How do i determine the final position?

[Philip Wood]

Don't think about it in terms of radius and pivot point.

You agree that y = \frac{L}{2}sin\theta ?

In that case \dot{y} = \frac{L}{2}cos\theta \dot{θ}.

The rotational KE must be calculated about the centre of mass (CM). The formula for rotational KE is \frac{1}{2}Iω². I, the moment of inertia of the rod about its centre, is \frac{1}{12}mL².

But ω = \dot{θ}, the rate of change of the angle θ shown on the diagram.

Hope it all makes sense now.

 

[cshum00]

You agree that y = \frac{L}{2}sin\theta ?

In that case \dot{y} = \frac{L}{2}cos\theta \dot{θ}.

Interesting. You derived velocity by calculating the derivative of y. I did it differently. I substituted the relationship between velocity, radius and angular velocity like this: \dot{y} = r \dot{\theta}. Then you can compare both results and see that that \frac{L}{2}cos\theta happens to be your radius for angular the velocity. And according to the physical geometry, it is the projection between point A and the center of mass in the x-axis.

The rotational KE must be calculated about the centre of mass (CM). The formula for rotational KE is \frac{L}{2}Iω².

I think it was a typo and you meant that the rotational KE is \frac{1}{2}Iω²

I, the moment of inertia of the rod about its centre, is \frac{1}{12}mL².

Let's define moment of inertia again: I = mr^2.
So, if I = m(\frac{L^2}{12}), means your radius is \frac{L}{\sqrt{12}}. How come?

 

[Philip Wood]

Sorry about the typo.

\frac{L}{\sqrt{12}} is what's called the 'radius of gyration' of the rod about its centre. The interpretation is this: if all the rod's mass were to be gathered together at one point, then if that point were to be at the radius of gyration from the rod's centre, then the rod would have the same moment of inertia about its centre as the original rod.

Do you, I wonder, need to learn up about moment of inertia: (a) its rôle in mechanics, and (b) how to calculate moments of inertia of a few simple bodies, like cylinders and rods? Sorry if this seems patronising, but you do seem very keen to get to the bottom of this stuff.

 

[rcgldr]

angular inertia of a rod rotating about one end.

That should have been rod rotating by it's center, since the end of the rod is not constrained horizontally (frictionless surface) and the center of mass moves and acceleates only vertically. Philip Wood already did the math using angular inertia for a rod rotating about it's center.

The 'radius of gyration' of the rod about its centre. The interpretation is this: if all the rod's mass were to be gathered together at one point ...

or all the mass located in a hollow cylinder at that radius of gyration, or any collection of point masses, as long as all of them are located at the radius of gyration.

 

[cshum00]

Sorry if this seems patronising, but you do seem very keen to get to the bottom of this stuff.

Of course i am. I am sure you didn't pick that value for radius of gyration randomly. I want to be able to understand the calculations myself.

\frac{L}{\sqrt{12}} is what's called the 'radius of gyration' of the rod about its centre.

I found how the number was obtained. It seems that i was just picking a radius away from my chosen pivot point; which is wrong? It seems that we have to calculate this radius by integrating the stick.
dm = \frac{M}{L}dr
I = \int_{\frac{-L}{2}}^{\frac{L}{2}} r^2 dm = \frac{M}{L} \int_{\frac{-L}{2}}^{\frac{L}{2}} r^2dr
I = \frac{M}{L} \frac{r^3}{3} |_{\frac{-L}{2}}^{\frac{L}{2}} = \frac{M}{3L} (\frac{L^3}{8} + \frac{L^3}{8}) = M \frac{L^2}{12}

However, i question that if finding the radius of gyration this way is really correct? Here is a simple example i am going to use. Assume that our problem, point A is a fixed pivot point. And i am going to try to find the torque of the stick when gravity is the only relevant force.
\tau = \alpha I = \frac{a_t}{r} mr^2 = F_t r

-First, let's calculate the inertia and radius of gyration; with the fixed pivot point A. So the points of integration are from 0 to L.
I = \int_{0}^{L} r^2 dm = \frac{M}{L} \int_{0}^{L} r^2dr = \frac{M}{L} \frac{r^3}{3} |_{0}^{L} = \frac{M}{3L} (L^3 - 0) = M \frac{L^2}{3}
-Therefore the radius of gyration when the pivot is at the end of the stick is: \frac{L}{\sqrt{3}}
-The tangent acceleration that can be obtained by the geometry is equals: a_t = -gcos(\theta)
-Then the torque equals: \tau = \alpha I = \frac{-gcos(\theta)}{\frac{L}{\sqrt{3}}}M \frac{L^2}{3} = \frac{-\sqrt{3}LMgcos(\theta)}{3}
-But the torque is also equals: \tau = F_t r = -Mgcos(\theta) \frac{L}{2}
-Or even if i use the radius of gyration instead of a distance between the pivot and the center of mass: \tau = F_t r = -Mgcos(\theta) \frac{L}{\sqrt{3}}
-Even if i do a mix and match on the torque that contains moment of inertia: \tau = \alpha I = \frac{-gcos(\theta)}{\frac{L}{2}}M \frac{L^2}{3} = \frac{-2LMgcos(\theta)}{3}

One last thing, it took me to try it with a different notation to actually find out that this is wrong:

You agree that y = \frac{L}{2}sin\theta ?

In that case \dot{y} = \frac{L}{2}cos\theta \dot{θ}.

This is what you are doing:
x = y^2
\frac{dx}{dz} = \frac{d}{dz}(y^2) = 2y \frac{dy}{dz} which is wrong

 

[Philip Wood]

Agree with your moment of inertia calculations. Can't (with only a few minutes to spare at present) follow your dynamics.

Don't see any connection between your stuff with y² at the bottom of your post and my differentiation of sinθ wrt t, using the chain rule.

Shall be off PF for some 48 hours starting in a few minutes.

 

[cshum00]

Don't see any connection between your stuff with y² at the bottom of your post and my differentiation of sinθ wrt t, using the chain rule.

Chain rule can't be applied such way because sinθ is not differentiable with respect to t. It is due to the fact that θ does not contain the time variable. Not even with the radius relation: \theta = \frac{d}{r}

Here is what your correct chain rule looks like:\frac{dy}{dt} = \frac{d}{dt}sin\theta\frac{dy}{dt} = \frac{d}{dt}(sin\theta)\frac{d}{dt}(\theta)\dot{y} = \dot{\theta}\frac{d}{dt}(sin\theta)Meaning that you can't get rid of the sinθ and suddenly change it to cosθ.

 

[Philip Wood]

Chain rule for this case :

\frac{dy}{dt} = \frac{dy}{d\theta} \times \frac{d\theta}{dt}.

 

[cshum00]

You are right.
y = sin\theta
\frac{dy}{dt} = \frac{d}{d\theta}(sin\theta) \frac{d\theta}{dt}
\dot{y} = cos\theta \dot{\theta}
And it can be confirmed by integration:
\frac{dy}{dt} = cos\theta \frac{d\theta}{dt}
\int{dy} = \int cos\theta \frac{d\theta}{dt}dt
y = sin\theta

Which is somewhat awkward because it means that my first example is also correct:
\frac{dx}{dz} = \frac{df(y)}{dy} \frac{dy}{dz}
x = y^2
\frac{dx}{dz} = \frac{y^2}{dy} \frac{dy}{dz}
\frac{dx}{dz} = 2y \frac{dy}{dz}
And confirmed by integration:
\int{dx} = \int 2y \frac{dy}{dz} dz
x = y^2