Why can I only resolve tangentially on a point particle?

[etotheipi]

For instance, in the case of a simple pendulum, it is quite acceptable to write down $-mg\sin{\theta} = ma_{tangential}$, and go from there.

However, if we introduce a rotating body which is not a particle, we may still calculate its torque from its centre of mass, however we can no longer resolve tangentially to solve for the tangential acceleration (presumably of the centre of mass). Instead, it is required to calculate the body's moment of inertia and proceed with $\tau = I\alpha$.

I'm not sure about how I could go about proving this mathematically; I wonder if it has something to do with the added rotation of the non point-like body during the motion (since, from an energy perspective, we will end up with some $\frac{1}{2} I \omega^{2}$, which will "reduce" the velocity of the centre of mass since we now have less $\frac{1}{2} m v^{2}$?). If so, how might I start to prove this with a force, as opposed to energy, approach? Thank you!

 

[BvU]

Hi $e^{i\pi}$

its torque from its centre of mass

You mean 'the torque wrt some axis of rotation' ?

go about proving this mathematically

For a point mass rotating about some axis kinetic energy $T$ is ${1\over 2} mv^2$ and $v =\omega r$ so that $T = {1\over 2} I\omega^2$ with $I = mr^2$

For an extended mass there is no single $r$ any more and we compute kinetic energy with $$
T = \int {\scriptstyle {1\over 2}} v^2 \,dm = {\scriptstyle {1\over 2}}\int \omega^2 r^2\, dm = {\scriptstyle {1\over 2} }\omega^2 \int r^2 \, dm \equiv {\scriptstyle {1\over 2}} I\omega^2$$ now with $I = \int r^2\, dm$

$\$

 

[etotheipi]

Hi $e^{i\pi}$

You mean 'the torque wrt some axis of rotation' ?
For a point mass rotating about some axis kinetic energy $T$ is ${1\over 2} mv^2$ and $v =\omega r$ so that $T = {1\over 2} I\omega^2$ with $I = mr^2$

For an extended mass there is no single $r$ any more and we compute kinetic energy with $$
T = \int {\scriptstyle {1\over 2}} v^2 \,dm = {\scriptstyle {1\over 2}}\int \omega^2 r^2\, dm = {\scriptstyle {1\over 2} }\omega^2 \int r^2 \, dm \equiv {\scriptstyle {1\over 2}} I\omega^2$$ now with $I = \int r^2\, dm$

$\$

Thank you, I'd never seen that derivation before so it makes a lot more sense now!

Do we then just say that only in the case of a point particle do the rotational equations reduce to the "tangential form"? For instance, for a point particle on a simple pendulum of length $l$$$mgl\sin{\theta} = ml^{2}\alpha \implies g\sin{\theta} = l\alpha = a$$whilst this wouldn't be the case for any bodies with non-$ml^{2}$ moment of inertia?

And if so, is the concept of "tangential acceleration" rigorous or just a consequence of this special case? Thanks for all your help!

 

[A.T.]

And if so, is the concept of "tangential acceleration" rigorous or just a consequence of this special case?

I vote for "rigorous consequence".

 

[A.T.]

I guess that works too!

The whole thing just still seems a little odd. Imagine if we constructed a pendulum by "hinging" a banana from its stalk. From Newton II, we know the acceleration of the centre of mass of a rigid body in any direction, instantaneously at least, is the resultant force in that direction divided by mass. If, at some point in the swing, we resolve in the direction of motion of the banana ($F = mg\sin{\theta}$), I would expect to be able to use this to find the tangential acceleration of the centre of mass of the banana.

However, since all of the bananas I've come across aren't point particles, if we could somehow find the moment of inertia of this particular banana wrt the axis of rotation we would apparently need to use the rotational form of Newton II, and I'm guessing this would throw out a different value for the linear acceleration.

It's not obvious why the first paragraph is wrong, and I certainly can't explain why it is so!

If your pendulum bob is an extended body that changes orientation, then the bob's attachment has to provide some torques around the bob's centre of mass. This can introduce tangential forces on the bob from the attachment.

 

[etotheipi]

I vote for "rigorous consequence".

I guess that works too!

I did a quick scribble of a scenario which I think sums up my confusion. Two spheres connected via a light rod, with the top sphere attached to the hinge by a light string.

With the first method, I get an angular acceleration of $\frac{3}{5}g\sin{\theta}$. If the centre of mass is at $1.5m$ from the pivot, its linear acceleration is $0.9g\sin{\theta}$.

However, when treating the two spheres/rod as a system and applying Newton II I get a different answer.

 

[etotheipi]

If your pendulum bob is an extended body that changes orientation, then the bob's attachment has to provide some torques around the bob's centre of mass. This can introduce tangential forces on the bob from the attachment.

Thank you, I'll have to think on it (just so I can clear things up for myself) for a little longer but that sounds correct!

So the extra external forces required for the rotation of the non point-like bob around the centre of mass also contribute to the linear acceleration. In the case of the scenario I just uploaded (sorry about the change, on second thoughts I thought the banana was a silly idea...), would we require an extra force to be added to the free body diagram at the connection between the string and upper sphere?

 

[A.T.]

So the extra external forces required for the rotation of the non point-like bob around the centre of mass also contribute to the linear acceleration.

The details will depend on the type of pendulum. But I would start simpler than the double sphere. Even a single sphere on a string is messy enough: It doesn't keep its CoM on the line extending the string, allowing the string to produce a torque around the CoM. But this also means that the center of mass doesn't actually move on a prefect circular arc.

 

[BvU]

start simpler

Still rather complicated. Wouldn't the rod pendulum be a better 'start simpler' candidate ?

 

[A.T.]

Still rather complicated. Wouldn't the rod pendulum be a better 'start simpler' candidate ?

Yes. Here the CoM moves on a perfect circle, but the force at the attachment is not always passing through the CoM, so it has a component parallel to the CoM's motion.

 

[etotheipi]

Yes. Here the CoM moves on a perfect circle, but the force at the attachment is not always passing through the CoM, so it has a component parallel to the CoM's motion.

Still rather complicated. Wouldn't the rod pendulum be a better 'start simpler' candidate ?

I'll have a go at analysing the rod pendulum now to see whereabouts the maths goes. Luckily the most important question in the thread has been really nicely answered (namely the origin of the discrepancy), which is attributed to contributions to the resultant force with components parallel to the direction of motion at the attachment. Thank you both for your patience!