've written $\tau= ma_{\theta }$.
Also, in the subsequent step, you've written $ma_{\theta }=mr^{2}(\ddot{\theta }+\frac{2\dot{r}\dot{\theta }}{r})$. I suppose, there's some error in this step. Please clarify these 2 points.

 

[PeroK]

Yes, you're right. I had two mistakes in mine. It was too early in the morning!

 

[NTesla]

Yes, you're right. I had two mistakes in mine. It was too early in the morning!

@PeroK, I had pointed out 2 errors(one of which was that the denominator is $d^2$ in your calculation) and had asked a question in post #21.
In your post#22, I'm not sure, which mistake are you referring to. Also, please let me know why you had written $\tau=ma_{\theta}$ in post#20.

 

[PeroK]

@PeroK, Also, please let me know why you had written $\tau=ma_{\theta}$ in post#20.

That was just a typo, I missed out an $r$. I've fixed both errors now. I should have had $\cos^3 \theta$. I'll type it up.

 

[PeroK]

FYI what I did was: $$\tan \theta = \frac{vt}{d}, \ \ (\sec^2 \theta)\dot \theta = \frac v d, \ \ 2(\tan \theta \sec^2 \theta) (\dot \theta)^2 + (\sec^2 \theta) \ddot \theta = 0$$ Hence $$\ddot \theta = - 2(\tan \theta )(\dot \theta)^2 = -2\tan \theta (\frac{v^2}{d^2}) \cos^4 \theta = -\frac{2v^2}{d^2} \sin \theta \cos^3 \theta$$
Which is the same as yours with $d = r\cos \theta$.

 

[NTesla]

@PeroK : Here's how I calculated $\tau$. However, something is wrong, which I don't understand what it is. Please help.

My calculation of $\tau$ is: $\tau =\overrightarrow{r} \times \overrightarrow{F} = m \overrightarrow{r} \times \overrightarrow{a} = m|\overrightarrow{r}||\overrightarrow{a}|Sin\theta =mra_{r}$.

However, you've written $\tau = mra_{\theta}$ in post#20.

 

[kuruman]

As an aside, when I have to deal with situations of this kind, I use the general form for the acceleration in polar coordinates $$\frac{d^2 \vec r}{\text{dt}^2}= \left(\ddot{r}-\dot{\theta }^2 r\right)\hat{r}+ \left(r \ddot{\theta }+2 \dot{\theta } \dot{r}\right)\hat{\theta }$$ and then apply it to the trajectory, in this case $\vec r=vt ~\hat x+y~\hat y$. To derive the above equation, write $\vec r=r~\hat r,$ note that $\dfrac{d\hat r}{dt}=\dot \theta~\hat \theta~ ;~~ \dfrac{d\hat\theta}{dt}=−\dot\theta~\hat r$ and differentiate twice.

 

[NTesla]

@kuruman, I would like to know your point of view of whether we can calculate moment of inertia in the case of the original question, i.e. when the particle is moving in straight line, but the axis about which we need to calculate torque is some distance apart from particle's line of motion.

 

[PeroK]

@PeroK : Here's how I calculated $\tau$. However, something is wrong, which I don't understand what it is. Please help.

My calculation of $\tau$ is: $\tau =\overrightarrow{r} \times \overrightarrow{F} = m \overrightarrow{r} \times \overrightarrow{a} = m|\overrightarrow{r}||\overrightarrow{a}|Sin\theta =mra_{r}$.

However, you've written $\tau = mra_{\theta}$

I don't understand what you've done there. In any case: $$|\vec r \times \vec a| = |\vec r \times (a_r \hat r + a_{\theta} \hat \theta)| = |ra_{\theta}|$$

 

[NTesla]

I don't understand what you've done there. In any case: $$|\vec r \times \vec a| = |\vec r \times (a_r \hat r + a_{\theta} \hat \theta)| = |ra_{\theta}|$$

I just figured out where I was going wrong with my calculation of $\tau$.

 

[kuruman]

@kuruman, I would like to know your point of view of whether we can calculate moment of inertia in the case of the original question, i.e. when the particle is moving in straight line, but the axis about which we need to calculate torque is some distance apart from particle's line of motion.

Why can't we have $I=mr^2$ in this particular case? It's just that the moment of inertia about the origin is not constant in time but there is nothing untoward about that. Watch this.

Start with the angular moment about the origin when the particle is at some arbitrary position $\vec r$ moving with arbitrary constant velocity $\vec v$. Then its angular momentum momentum about the origin is given by $\vec L= \vec r\times(m\vec v).$ Using results posted in #27, $\dfrac{d\vec r}{dt}=\vec v=\dot r~\hat r+r \dot \theta ~\hat \theta.$ Substitute,$$\vec L= r~\hat r\times[m(\dot r~\hat r+r \dot \theta ~\hat \theta)]=m[(r\dot r~(\hat r\times\hat r)+mr^2\dot \theta~(\hat r\times \hat \theta)]=mr^2 \dot\theta~\hat z.$$With the definitions $I=mr^2$ and $\dot \theta\hat z=\vec \omega$, you have the well known result, $\vec L = I\vec \omega$.

Note, that both $I$ and $\vec \omega$ change with respect to time, nevertheless their product is constant because angular momentum about the origin is conserved since the torque about the origin is zero.

On edit:
If there is a torque about the origin, then you can find an expression for it simply by using $$\vec \tau=\frac{d\vec L}{dt}=\frac{d}{dt}\left(mr^2 \dot\theta~\hat z\right).$$ The Cartesian unit vector is constant in time, so you only have to use the product rule on $r^2$ and $\dot\theta$.

 

[NTesla]

@kuruman , @PeroK, @haruspex,
As there is a general form of $\alpha =\ddot{\theta }+\frac{2\dot{r}\dot{\theta }}{r}$, Is there any general form for $\omega$ ? Or is it just $\frac{d\theta}{dt}$ ?

 

[PeroK]

@kuruman , @PeroK, @haruspex,
As there is a general form of $\alpha =\ddot{\theta }+\frac{2\dot{r}\dot{\theta }}{r}$, Is there any general form for $\omega$ ? Or is it just $\frac{d\theta}{dt}$ ?

https://en.wikipedia.org/wiki/Angular_acceleration

 

[kuruman]

@kuruman , @PeroK, @haruspex,
As there is a general form of $\alpha =\ddot{\theta }+\frac{2\dot{r}\dot{\theta }}{r}$, Is there any general form for $\omega$ ? Or is it just $\frac{d\theta}{dt}$ ?

You can easily derive the expression. From what has been said above, $$\vec \omega=\frac{\vec L }{I}=\frac{\vec r \times (m \vec v)}{mr^2}=\frac{\vec r\times \vec v}{r^2}.$$I don't know if it qualifies as a "general" form, but the same expression is given in the Wikipedia link provided by @PeroK in post #33.