# I The time for a tilted brick to hit the ground

#### GreenWombat

Summary
My theory/computer program gives a bigger result than my experiment. I may have the "moment of inertia" wrong.
I am interested in climate change and thereby interested in tipping points. So for the last few months, I have been investigating the dynamics of a toppling brick. I derived a differential equation of for the motion and wrote a computer program to solve this from initial conditions.

I have more detail and a diagram on
https://www.feedbackreigns.net/tipping-points/

Trouble is that my computer output suggests a longer toppling time than my experiment with a brick.

I suspect my moment of inertia may be wrong.
Inertia = M * (W * W + H * H) / 12 + M * R * R

I did a primitive experiment. I used some cut cardboard to give my red brick an initial tilt of 35 degrees from the vertical. Then I let it rotate from rest to the ground (tilt = 90 degrees), recording the sound of this on a mobile phone. It all happens so fast.
The experimental toppling time is about 0.4 or 0.5 seconds.
My calculation is out by about a factor of 6.

My program gives a much larger result.
Block Mass: kilogram = 4.06
Block Height: metre = 0.225
Block Width: metre = 0.104
Brick Depth = 0.075 is not used in my calculations
Initial Tilt from vertical: degrees = 35.0
Initial tilt speed: degrees/sec = 0.0

Radius of motion of Centre of mass: metre = 0.12394
Tipping point Angle: degrees = 24.80745
Initial Tilt past tipping point: degrees = 10.19255
Acceleration of gravity = 9.8
Moment of inertia: kg metre^2 = 0.08315
Acceleration factor = 59.30457
Maximum Tilt: degrees 90.00010
Topple Time: seconds = 2.52319
TiltStep for numerical calculation = 1 degree

Can anyone suggest why I get this discrepancy between theory and experiment?

Last edited:
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#### mfb

Mentor
What is the "acceleration factor"? It looks like all the interesting things happen there and you don't show them.

How did you calculate the moment of inertia?

Cross check: Initial acceleration of the center of mass should be of the order of sin(10.2 degree)*g = 1.7 m/s2 A bit lower as the block rotates instead of sliding, let's say 1.2 m/s2. To move a distance of 18 cm at uniform acceleration it would need sqrt(2s/a) = 0.55 s, as the acceleration increases the time will be a bit shorter. Your measured 0.4 to 0.5 seconds fit.

#### A.T.

Science Advisor
I used some cut cardboard to give my red brick an initial tilt of 35 degrees from the vertical.
If that is close to the tipping point, then even small deviations in the initial conditions can result in great differences of the duration.

• sophiecentaur

#### anorlunda

Mentor
Gold Member
I think it is common for the contact point of the brick with the table to slide horizontally as the brick tips. That can influence the results.

I think that the tipping of dominoes has been extensively modeled and compared with experiments. If you google that, you may find some studies.

#### GreenWombat

Apologies for not including more info. Here is more detail including my code for various calculations like the acceleration factor.

I realised that I used the wrong width. I rotated the brick on the edge with length 0.104 m, so this is not the width, it is the depth. The width is the 0.075 m side.

My program now gives a smaller toppling time = 2.1 seconds
Still four times greater than the rough experimental result of 0.5 sec

We now have:
M = Block Mass: kilogram = 4.06
H = Block Height: = 0.225 m
W = Block Width: m = 0.075
Brick Depth = 0.104 m (not used in calculations)
Initial Tilt from vertical: degrees = 35.0
Initial tilt speed: degrees/sec = 0.0
Tilt Step: degrees = 1.0

As the brick crashes, it rotates about the center of mass (COM), traveling in a part circle around the grounded edge. Given uniform density, the COM is halfway up the diagonal of the brick.
R = Radius of motion of COM: m = 0.11859
R = 0.5 * Sqrt(H * H + W * W)

The tipping angle is the angle between the long side of the brick and the diagonal that runs through the axis of rotation.
Tipping Angle: degrees = 18.43496
Tipping Angle = Atan(W / H) * 180 / MyPi

Initial Tilt past tipping point: degrees = 16.56504
= Initial Tilt – Tipping Angle

I = Moment of inertia: kg m^2 = 0.07613
I = M * (W * W + H * H) / 12 + M * R * R
Wikipedia gave the M * (W * W + H * H) / 12 for the moment of inertia for a brick rotated about its centre of mass (COM)

As I rotate the brick on an edge R away from the COM, I believe you have to add the M * R * R to the inertia

This moment of inertia may be incorrect.

