MHB Useful Derivation for Labs Involving Rolling Balls Down an Inclined Plane

[malawi_glenn]

I just wanted to solve the problem somehow

I would say mission accomplished.

The title of the thread is "Useful derivation". And this what we are arguing about here. And what didactical advantages this integration method is compared to differentiation method + SUVAT.

Imo, it is more like "Somewhat cumbersome derivation", because the differentiation method is quicker and that its easier to differentiate than integrate.

 

[Ackbach]

The title of the thread is "Useful derivation". And this what we are arguing about here. And what didactical advantages this integration method is compared to differentiation method + SUVAT.

Imo, it is more like "Somewhat cumbersome derivation", because the differentiation method is quicker and that its easier to differentiate than integrate.

The derivation is useful in the lab. That is, it is useful for the derivation to have been accomplished so that, in the lab, you get far less experimental error. The method of deriving the result is basically not the point of the OP at all. I derive it so that the student has confidence in the result, but that's all. The point of the OP is that having the correct formula reduces your experimental error.

 

[vanhees71]

I'd also say for 1D problems the most simple way to derive the equation of motion is to use energy conservation and derive the corresponding equation once by time. In your case you have
$$\frac{m}{2} \dot{x}^2 + \frac{I}{2 R^2} \dot{x}^2 - m g x \sin \theta= \frac{M_{\text{eff}}}{2} \dot{x}^2 - m g x \sin \theta=E=\text{const}, \quad M_{\text{eff}}=m+\frac{I}{R^2}=\frac{7}{5}m.$$
Taking the time derivative gives
$$\dot{x} (M_{\text{eff}} \ddot{x}-m g \sin \theta)=0.$$
Now you can have either $x=\text{const}$, but it's obvious that this is not a solution, or
$$M_{\text{eff}} \ddot{x}=m g \sin \theta \; \Rightarrow \; x(t)=x_0 + v_0 t +\frac{m g}{2M_{\text{eff}}} t^2.$$

 

[Ackbach]

I have re-framed the OP to focus on what was the main event, in my mind. The main event is the lab, and trying to reduce experimental error. My apologies if that wasn't clear. I'm not interested in continuing any discussions about my method of solving the theoretical problem. I would argue it's a valid way, and that's what's needed.

 

[vanhees71]

Well, but also in a lab the theoretical foundation should be made as clear as possible, and an overly complicated derivation for sure doesn't help the understanding of the physics. When I was a student, we had to prepare the theory underlying the experiment in the lab beforehand, and for mechanics labs I always used Hamilton's principle to prepare. Once the "colloquium" checking whether we were prepared took only 5 minutes, because instead of being a complicated derivation with the Lagrangian it were 3 lines. A good theoretical methodology always helps also the experiments!

 

[PeroK]

My preferred solution would be:

We have an accelerating force of $F = mg\sin \theta$ down the slope. The work done by this force over a distance $x$ is $Fx$ and this must equal the total linear and rotational KE. Hence$$mgx \sin \theta = \frac 1 2mv^2 + \frac 1 2 Iw^2 = \frac 1 2 mv^2 + \frac 1 2 kmr^2\frac {v^2}{r^2} = \frac 1 2 mv^2(1 +k)$$This gives us $$v^2 = 2\frac{g\sin \theta}{1 +k}x$$And using $$v^2 = 2ax$$gives$$a = \frac{g\sin \theta}{1 +k}$$Finally, taking $k = \frac 2 5$ and using $$x = \frac 1 2 at^2$$ we get$$x = \frac{5g\sin \theta }{14}\ t^2$$And that only uses what I believe to be high school physics and maths.

 

[vanhees71]

This has the disadvantage that it assumes $v_0=0$, i.e., it's not the general solution. Instead of forcing people memorize SUVAT equations one should teach them how to derive them from the general Newtonian laws of motion, and this indeed can be done with high-school mathematics only as demonstrated in my previous posting above.

 

[PeroK]

This has the disadvantage that it assumes $v_0=0$, i.e., it's not the general solution.

A change of reference frame takes care of that.

