Fall Speed Based on Angle: 0°, 45°, 90°

[adjacent]

Homework Statement

Let the angle of the block(slope) to be 30°.
height=0.6m, length=1m. Breadth=1m
Smallest height on diagram is 0m

Let the red line to divide the block by 2(when viewed from the top)
The two wheels are of equal diameter and is connected by a rod.
The total mass of two wheels + rod is 1kg.
Let the angle be measured from the red line.(At the image,the wheel system is 90° to the line)
1-Rotate the wheel to 0°.
It will not move.
2-Rotate to 90°
it will fall at maximum speed.
3- Rotate to 45°
will be slower than 90°.

So I am thinking of the relation between the angle and the speed of fall.

Homework Equations

$F=ma$
...

The Attempt at a Solution

Sorry I am at Gr.10. So I haven't studied vectors,forces or speeds at angles.
-Just curious-

 

[voko]

A useful property here is the conservation of energy. When the system moves on the incline, its centre of mass gets lower. That means its potential energy is reduced. Which in turn means its kinetic energy is increased. Clearly when it goes "sideways" it needs to go a longer distance to gain the same amount of energy as compared with the case when it goes "straight".

Alternatively, you can look at the system so that your line of sight is always parallel to the axis of the system. Then you will see that it effectively rolls on an incline with some angle that depends on the "real" incline angle and "sideways" angle.

 

[adjacent]

Thank you.
Any specific equations for these kinds of "incline" things?

 

[voko]

I think you could try figuring out the "effective" incline angle. This is pure geometry.

 

[adjacent]

I don't understand what you mean by the axis of the system.
Anyway,I want to find the relationship.
So I have to make an equation relating incline angle and wheel angle.

 

[voko]

Let the top point of the red line be A. The lowest point of the red line is B. The point on the ground under A is C, so ABC is a right triangle, and the angle ABC is $\alpha$.

The system starts at A and rolls down the incline at some angle with the red line and it touches the ground at D. The angle DAB is the "sideways" angle $\beta$. The "effective" incline angle is ADC. Express angle ADC in terms of $\alpha$ and $\beta$.

 

[adjacent]

Let the top point of the red line be A. The lowest point of the red line is B. The point on the ground under A is C, so ABC is a right triangle, and the angle ABC is $\alpha$.

The system starts at A and rolls down the incline at some angle with the red line and it touches the ground at D. The angle DAB is the "sideways" angle $\beta$. The "effective" incline angle is ADC. Express angle ADC in terms of $\alpha$ and $\beta$.

Isn't D the same as B?I said in the OP that lowest point of the red line and the ground has a distance of 0m

 

[voko]

Isn't D the same as B?

How? The system rolls away from the red line, so it cannot end up at B.

I said in the OP that lowest point of the red line and the ground has a distance of 0m

Do you think I indicated otherwise?

 

[adjacent]

How? The system rolls away from the red line, so it cannot end up at B.

Oh sorry.I thought B as the point of ground.
Ok.Its a point on the red line and D is a point on ground(Where it touches)

How's that going to relate wheel angle and speed?

 

[voko]

Oh sorry.I thought B as the point of ground.
Ok.Its a point on the red line and D is a point on ground(Where it touches)

I do not think you understood me. Points B, C, D are all on the ground. Point B in particular is where the red line touches the ground. Point D is where the rolling system touches the ground. These two points can also be said to lie on the edge of the incline.

How's that going to relate wheel angle and speed?

I do not know what the "wheel angle" is and I would like you to answer my question about angle ADC first, anyway.

 

[adjacent]

Let the top point of the red line be A. The lowest point of the red line is B. The point on the ground under A is C, so ABC is a right triangle, and the angle ABC is $\alpha$.

The system starts at A and rolls down the incline at some angle with the red line and it touches the ground at D. The angle DAB is the "sideways" angle $\beta$. The "effective" incline angle is ADC. Express angle ADC in terms of $\alpha$ and $\beta$.

I would like you to answer my question about angle ADC first, anyway.