Let g = Acceleration of gravity = 9.80000
The brick’s weight force = M * g

You can take the weight force as acting through the COM
Maximum torque = R * M * g
Let A = angular acceleration
Torque = I * A

Factor = R * M * g / Inertia = 61.98064
A = Acceleration = Factor * Sin(Tilt past tipping point)

Maximum Tilt where block hit the ground: degrees 90.0001
TiltStep for numerical calculation = 1 degree

I wrote a computer program to solve this equation for given initial conditions of Time, Tilt, and Tilt Speed. Knowing this tilt, the acceleration equation then determines the angular acceleration.

Then you can calculate what happens each time the block tilts say 1 degree further. For each step change of the tilt you know the start time, start tilt, start tilt-speed, start tilt-acceleration (via the acceleration equation), the next tilt and the next tilt-acceleration (via the acceleration equation).

From this you can calculate the average tilt-acceleration over the step. As an approximation, you can take this “average tilt acceleration” as the “constant acceleration” over the step, and so calculate the next tilt-speed
AvAccel = 0.5 * (Accel + NextAccel)
NextSpeed = Sqrt(Speed * Speed + 2 * AvAccel * TiltStep)
( Like the standard constant acceleration equation with v = final velocity, u = initial velocity, a = linear constant acceleration, t = time step.
Equation: v^2 = u^2 + 2 * a * s )

Then you calculate the time taken for the step.
TimeStep = (NextSpeed - Speed) / AvAccel
( Like standard constant acceleration equation: v = u + a * t )

Topple Time: seconds = 2.09873
Experiment gives about 0.5 seconds

I hope this is not too much detail.

To mfb, I am not sure of your “cross check” as this is angular motion through 90 – 35 = 55 degrees rather than linear motion through 18 cm.

#### metastable

Can we think of the corner of the brick as a marble rolling down a portion of half-circular “ramp” (shaped like a half circle from the side profile) with the flat edge resting on the ground? The acceleration would be calculated based on the acceleration of a marble down the ramp. In the brick the “curved ramp” is analogous to the path taken by the corner of the brick as it falls.

#### metastable

If the brick is a meter tall will it fall over in about 1 second? #### jbriggs444

Science Advisor
Homework Helper
Can we think of the corner of the brick as a marble rolling down a portion of half-circular “ramp” (shaped like a half circle from the side profile) with the flat edge resting on the ground? The acceleration would be calculated based on the acceleration of a marble down the ramp. In the brick the “curved ramp” is analogous to the path taken by the corner of the brick as it falls.
Close.

The trajectory of the marble can match the trajectory of the center of mass of the brick. The potential energy of the marble then automatically tracks with the potential energy of the brick. But the kinetic energy of the marble (as a function of velocity) does not match up with the kinetic energy of the brick (as a function of velocity).

There is a constant scaling factor between the two that will depend on the moment of inertia of the marble and the moment of inertia of the brick.

Since OP is concerned with moments of inertia already, it is probably best not to complicate matters with a scaling factor that needs to be computed using those moments.

#### metastable

My program now gives a smaller toppling time = 2.1 seconds
Still four times greater than the rough experimental result of 0.5 sec

We now have:
M = Block Mass: kilogram = 4.06
H = Block Height: = 0.225 m
$(1/2)*0.95=0.47$ $seconds$

Your measured result is fairly close to 1/2 the pendulum period for that length... #### mfb

Mentor
If that is close to the tipping point, then even small deviations in the initial conditions can result in great differences of the duration.
But the experimental result is fine, it's the theory that has a problem.
I think it is common for the contact point of the brick with the table to slide horizontally as the brick tips. That can influence the results.
That is an issue in the very last moments but shouldn't have a relevant impact on the overall time.
I wrote a computer program to solve this equation for given initial conditions of Time, Tilt, and Tilt Speed. Knowing this tilt, the acceleration equation then determines the angular acceleration.
Up to here everything looks fine.
Keep in mind that the acceleration will be in rad/s2, if you want the acceleration in degrees/s2 you need to convert it. Maybe some error was introduced here?

Instead of making steps in angles it is easier to make steps in time (and make smaller steps). Just track angle and angular velocity as function of time.
Alternatively you can calculate angular velocity as function of angle based on conservation of energy.
To mfb, I am not sure of your “cross check” as this is angular motion through 90 – 35 = 55 degrees rather than linear motion through 18 cm.
It works. It doesn't matter that the 18 cm are not in a straight line.