Instead of forcing people memorize SUVAT equations one should teach them how to derive them from the general Newtonian laws of motion, and this indeed can be done with high-school mathematics only as demonstrated in my previous posting above.

A physics student needs to be able remember some things. Do you advocate deriving the equations of electromagnetism from first principles every time you need them?

 

[Ackbach]

Different people have different ideas on what is clear, what is straight-forward, etc. Anyone should feel free to use their own derivation in place of mine.

 

[PeroK]

PS if the student can't "memorise" that $KE = \frac 1 2 mv^2$ etc. then physics will be a long haul!

 

[vanhees71]

A change of reference frame takes care of that.

A physics student needs to be able remember some things. Do you advocate deriving the equations of electromagnetism from first principles every time you need them?

Of course not, but just rote learning some equations without understanding them is also contrary to what we really want, right?

 

[malawi_glenn]

The derivation is useful in the lab. That is, it is useful for the derivation to have been accomplished so that, in the lab, you get far less experimental error. The method of deriving the result is basically not the point of the OP at all. I derive it so that the student has confidence in the result, but that's all. The point of the OP is that having the correct formula reduces your experimental error.

The result is useful of course.

This has the disadvantage that it assumes $v_0=0$, i.e., it's not the general solution. Instead of forcing people memorize SUVAT equations one should teach them how to derive them from the general Newtonian laws of motion, and this indeed can be done with high-school mathematics only as demonstrated in my previous posting above.

Will be a more cumbersome lab to conduct, you need an additional setup of speed measurement if you do not release the cylinder/ball/thing at rest. I hate photogates!

 

[PeroK]

Of course not, but just rote learning some equations without understanding them is also contrary to what we really want, right?

Yes, but just because the student remembers something doesn't mean they don't understand what they are doing.

 

[PeroK]

For example, I remember that $v^2-u^2 = 2as,$ is really useful in some cases that may otherwise be problematic. I remember that that is a formula not to forget! Memory is fundamental to thinking IMHO.

 

[malawi_glenn]

For example, I remember that $v^2-u^2 = 2as,$ is really useful in some cases that may otherwise be problematic. I remember that that is a formula not to forget! Memory is fundamental to thinking IMHO.

I refer to this forumlas as the "ass" formula, helps the students remember it

 

[PeroK]

I refer to this forumlas as the "ass" formula, helps the students remember it

In case I forget it a derivation is:
$$v^2-u^2 = (v - u)(v + u)= at(v +u)= 2at\frac{v +u} 2 = 2a(tv_{avg}) = 2as$$It's simpler, of course, once you know about energy!

 

[vela]

I suspect @malawi_glenn prefers the derivation
$$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx} \quad \Rightarrow \quad \int_{v_0}^v v\,dv = \int_{x_0}^x a\,dx \quad \Rightarrow \quad v^2 - v_0^2 = 2a(x-x_0)$$

 

[malawi_glenn]

Nah as long as you know that a is constant, peroK derivation is preferrable since it does not require calculus. My students however prefers your derivation because they think using (u+v)/2 = vtavg is magic

Earlier, I did teach that if we have $v^2 = kx$ then we can just use $v^2 - v_0{}^2 = 2a \Delta x$ formula to figure out $a$. But students did not perform well, and there was lot of debate why we could do that, since $v^2 - v_0{}^2 = 2a \Delta x$ is only valid for constant acceleration and it was not evident for them why $v^2 = kx$ is such case.

Same with the $s = v_0t + \dfrac{at^2}{2}$ formula, before they learned how to differentiate and integrate, this formula was magic to them. Counting areas in a vt-diagram they thought was very tricky.

 

[vanhees71]

To avoid calculus is misleading since calculus makes the subject easier. That's why nowadays we don't use the methods of Newton's principia anymore but Euler's formulation in terms of calculus.

 

[malawi_glenn]

To avoid calculus is misleading since calculus makes the subject easier. That's why nowadays we don't use the methods of Newton's principia anymore but Euler's formulation in terms of calculus.

Yeah but in ours, and most schools, syllabus, physics courses starts prior to calculus based math classes...

 

[PeroK]

In any case, basic algebra, geometry and trigonometry must come before calculus.