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Tumbling Block on Inclined Plane

Introduction

Here we will use shorten the phrase "inclined plane" to the term "ramp".

I want to apologize in advance for the large amount of math that I've used here. It is necessary if we are going to see that total energy, kinetic+potential, is conserved. Without energy conservation, the animation would just be a cartoon and that is not acceptable for a serious physics animation.

In this animation, if the ramp angle is more than `pi/4`, then the square block starts to tumble down the ramp. The speed of periodic tumbling increases as the block descends the ramp since potential plus kinetic energy must be conserved. To understand the physics of this animation, you will need to learn about angular momentum which has the symbol `vec L` and the inertia to angular acceleration of a rigid body which is called the Moment of Inertia and has the symbol `I`.
Note that symbols with arrows over them are vectors which have both direction and magnitude.

Angular Momentum of a Point Particle

To proceed further we need to introduce the concept angular momentum which is usually expressed as the vector `vec L`. Angular momentum of a point particle is defined as

`vec L=m vec r times vec v \ \ \ \ (1)`

where `m` is the particle mass, `vec r` is its distance from its rotation axis, and `vec v` is its linear speed. The `times` symbol here denotes the cross product of two vectors. If the vectors are at angle `psi` with respect to each other, then the magnitude of the cross product is `|vec r| |vec v| sin psi`, where `psi` is the angle between the vectors and a vector expressed as `|vec a|` represents the magnitude of `vec a`. The direction of `vec L` is mutually perpendicular to both `vec r` and `vec v`. Similarly to linear momentum, angular momentum is constant unless an outside force, here called a torque, is applied.

Angular Momentum of a Rigid Body

To explain this best, first assume that the rigid body rotates at rate `(dphi)/dt` about one of its axes of symmetry. We would like to be able to writes its angular momentum as the following expression:

`vec L=I vec(dphi)/dt\ \ \ \ \ \ (2)`

where `I` is called the moment of inertia and `vec(dphi)/dt` is a vector that points along the rotation axis. So now we have to use equation 1 to find an expression for `I`.

Moment of Inertia `I`

Equation 1 provides the angular momentum for a point mass. Here we'll first let the value of the mass, `delta m`, be finite but very small compared with its distance, `r`, from the axis of rotation. For a rigid rotating body

`vec r times vec v=r^2 vec(dphi)/dt\ \ \ \ \ (3)`

We may rewrite equation 1 using `r` and `vec(dphi)/dt`

`vec deltaL=delta m vec r times vec v = delta mr^2 vec(dphi)/dt\ \ \ \ (4)`

So now we have to sum over the masses that are at distance `r` from the axis. To keep this simple, we'll first do the example of a circular cylinder rotating about its axis. Mass elements of the cylinder can be written:

`dm= 2pirhorl dr\ \ \ \ (5)`

where `rho` is the cylinder density (e.g. grams/`"centimeter"^3`), `dr` is the thickness of a very thin cylindrical shell, and `l` is the cylinder length. Now we have to integrate the expression in equation 1a over `dr` from 0 to `R` which is the outer radius of the cylinder:

`vec L = 2pirholvec(dphi)/dt int_0^R r^3 dr =2pirholvec(dphi)/dt R^4/4\ \ \ \ (6)`

But we know that the volume of the cylinder is:

`V= l pi R^2\ \ \ \ \ (7)`

and the mass of the cylinder is `M=rho V` so we can simplify equation 7 to:

`vec L=MR^2/2 vec (dphi)/dt=I vec (dphi)/dt (8) `

So we may conclude that our moment of inertia is

`I_("circular cylinder")=MR^2/2 \ \ \ \ (9)`

Wikipedia contains a long list of values of moments of inertia for object shapes similar to the circular cylinder. For the square shape that we use here, the moment of inertia for rotation about its center is:

`I_("square cylinder")=Mw^2/4\ \ \ \ (10)`

where `w` is the length of a side of the square.