That seems to be a 3d problem.How am I going to do that?I haven't studied that yet.Can you give me a hint?

 

[voko]

You solve a 3D problem by disassembling it into 2D subproblems.

In this case, you have three triangles: ABC, ABD, and ADC. These are right triangles and they are all related to each other in some way. Triangle ABD and triangle ADC share hypotenuse AD. Triangle ABC and triangle ADC share cathetus AC. This is enough to express angle ADC via angle ABC ($\alpha$) and angle DAB ($\beta$), it takes just a bit of trigonometry.

 

[adjacent]

Is it OK to include other variables like BD,CB and AC?

 

[voko]

There is an answer that only has $\alpha$ and $\beta$. But we can discuss your approach even if you do not have that answer yet. Show what you have so far.

 

[adjacent]

OK.

I've got
$\alpha=tan^{-1}(\frac{h}{b})$
$\beta=tan^{-1}(\frac{S}{\sqrt{h^2+b^2}})$
$ADC=tan^{-1}(\frac{h}{\sqrt{s^2+b^2}})$

Not sure how to link $\alpha$,$\beta$ and ADC in one equation

EDIT:Latex problem.How do I fix that?

 

[voko]

You meant those:
$\alpha=tan^{-1}(\frac{h}{b})$
$\beta=tan^{-1}(\frac{S}{\sqrt{h^2+b^2}})$
$ADC=tan^{-1}(\frac{h}{\sqrt{s^2+b^2}})$

(one closing curly brace was missing)

 

[adjacent]

You meant those:
$\alpha=tan^{-1}(\frac{h}{b})$
$\beta=tan^{-1}(\frac{S}{\sqrt{h^2+b^2}})$
$ADC=tan^{-1}(\frac{h}{\sqrt{s^2+b^2}})$

(one closing curly brace was missing)

Yeah,it's that.
How do I link those three together?

 

[voko]

As I said:

Triangle ABD and triangle ADC share hypotenuse AD. Triangle ABC and triangle ADC share cathetus AC.

So, can you express AD via AB and $\beta$? Then, can you express AB via AC and $\alpha$? And, finally, find the ratio of AC with AD?

 

[adjacent]

So, can you express AD via AB and $\beta$?

$AD=\frac{AB}{cos\beta}$

Then, can you express AB via AC and $\alpha$?

$AB=\frac{AC}{sin\alpha}$

And, finally, find the ratio of AC with AD?

How do I do that?I don't understand what you said.

 

[voko]

You are required to find a certain angle, let's call it $\gamma$. Is it related in any way with $$AC \over AD$$ ?

 

[adjacent]

You are required to find a certain angle, let's call it $\gamma$. Is it related in any way with $$AC \over AD$$ ?

Now I get
$\gamma=\frac{AB\sin\alpha}{\frac{AB}{\cos\beta}}$
Now I need to cancel AB somehow.
So ans is :(I always have trouble when 3 divisions are involved.Can you explain that as well?)

$\frac{\sin\alpha}{\cos\beta}$ = effective incline?

 

[voko]

Now I get
$\gamma=\frac{AB\sin\alpha}{\frac{AB}{\cos\beta}}$

Close, but not quite. Is it the angle that is a ratio of sides, or some function of the angle?

Now I need to cancel AB somehow.
So ans is :(I always have trouble when 3 divisions are involved.Can you explain that as well?)

Oh, that is very simple. When you have $$ {\frac a b \over \frac c d}$$ you eliminate the inner denominators by pulling $d$ up and pulling $b$ down, so that results in $$ ad \over bc $$ Note in your case $b = 1$.

$\frac{\sin\alpha}{\cos\beta}$ = effective incline?

Is it?

 

[adjacent]

Now I get
$\gamma=\frac{AB\sin\alpha}{\frac{AB}{\cos\beta}}$

Close, but not quite. Is it the angle that is a ratio of sides, or some function of the angle?

