Finding angular velocity when the CM is directly below the pivot

[jisbon]

Homework Statement:

A triangular plate of a mass of 150kg is shown as below. Calculate the angular velocity of the plate when CM is directly below the pivot.

Relevant Equations:

-

So first of all, I will have to find the centre of mass.
$X_{cm} = \frac{1}{M}\int x dm$
likewise for Y.

$M$ =150kg
From the above-given points, I can find the equation of a line to be
$y =-\frac{3}{4}x +3$ .
Area density = $150kg/(0.5*4*3) = 25kg/m^2$

$X_{cm} = \frac{1}{M}\int x dm = \frac{1}{150}\int x \mu y dx = \frac{1}{150}\int x (25) (-\frac{3}{4}x+3) dx =\frac{1}{150}\int_{0}^{4} x (25) (-\frac{3}{4}x+3) dx = 4/3$

$Y_{cm} = \frac{1}{M}\int y dm = \frac{1}{150}\int y \mu x dy = \frac{1}{150}\int y (25) (-\frac{4}{3}y+4) dy =\frac{1}{150}\int_{0}^{3} y(25) (-\frac{4}{3}y+4) dy = 1$

After getting the coordinates, what concepts should I apply next? Moment of inertia/rotational motion? Not sure what formulas/theories to apply here, any guidance will be appreciated. Thanks

 

[jbriggs444]

So first of all, I will have to find the centre of mass.
$X_{cm} = \frac{1}{M}\int x dm$
likewise for Y.

$M$ =150kg
From the above-given points, I can find the equation of a line to be
$y =-\frac{3}{4}x +3$ .
Area density = $150kg/(0.5*4*3) = 25kg/m^2$

$X_{cm} = \frac{1}{M}\int x dm = \frac{1}{150}\int x \mu y dx = \frac{1}{150}\int x (25) (-\frac{3}{4}x+3) dx =\frac{1}{150}\int_{0}^{4} x (25) (-\frac{3}{4}x+3) dx = 4/3$

$Y_{cm} = \frac{1}{M}\int y dm = \frac{1}{150}\int y \mu x dy = \frac{1}{150}\int y (25) (-\frac{4}{3}y+4) dy =\frac{1}{150}\int_{0}^{3} y(25) (-\frac{4}{3}y+4) dy = 1$

After getting the coordinates, what concepts should I apply next? Moment of inertia/rotational motion? Not sure what formulas/theories to apply here, any guidance will be appreciated. Thanks

So after much calculation, the conclusion is that the center of mass of the triangle at its starting position is 1/3 of the way from the flat side on the bottom to the point at the top. That is a correct result. And one possibly worth remembering.

It is also worth noting the numbers for mass, area and length were largely irrelevant. It would have been more efficient to have left them as symbols and saved a bunch of writing and arithmetic.

Nice presentation, by the way. Clear drawing, typeset equations and a description of what it is you are trying to do.

But your question is not about the tactics for determining a center of mass. You ask about a strategy for solving the problem. Practice in these forums is to make questioners work for their answers. So rather than handing you a strategy, let me offer some suggestions to pick through.

A good starting point is often to pick a conservation law and see whether it applies and where it might lead. For each of the following quantities, try to identify why it is not helpful or why it is difficult to use for this situation. Perhaps one of them is more viable than the rest.

1. Angular momentum.
2. Rotational kinetic energy.
3. Linear momentum.
4. Linear kinetic energy.

 

[jisbon]

So after much calculation, the conclusion is that the center of mass of the triangle at its starting position is 1/3 of the way from the flat side on the bottom to the point at the top. That is a correct result. And one possibly worth remembering.

It is also worth noting the numbers for mass, area and length were largely irrelevant. It would have been more efficient to have left them as symbols and saved a bunch of writing and arithmetic.

But your question is not about the tactics for determining a center of mass. You ask about a strategy for solving the problem. Practice in these forums is to make questioners work for their answers. So rather than handing you a strategy, let me offer some suggestions to pick through.

A good starting point is often to pick a conservation law and see whether it applies and where it might lead. For each of the following quantities, try to identify why it is not helpful or why it is difficult to use for this situation. Perhaps one of them is more viable than the rest.

1. Angular momentum.
2. Rotational kinetic energy.
3. Linear momentum.
4. Linear kinetic energy.

1. Angular momentum - since I have to find out angular velocity, in the end, this will probably be useful?
2. Rotational kinetic energy - I don't think energy is related in this question is it?
3. Linear momentum - clearly not a linear system
4. Linear kinetic energy - same as reason 2.

Hence using COAM,

$L = I \omega$
However, in this case, will I need to find out/derive the inertia of this triangle? Or is this method already wrong?
Thanks

 

[jbriggs444]

Hence using COAM,

$L = I \omega$
However, in this case, will I need to find out/derive the inertia of this triangle? Or is this method already wrong?