The pendulum calculator doesn't help here.

#### jbriggs444

Science Advisor
Homework Helper
Your measured result is fairly close to 1/2 the pendulum period for that length...
The distinction between a pendulum and an inverted pendulum involves a factor of $\sqrt{-1}$ in addition to the constant factor that is being glossed over.

#### GreenWombat

Thank you all for considering this.

It seems that I have problems with both my theory and my experiment.

I redid my experiment. This time I used slow-motion video on my mobile rather than a stopwatch. I got far smaller results. My previous stopwatch measurements were wrong. The toppling time is about 0.1 seconds, not 0.5 seconds.

This makes my theoretical toppling times of about 2 seconds in error by about a factor of twenty! After more reading, my moment of inertia seems correct. I would really like to find out where my error is.

All my filmed brick drops seemed to occur within or about 0.01 on the video timer, i.e. about one-tenth of a second. This was true for of the toppling from 35 degree-tilt for both the 0.075 width and the 0.104 m width.

For these the brick experiments, I need more accurate time measurement, perhaps timing accuracy to 100th, not to a tenth of a second.

I might find a large woodblock and try with this - but then measuring and handling weight might become a problem. My kitchen scales only go up to 5 kg.

I will look for some data on the web.

Below I also include a little more detail of one toppling time calculation.

Note: I found a surprise with my mobile slow-motion video. When I videoed my watch, I found that after 20 seconds on the watch, video timer had advanced 2.03, so 1 second was 0.105 on the video timer.

*** A.T.
AT asked if the initial tilt was close to the tipping point. A good point. The closer you get to the tipping point the more the butterfly effect comes into play.
Initial tilt = 35 degrees
Tipping point angle = 18.4 degrees
Initial tilt past tipping point = 16.6 degrees

*** Anorlunda

A mentioned domino studies. I searched for these but they mostly seem to concern multiple dominos. No joy.
One study (Koellhoffe et al) only mentions the moment of inertia around the center of mass. They put this as M * (H*H + W*W)/12 which is what I have used. They did not add the M*R*R to consider inertia around the axis of rotation. Seems strange.

**** Here I include the first and last time steps of my solution

Solve the differential equation for the Toppling Block

Toppling Red Brick
Block Mass: M kilogram = 4.06
Block Height: H meters = 0.225
Block Width: W meters = 0.075
Narrow Width
Initial Tilt from vertical (degrees) = 35.0
Initial Speed (degrees/sec) = 0.0
Tilt Step (degrees) = 1.0
Maximum Tilt (degrees) 90.00010
Acceleration of gravity: g = 9.8

Radius of motion of COM: R meters = 0.11859
R = 0.5 * Sqrt(H * H + W * W)

Tipping Point Tilt (degrees) = 18.43495
Tipping Point Tilt = Atan(W / H) * 180 / PI

Initial Tilt past tipping point (degrees) = 16.56505

Moment of inertia: I = 0.07613
I = M * (W*W + H*H) / 12 + M * R*R

Factor = R * M * g / I = 61.98064
Accel = Factor * Sin( Tilt Past Tip Point )
Initial acceleration (degree/s^2) 17.67091

AvAccel = 0.5 * ( Accel + NextAccel)
NextSpeed = Sqrt(Speed^2 + 2 * AvAccel * TiltStep

Step, Time, Tilt, Speed, Instantaneous Accel,
00 , 0.00000 , 35.00000 , 0.00000 , 17.67091 ,
01 , 0.33161 , 36.00000 , 6.03125 , 18.70504 ,
02 , 0.46784 , 37.00000 , 8.64954 , 19.73347 ,
03 , 0.57100 , 38.00000 , 10.73796 , 20.75588 ,

53 , 2.06873 , 88.00000 , 65.78597 , 58.08015 ,
54 , 2.08383 , 89.00000 , 66.66576 , 58.44898 ,
55 , 2.09873 , 90.00000 , 67.53941 , 58.80000 ,
Toppling Time (seconds) = 2.09873

*** Metastable
Metastable raised the simple pendulum calculator.
I see that the acceleration of a simple pendulum, like the toppling brick, involves a sine
a = minus amplitude * w^2 * sin(wt)
However, I am not clear about how the pendulum helps as the brick has width and so does not have a tipping point at vertical. You put the pendulum length equal to the brick height of 0.225 m and got a pendulum period of 0.95 seconds. Is this just a rough check on the expected toppling time from 35-degree tilt?

