

Wall Tipping
Name: Nicole
Status: educator
Grade: 912
Location: VT
Country: USA
Date: Fall 2013
Question:
I have been discussing the following question with one my students. "How long will it take a wall that is 2 m tall to tip over (go from vertical to horizontal)?"(I am looking for a general solution that would work for a wall of any height)
It is easy to find the angular velocity or the velocity of the wall just before it hits the ground using conservation of energy. Since I do not know the force of friction between the wall and the ground I cannot find the net force. Also I feel like them net would not be constant (since friction would not be constant) so the acceleration is not constant which means I cannot use the equations of kinematics to go from velocity to time. Another idea I considered was to think about a pendulum. I thought the time it takes the wall to go from vertical to horizontal would be the same as the time it takes the wall to go from horizontal to vertical like a physical pendulum and the time to fall would be 1/4 of the period. The problem with that idea is that 90 degrees is not a small angle so I could not make the small angle approximation and get a nice solution. I am just now thinking maybe I should go back and think about the problem without friction. Actually, I think I may have considered this but still got stuck somewhere. Any suggestion about how to approach this problem ?
Replies:
Hi Nicole,
Thanks for the question. You are correct in that the force of friction does vary since the normal force varies. Depending on how the wall is anchored to the ground, there may be slippage. Additionally, one needs to know the moment of inertia for the wall. You are correct in that the acceleration is not constant, so you cannot use the familiar kinematics equations. But you can derive a set of differential equations (which can be numerically solved using a spreadsheet) to get answer.
The use of a pendulum to model the system shows your insight. You can model the system without making the small angle approximation (sin theta = theta for small theta). You will either need to lookup the solution for the general pendulum which involves a power series solution or use a spreadsheet to solve the differential equation numerically.
I hope this helps. Please let me know if you have more questions.
Thanks
Jeff Grell
Hi Nicole,
I think your idea of looking at this as a pendulum is a good one.
If the wall is thin, uniform, rigid, and strong, and the bottom point
stays fixed as it falls over, we can treat it as physical pendulum.
As you note, however, we are nowhere near a smallangle
approximation, so we can just guess that this is not going to end well.
However, the equation of motion is pretty simple. If we write down
the torque about the fixed point at the bottom of the wall, we do not
have to worry about the force there, because a force there has a zero
moment arm to act over, and so will produce zero torque.
T = I a
where T is the torque, I is the moment of inertia, and a is the angular
acceleration. The only force is that of gravity acting on the wall's center
of mass, so we can just write the torque down:
T = m g r sin(s)
where m is the mass of the wall, r is height of the wall's center of mass,
and s is the angle of the wall. (When s=0, the wall is vertical.)
The moment of inertia of a rod about one end is
I = (1/3) m l^ = (4/3) m r^2
where l is the length, which I would rather have in terms of r.
Then T = I a yields the following
a = (3/4) (g/r) sin(s) = C sin(s)
This is a simple looking differential equation, but it is very hard to solve
analytically. Too hard for me, in fact, so I would solve it numerically. But
there are some things we can say about the motion just by inspection:
1. If s=0, then a=0, and the wall will take forever to fall over. Yes, we
already knew that, but it is comforting to see nevertheless. If we want
the time to fall, we are going to have to specify initial conditions for which
a is nonzero, or else give the wall a shove.
2. a is inversely proportional to (g/r), so a tall wall will fall more slowly than a
short wall.
It is pretty easy to set up the problem of calculating the tipping time from
given initial conditions. Suppose we start with the wall motionless and tipped
at 10 degrees, and calculate the angular acceleration, a, the angular velocity,
w, the angle, s, and the time, t, for each angular interval. If we take each
angular interval to be 1 degree, a will hardly change while the wall rotates
through the interval, and we can approximate it as constant. In that
approximation, the angular velocity, w, will increase linearly with time from
some initial value to some final value, as s increases by 1 degree.
For each interval, s1 = s0 + w0 (t1  t0) + (1/2) a0 (t1  t0)^2
We can solve for t1 = 2 sqrt((s1s0)/C sin(s0)), and this allows us to
calculate w1 = w0 + a0 t1^2. The remaining intervals will be slightly
more complicated, because w0 will not be zero for them.
If we keep this up until s=90 degrees, we will have the answer. I am not
going to keep it up, so I am just going have to remain in the dark.
Tim Mooney
Click here to return to the Engineering Archives
 
Update: November 2011