Let us try to work based on angular momentum and see where it leads. [No, this is not the strategic choice I would have made]

Yes, we will need the moment of inertia of the triangle. But before we can calculate that, we need to choose an axis of rotation to work with. There are two obvious choices:

1. The fixed point at the tip of the triangle about which it rotates.
2. The center of mass of the triangle.

We will be looking to write an equation involving torques and a resulting angular acceleration. One of the choices will make for a simpler equation than the other. Which will you pick?

[We will need the moment of inertia of the triangle -- this is not wasted effort]

 

[jisbon]

Let us try to work based on angular momentum and see where it leads. [No, this is not the strategic choice I would have made]

Yes, we will need the moment of inertia of the triangle. But before we can calculate that, we need to choose an axis of rotation to work with. There are two obvious choices:

1. The fixed point at the tip of the triangle about which it rotates.
2. The center of mass of the triangle.

We will be looking to write an equation involving torques and a resulting angular acceleration. One of the choices will make for a simpler equation than the other. Which will you pick?

[We will need the moment of inertia of the triangle -- this is not wasted effort]

I will say the centre of mass? Not entirely sure about that.

About the moment of inertia (it's going to be long but let's go):

$I_{x} =\int y^2 dA = \int_{0}^{3} y^2 (4-\frac{4}{3}y)dy = 4\int_{0}^{3} y^2 dy -\frac{4}{3}\int_{0}^{3} y^3 dy = 36-27=9$

I suppose I won't need to find about the y-axis since it's swinging by the x axis?

 

[jbriggs444]

I will say the centre of mass? Not entirely sure about that.

I prefer the pivot point at the tip. To see why, let us take the next step and write down torque equations.

Let us first try using the center of mass. We are going to equate moment of inertia times angular acceleration to the sum of all external torques. What external torques are there?

There is no external torque from gravity. That is because we chose our axis of rotation at the center of mass. But there is an external torque from the supporting force of the pivot. How much external torque? That is the problem right there. We do not know [at least not immediately] how much external torque the pivot point is supplying. It is not equal to the weight of the triangle because the triangle is accelerating downward.

Now let us try writing the same equation with the axis at the triangle's tip. Again, we are equating moment of inertia times angular acceleration to the sum of the external torques. This time there is zero torque from the pivot point at the tip. There is a torque from gravity. But we can easily write down a formula for that torque in terms of the mass of the triangle, the acceleration of gravity and the offset of the center of gravity from the pivot point.

One heuristic is that if you have an unknown force and you do not otherwise need to solve for that force, put your axis of rotation where that force acts. It will then cancel out of your torque equation and you can safely ignore it.

So let's use the moment of inertia about the tip.

About the moment of inertia (it's going to be long but let's go):

$I_{x} =\int y^2 dA = \int_{0}^{3} y^2 (4-\frac{4}{3}y)dy = 4\int_{0}^{3} y^2 dy -\frac{4}{3}\int_{0}^{3} y^3 dy = 36-27=9$

I suppose I won't need to find about the y-axis since it's swinging by the x axis?

If I understand the question, you do need to worry about both axes. A wide triangle has a greater moment of inertia than a narrow triangle even if their heights are the same.

Personally, I cheated for this part. Wikipedia has a list of moments of inertia all worked out. Including one for a thin triangle rotating in its own plane about its tip.

We can proceed further with this strategy if you like. Though it may be time to look at the approach using rotational kinetic energy instead.

 

[jisbon]

I prefer the pivot point at the tip. To see why, let us take the next step and write down torque equations.

Let us first try using the center of mass. We are going to equate moment of inertia times angular acceleration to the sum of all external torques. What external torques are there?

There is no external torque from gravity. That is because we chose our axis of rotation at the center of mass. But there is an external torque from the supporting force of the pivot. How much external torque? That is the problem right there. We do not know [at least not immediately] how much external torque the pivot point is supplying. It is not equal to the weight of the triangle because the triangle is accelerating downward.

Now let us try writing the same equation with the axis at the triangle's tip. Again, we are equating moment of inertia times angular acceleration to the sum of the external torques. This time there is zero torque from the pivot point at the tip. There is a torque from gravity. But we can easily write down a formula for that torque in terms of the mass of the triangle, the acceleration of gravity and the offset of the center of gravity from the pivot point.

One heuristic is that if you have an unknown force and you do not otherwise need to solve for that force, put your axis of rotation where that force acts. It will then cancel out of your torque equation and you can safely ignore it.

So let's use the moment of inertia about the tip.

If I understand the question, you do need to worry about both axes. A wide triangle has a greater moment of inertia than a narrow triangle even if their heights are the same.

Personally, I cheated for this part. Wikipedia has a list of moments of inertia all worked out. Including one for a thin triangle rotating in its own plane about its tip.

We can proceed further with this strategy if you like. Though it may be time to look at the approach using rotational kinetic energy instead.