#### hutchphd

Does your program give you the correct result for the brick swinging from 3pi/4 to pi? You know the answer to that from the linearized physical pendulum solution.
Also you realize that the solution here has to be independent of the mass of the block and the resulting time should scale with √length /g
If your calculation does not do this then it is incorrect. These might provide useful clues.

#### mfb

Mentor
0.1 seconds is too short. In free-fall a block would just fall down 5 cm. Your center of mass moves downwards by more than 5 cm, and it is not in free fall.

You don't need a scale. The weight of the block doesn't matter, it cancels out in the calculation as long as the distribution is uniform. Torque and angular momentum are both proportional to it.
Initial acceleration (degree/s^2) 17.67091
This is what I suspected in the last post: This should be rad/s2. Fixing it will shorten your time by a factor sqrt(180/pi) ~ 8. From 2.1 seconds to ~0.25 seconds.
NextSpeed = Sqrt(Speed^2 + 2 * AvAccel * TiltStep
Same problem here. You are mixing rad and degrees.

#### GreenWombat

I worked out that I was mixing radians and degrees - and then saw your post.
MFB, you are quite right. Thanks.

I have made changes and now as a check, the lost potential energy = the acquired kinetic energy.
For width 0.075 m
initial tile = 35 degrees
Tipping time now predicted = 0.277 seconds

What you say about the free falling also seems spot on. Thanks.
the center of mass drops from h = R Cos(16.6) = 0.1136 m to h = 0.5 * W = 0.038 m
total drop = 0.0756 m
a free falling body would take 0.12 sec to fall that distance.
So my video measurments of about 0.1 sec seem on the small side.

That is good news for the above prediction of 0.277 sec.

If only I had some good experimental data.

• mfb

#### mfb

Mentor
Good. Now with the prediction side fixed let's look at the experimental side. How do you determine the starting time of the fall? How accurate is the starting angle?

Here is how I would set up the experiment, the quick retraction of the element stopping the block gives a good starting time measurement: #### GreenWombat

Thanks for your help. I think I have the theory right now.
The output shows that the potential energy lost in the toppling = the kinetic energy gained. That gives me confidence.

Thanks for the good ideas for the experiment – but I will leave that for now.
Here are the edited results from my program now.

Solve the differential equation for the Toppling Block

Output in degrees, deg/sec & deg/sec^2
Calculations done in radians, rad/sec & rad/sec^2

Toppling Red Brick
Block Mass: M kilogram = 4.06
Block Height: H meters = 0.225
Block Width: W meters = 0.07500 Note: Narrow Width
Initial tilt from vertical (degrees) = 35.0
Initial Speed (degrees/sec) = 0.0
Tilt Step (degrees) = 1.0
Maximum Tilt (degrees) 90.00010
Acceleration of gravity: g = 9.8

Radius of rotation of COM: R meters = 0.11859
R = 0.5 * Sqrt(H * H + W * W)

Tipping Point Tilt (degrees) = 18.43495
Tipping Point Tilt Deg = Atan(W / H) * 180 / PI

Initial Tilt past tipping point (degrees) = 16.56505

Moment of inertia: I = 0.07613
I = M * (W*W + H*H) / 12 + M * R*R

Factor = R * M * g / I = 61.98064
Accel = Factor * Sin( Tilt Past Tip Point )
Initial acceleration (degrees/s^2) 1012.46881

AvAccel = 0.5 * ( Accel + NextAccel)
NextSpeed = Sqrt(Speed^2 + 2 * AvAccel * TiltStep

Potential energy lost by toppling = 3.03040
Lost PE = M * g * (R * Cos(TiltPastTP) - 0.5 * W

Step, Time, Tilt, Speed, Accel, Kinetic Energy
00 , 0.00000 , 35.00000 , 0.00000 , 1012.46881 , 0.00000
01 , 0.04381 , 36.00000 , 45.65291 , 1071.71983 , 0.02417
02 , 0.06181 , 37.00000 , 65.47177 , 1130.64440 , 0.04970

53 , 0.27330 , 88.00000 , 497.96018 , 3327.74750 , 2.87502
54 , 0.27530 , 89.00000 , 504.61963 , 3348.87972 , 2.95243
55 , 0.27727 , 90.00000 , 511.23267 , 3368.99184 , 3.03033

Toppling Time (seconds) = 0.27727
final KE = 3.03003 = Lost PE

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