Parallel Axis Theorem

The motion of our square tumbler is not a rotation about its center. Instead its motion can be described as rotation about its lower right corner until it tumbles an angle of `pi/2` to its next location and then begins another `pi/2` rotation about the new lower right corner. So its axis of rotation is really displaced by `b=w/sqrt(2)` from its center. The moment of inertia for this axis of rotation is larger than that about the square center. Wikipedia has a page that explains how to compute the new moment of inertia. The result for the displacement of `b` that we have here is

`I_("offset")=I_("square cylinder")+Mh^2=Mw^2/4+Mb^2`
`I_("offset")=Mw^2/4+Mw^2/2=3/4 Mw^2\ \ \ \ (11)`

Accelerating the Angular Rotation Rate

We have the equation for the angular momentum:

`vec L=I vec(dphi)/dt\ \ \ \ \(12)`

Since `I` is a constant, the equation for change in the angular momentum is:

`(dvec L)/dt=I (d^2vec phi)/dt^2\ \ \ \ \(13)`

The "force" that causes change in angular momentum has the symbol `vec tau` and is called torque. So we have the equation:

`(dvec L)/dt=I (d^2vec phi)/dt^2=vec tau\ \ \ \ (14)`

The torque is defined as the following:

`tau=vec r times vec F\ \ \ \ \(15)`

where `|vec r|` is the distance from the axis where a force `vec F` is applied and the `times` symbol denotes a cross product. The cross product is perpendicular to both `vec r` and `vec F`. The magnitude of the cross product is

`|vec r times vec F|=|r||F|sin psi \ \ \ \ (16)`

where `|vec a|` denotes the magnitude of `vec a` and `psi` is the angle between the vectors `vec r` and `vec F`. For our "hinged" square, the magnitude of `|vec r|` is the distance, `w/sqrt(2)`, from the hinge to the center of the square (which is the center of mass, CM).

As an aside, we must explain the importance of the center of mass (CM) of a rigid body. Gravity forces acting on the rigid body can be considered to be acting at the CM so, in that way gravity acts on an extended rigid body in the same way that it acts on a point particle.

And `vec r` points from the hinge to the center of the square. The force `vec F` is the force of gravity, `|vec F|=Mg`, on the mass of the square cylinder and it points downward from the CM. The value of the torque usually changes from very small at the start of each `pi/2` rotation period to maximum at its end. Therefore the angular acceleration of the block rotation rate also changes during each `pi/2` rotation period. The starting angle `psi_0` depends on the slope of the ramp, `theta`.

`psi_("start")=theta-pi/4\ \ \ \ \(17)`

and the ending angle, `psi_("end")`, after the block has tumbled `pi/2`, is `pi/2` larger than `psi_0`:

`psi_("end")=theta+pi/4 \ \ \ \ \(18)`

Between `psi_("start")` and `psi_(end)` the value of `psi` depends on the rotation angle, `phi`, as

`psi(phi)=psi_("start")+phi\ \ \ \ \(19)`

where `phi` ranges from 0 to `pi/2`.

Computing the Increase in Angular Speed `(dphi)/dt`

To compute the increase of `(dphi)/dt` during a single `pi/2` tumble, we must integrate the change in angular momentum over the range of `phi`. In order to find the increase in angular speed, `(dphi)/dt`, it is most convenient to make a change of variable name:

`omega=(dphi)/dt`

Then we have

`(domega)/dt=(d^2phi)/dt^2\ \ \ \ \(20)`

Using the chain rule of calculus, then we can re-write the expression for `(d^2phi)/dt^2`

`(d^2phi)/dt^2=(domega)/dt=omega (domega)/(dphi)\ \ \ \ (21)`

Then we have the expression for the rate of change of `omega`:

`Iomega (domega)/(dphi)=`r xx F`=Mgw/sqrt(2)sinpsi=w/sqrt(2)sin(phi+theta-pi/4)\ \ \ \ \(22)`

Both sides of this equation can be easily integrated and the result is:

`Iomega^2/2=Mgw/sqrt(2)sinpsi=Mgw/sqrt(2)(cos(theta-pi/4)-cos(phi+theta-pi/4))\ \ \ \ \(23)`

Rewriting this equation we have:

`Iomega^2/2=Mgw/sqrt(2)sinpsi=Mgw(cos(theta+sintheta-cos(phi+theta)-sin(phi+theta)))\ \ \ \ \ \(24)`

where the trigonometric identities `cos(a-b)=cosacosb+sinasinb` and cos(a+b)=cosacosb-sinasinb have been used. Note that `I omega^2/2` is the kinetic energy gained at angle `phi` during the first tumble. To compute the increase of `(dphi)/dt` during a single `pi/2` tumble, we must integrate the expression in equation 24. At the end of the first tumble, where `phi=pi/2`, the energy gained is:

`deltaKE=Iomega^2/2=Mgw(sintheta)\ \ \ \ \(25)`

Where it has been assumed that the initial `phi` and `omega` are zero. One can easily verify that the result in equation 25 is the negative of the change in potential energy over a single tumble.
For each subsequent tumble, the increase in energy is the same as equation 25. Using a computer, it is also easy to numerically iterate the value of `omega` and `phi` which is what is done in this animation.