I think the energy part will certainly be easier:

Is it true to say that:

the gain in rotational ke = lost in gpe (from centre of mass)

$\frac{1}{2}I\omega^2 = 150(9.8)(-1) - 150(9.8)(1)$

 

[jbriggs444]

I think the energy part will certainly be easier:

Is it true to say that:

the gain in rotational ke = lost in gpe (from centre of mass)

Yes.

$\frac{1}{2}I\omega^2 = 150(9.8)(-1) - 150(9.8)(1)$

Have you calculated where the center of mass is when the triangle hanging downward in its final position?

 

[jisbon]

Yes.

Have you calculated where the center of mass is when the triangle hanging downward in its final position?

initial com of y-axis is 1.
how do I exactly calculate the position of com when its hanging ?

 

[jbriggs444]

initial com of y-axis is 1.
how do I exactly calculate the position of com when its hanging ?

The easy way is to refer back to post #2. The center of mass is 1/3 of the way from the flat side on the bottom to the point at the top.

[I did say that it might be worth remembering :-)]

 

[jisbon]

The easy way is to refer back to post #2. The center of mass is 1/3 of the way from the flat side on the bottom to the point at the top.

[I did say that it might be worth remembering :-)]

I drew the diagrams, and I found out that the height at which when the CM is below the axis is simply using Pythagoras theorem, finding the length to be -5/3.

So now my equation looks like:

$\frac{1}{2}I\omega^2 = 150(9.8)(-5/3) - 150(9.8)(1) = -3920$
However, subsututing I as 400/9 does not give me my answer for omega correctly.
In this case, what is the inertia I'm looking at here exactly?

Thanks for the guidance

 

[jbriggs444]

However, subsututing I as 400/9 does not give me my answer for omega correctly.

Can you share your calculation for the moment of inertia?

It is worth doing a quick sanity check on the claimed result of 400/9 kilogram meter squared. That result is approximately 45 kg m². That would be the moment of inertia of a 45 kilogram point mass located one meter from the origin. Our object is a 150 kilogram extended mass whose center is more than one meter from the origin. Its moment of inertia will be more than 150 kg m². [A lot more].

You say that you know your result is incorrect. What result are you trying to shoot for?

 

[jisbon]

Can you share your calculation for the moment of inertia?

It is worth doing a quick sanity check on the claimed result of 400/9 kilogram meter squared. That result is approximately 45 kg m². That would be the moment of inertia of a 45 kilogram point mass located one meter from the origin. Our object is a 150 kilogram extended mass whose center is more than one meter from the origin. Its moment of inertia will be more than 150 kg m². [A lot more].

You say that you know your result is incorrect. What result are you trying to shoot for?

What I meant was
$I_{y} = 400$ and $I_{x} = 9$

I'm not sure which inertia to use in this case, but my guess will be $I_{x} = 9$?

 

[jbriggs444]

What I meant was
$I_{y} = 400$ and $I_{x} = 9$

I'm not sure which inertia to use in this case, but my guess will be $I_{x} = 9$?

Could you explain what you mean by $I_y$ and $I_x$?

The moment of inertia of an object about an axis is the integral of the object's density at each point times the squared distance of that point from the axis of rotation. There is no $I_x$ or $I_y$. There is only $I$.

Edit to add...

Possibly you are attempting to invoke the perpendicular axis theorem. To use the theorem, you would consider the triangle as a planar figure in the x-y plane. You would first compute its moment of inertia about the y axis, obtaining $I_y$. Then you would compute its moment of inertia about the x axis, obtaining $I_x$. The sum of those would be the moment of inertia about the z axis, $I_z$.

However, the figures you quote for $I_x$ and $I_y$ are suspect.

 

[jisbon]

Could you explain what you mean by $I_y$ and $I_x$?

The moment of inertia of an object about an axis is the integral of the object's density at each point times the squared distance of that point from the axis of rotation. There is no $I_x$ or $I_y$. There is only $I$.

Edit to add...

Possibly you are attempting to invoke the perpendicular axis theorem. To use the theorem, you would consider the triangle as a planar figure in the x-y plane. You would first compute its moment of inertia about the y axis, obtaining $I_y$. Then you would compute its moment of inertia about the x axis, obtaining $I_x$. The sum of those would be the moment of inertia about the z axis, $I_z$.

However, the figures you quote for $I_x$ and $I_y$ are suspect.

Ah got it, so in this case, the I simply equals to$I = m((1^2+(4/3)^2)$

With that I can use:

$\frac{1}{2}I\omega^2 = 150(9.8)(-5/3) - 150(9.8)(1) = -3920$

to solve the problem?

 

[jbriggs444]

Ah got it, so in this case, the I simply equals to$I = m((1^2+(4/3)^2)$

The moment of inertia of an extended object is not the same as that of a point object with the same mass and center of mass.

For instance, the moment of inertia of a 2 meter stick with mass 1 kg about its end is $\frac{1}{3}ml^2 = \frac{4}{3}$ while the moment of inertia of a 1 kg point mass orbitting at 1 meter from the origin is $mr^2 = 1$