AC=$AB\sin\alpha$
AD=$\frac{AB}{\cos\beta}$

So $AC \over AD$ = $\frac{AB\sin\alpha}{\frac{AB}{\cos\beta}}$


So,$\gamma$=$\sin\alpha\cos\beta$ ?

 

[voko]

So, is it $$ \gamma = {AC \over AD} $$ or $$ \mathrm{function}(\gamma) = {AC \over AD} $$ ? And what $\mathrm{function}$ is that?

And please simplify that ratio!

 

[adjacent]

So, is it $$ \gamma = {AC \over AD} $$ or $$ \mathrm{function}(\gamma) = {AC \over AD} $$ ? And what $\mathrm{function}$ is that?

And please simplify that ratio!

Ah.You said just a ratio.From where did this FUNCTION come.
Anyway,I know what you mean.
It's $\gamma=tan^{-1}(\sin\alpha\cos\beta)$

 

[voko]

Ah.You said just a ratio.From where did this FUNCTION come.

Ratios of sides in a right triangle are always trigonometric functions of its angles.

Anyway,I know what you mean.
It's $\gamma=tan^{-1}(\sin\alpha\cos\beta)$

Why is that a $\tan^{-1}$? Can the ratio of a cathetus with the hypotenuse be a $\tan$?

 

[adjacent]

Ratios of sides in a right triangle are always trigonometric functions of its angles.



Why is that a $\tan^{-1}$? Can the ratio of a cathetus with the hypotenuse be a $\tan$?

Oh.I'm sorry it's $\sin^{-1}$

So $\gamma=\sin^{-1}(\sin\alpha\cos\beta)$

Then what does this have to do with the speed of the wheel?

 

[voko]

Very well. So now the problem is reduced from 3D to 2D. The system rolls along a straight line at angle $\gamma$ with the horizon.

Now go back to the beginning of the thread. Can you apply conservation of energy?

 

[adjacent]

$E_p=mgh$
h=0.6m here.
so $E_p=6J$
$E_k=\frac{1}{2}mv^2$
$v=\sqrt{12}$
Ah how does this relate to AD?

 

[voko]

$E_p=mgh$
h=0.6m here.
so $E_p=6J$

That is potential energy at the top of the incline. We need it at other points, too.

Let $x$ be the distance from point D (in the direction of A) where the system is located. Can you find the potential energy of the system at $x$?

$E_k=\frac{1}{2}mv^2$

That is correct for a point mass. But we have a rod and wheels, which rotate. Their rotation also has kinetic energy. Are you familiar with that?

 

[adjacent]

:zzz: just got free

Let $x$ be the distance from point D (in the direction of A) where the system is located. Can you find the potential energy of the system at $x$?

yes.
$P_E=mg\sin\gamma x$

That is correct for a point mass. But we have a rod and wheels, which rotate. Their rotation also has kinetic energy. Are you familiar with that?

Rotational kinetic energy?It's an A level subject.

 

[voko]

Have you studied moments of force (torques) and moments of inertia?

 

[adjacent]

I have studied moment of force but not inertia(That's also an A-level topic)

You just tell me how moment of inertia is used(I am a fast learner)

 

[voko]

For a point mass at distance $d$ away from the axis of rotation, the moment of inertia is $I = md^2$. When it is rotating with angular speed $\omega$, its linear speed is $\omega d$, so its kinetic energy is $\frac {mv^2} 2 = \frac {m (\omega d)^2 } 2 = \frac {I \omega^2} 2$. The same is true for a system of point masses $m_i$ and distances $d_i$, so the kinetic energy is given by the same formula, where $I = \sum m_i d_i^2$. For solid bodies, we replace the sum with an integral. See http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html for more details.

This particular system has two wheels and one rod. I suggest that we treat the rod as very thin and very light, in which case its moment of inertia is negligible, so the total moment of inertia is twice the moment of inertia of a wheel. The wheel is a cylinder, whose moment of inertia about its axis of symmetry is given by $\frac {mR^2} 2$, where $m$ is the mass and $R$ is the radius. In this case $m = \frac M 2$, so the total moment of inertia is $MR^2 \over 2$, and the rotational kinetic energy is then $\frac {MR^2 \omega^2} 4 = \frac {MR^2} 4 \left( \frac v R \right)^2 = \frac {Mv^2} 4$.