Change in Potential Energy of the Block

The potential energy, `V`, of the block is its mass times gravity times the vertical height,`h`, of its CM above the ramp:

`V(h)=Mgh\ \ \ \ \(26)`

The CM height of the block on the plane between the beginning of the `pi/2` tumble and the end is reduced by

`deltah=w sin(theta)\ \ \ \ \ \(27)`

so the change in potential energy for a single tumble is:

`deltaV=-Mgw sin(theta)\ \ \ \ \ \(28)`

Note that `deltaV` is the negative of the kinetic energy gained (equation 25) in a single tumble. This demonstrates conservation of total energy. During the range of the `pi/2` tumble the block CM goes through height changes.
First we have the variation of CM position parallel to the ramp:

`s(phi)=(w/2+w/sqrt(2)sin(phi-pi/4)\ \ \ \ (29)`

where we've chosen to make the CM start at `s(phi)=0` when `phi=0`. The parallel motion along the ramp causes the CM to descend by

`deltay_("parallel")=-s(phi)sintheta=-(w/2+w/sqrt(2)sin(phi-pi/4))sin theta\ \ \ \ (30)`

Then we have the y component of the rise of the CM perpendicular to the ramp:

`deltay_("perpendicular")=(w/sqrt(2)cos(phi-pi/4)-w/2)cos theta`

The change in height during a `pi/2` tumble is then:

`delta y(theta,phi)=deltay_("perpendicular")+deltay_("parallel")`
`=(w/sqrt(2)cos(phi-pi/4)-w/2)cos theta-(w/2+w/sqrt(2)sin(phi-pi/4))sin theta\ \ \ \(31)`

This result can be simplified by some more trigonometric identities:

`w/sqrt(2)cos(phi-pi/4)=w/sqrt(2)cosphicospi/4+sinphisinpi/4=w/2(cosphi+sinphi)\ \ \ \ \(32a)`
`w/sqrt(2)sin(phi-pi/4)=w/sqrt(2)sinphicospi/4-cosphisinpi/4=w/2(sinphi-cosphi)\ \ \ \ (32b)`

So now we have:

`delta y(theta,phi)=w/2[(cosphi+sinphi-1)cos theta-(sinphi-cosphi+1)sin theta]\ \ \ \ (33)`

Grouping like terms in `phi` we have:

`delta y(theta,phi)=w/2[(cosphi(costheta+sintheta)+sinphi(costheta-sintheta)]\ \ \ \ \ (34)`

Then using the above in equation 33 we have:

`delta y(theta,phi)=w/2[(cosphi(costheta+sintheta)+sinphi(costheta-sintheta)-costheta-sintheta]`
`=w/2[(cosphicostheta-sinphisintheta+sinphicostheta+cosphisintheta-costheta-sintheta]\ \ \ \ \ (35)`

Again using trig identities we have:

`delta y(theta,phi)=-w/2[cos(phi+theta)+sin(phi+theta)-costheta-sintheta]\ \ \ \ \(36)`

So then the variation of the potential energy with `phi` is:

`V(phi,theta)=Mgdeltay(theta,phi)=-Mgw[cos(phi+theta)+sin(phi+theta)-costheta-sintheta]`

which, as required by conservation of energy, is the negative of the expression for the variation of kinetic energy in equation (23) At the end of the ith tumble, the height of the block CM decreases by

`delta y_i=-i w sin theta\ \ \ \ \(37)`

so this is to be added to intermediate `deltay(theta,phi)` to obtain the total change in potential energy after the nth tumble.

`V_n(theta,phi)=-Mgw(sum_(i=1)^(i=n)sin theta-Mgw[cos(phi+theta)+sin(phi+theta)-costheta-sintheta)]\ \ \ \ \(39)`

Remember, the value of `phi` resets to zero after each tumble.

Discussion of Graphics and Results

In addition to the ramp angle, `theta`, being adjustable, I've included a slider that adjusts the block initial tilt angle, `phi_0`. Therefore the criterion for the block to start moving is that `theta+phi_0 gt pi/4`. Most of the math assumes that `phi_0=0` but the animation starts fine with a non-zero `phi_0`. The tumbling block accelerates as expected from the equations above. I've included a plot of the value of `domega=(dphi)/dt` and you see that the periods of variation of angular speed get closer together as the block tumbles down the ramp.