 

[adjacent]

Ok.Now the wheel system has kinetic energy of linear kinetic energy + rotational kinetic energy?

 

[voko]

Yes.

 

[adjacent]

So the total kinetic energy = potential energy at the top?
But that does not give any increase in speed at any given point on the path AD

 

[voko]

No. Total mechanical energy at any point = total mechanical energy at the top.

 

[adjacent]

No. Total mechanical energy at any point = total mechanical energy at the top.

Really?
Shouldn't it be total kinetic+ total potential energy (At that position)= Total potential energy at the top.

 

[voko]

What is total mechanical energy? And what is it at the top?

 

[adjacent]

Mechanical energy is sum of Potential and kinetic energy
(I will be learning those names in Gr.11)

Now back to the question
$P_E=mg\sin\gamma x$
$E_K=\frac{1}{2}mv^2+\frac{Mv^2}{4}$
so,
$mg0.6=mg\sin\gamma x+\frac{1}{2}mv^2+\frac{Mv^2}{4}*2$

 

[voko]

If I say something you do not know, say that right away :)

I think you are using two variables for the same thing, mass of the system: $m = M$. Use just one for simplicity. Then, why do you have *2 in the final equation, which you did not have in the equation for $E_K$? $E_K$ already includes the energy of both wheels.

 

[adjacent]

If I say something you do not know, say that right away :)

I think you are using two variables for the same thing, mass of the system: $m = M$. Use just one for simplicity. Then, why do you have *2 in the final equation, which you did not have in the equation for $E_K$? $E_K$ already includes the energy of both wheels.

m=M/2 So linear kinetic energy + rotational kinetic energy
Rotational kinetic energy is $\frac{Mv^2}{2}$
And you said m =M/2
So mass of two wheels should be 2M

 

[voko]

m=M/2 So linear kinetic energy + rotational kinetic energy
Rotational kinetic energy is $\frac{Mv^2}{2}$
And you said m =M/2

I could have been sloppy in my notation. I used $m$ as a dummy variable when talking about the moment of inertia of an abstract cylinder. So it is not the $m$ that we have in the linear kinetic energy. I used $M$ to denote the total mass of the system, which is the $m$ in the linear kinetic energy. It is the mass of the entire system, i.e., the two wheels and the infinitely light rod. The total rotational kinetic energy is $mv^2 \over 4$ or $Mv^2 \over 4$ depending on whether you prefer $m$ or $M$ for the total mass of the system; whichever your preference is, use it in the linear kinetic energy and the potential energy.

 

[adjacent]

Oj.Then it's $mg0.6=mg\sin\gamma x+\frac{1}{2}mv^2+\frac{Mv^2}{4}$?

 

[voko]

Like I said, $m = M$. Just choose one of them and use it everywhere! Then simplify.

 

[adjacent]

Oh.My problem is solved now.Did you have all this in mind from the beginning? or Did you solve the problem at first?
-Just curious-
Time to thank you.

 

[voko]

Oh.My problem is solved now.Did you have all this in mind from the beginning? or Did you solve the problem at first?

Let's put it this way: I knew the general character of motion (uniformly accelerated motion in a straight line - did you get that?) when I initially responded. Then as you were getting stuck at various points I made sure that I had a clear picture of the relevant details.

I suggest that you pay attention to the following: how the 3D problem was reduced to the 2D problem. That is very important in physics! Then how it was simplified further by the choice of the coordinate $x$. And finally, how conservation of energy was used instead of the tedious force/torque analysis. Reduction of dimensionality, convenient coordinates and conservation laws are the most powerful tools in physics, master them.

 

[adjacent]

Ok.Thank you.Since you are so intelligent, What's your education level?

 

[voko]

What's your education level?

Masters in maths.