Ladder falling back puzzle.

Originally posted by sonhouse
I didn't say I totally ignored angular acceleration, just for the first couple of degrees of movement of the ladder. I stated there would not be much in the way of angular acceleration in that time frame, I think the velocity does not follow a freefall since it is clear there is not much acceleration in the first few degrees. Later, 10 degrees, 20 degrees, and put the ladder back in place to continue the attack on my mythological middle age castle.

well after some thought, the real ( or should I say more precise ) model for the situation of the man riding the radius down to the ground would be modeled by a second order non-linear differential eqyation of the form

-.7357*cos(A) = d^2(A)/dt^2

unluckily for me,... this is not easy to solve for A(t)

So Im going to "buy" your "ignore angular acceleration approach for the first 10 degrees or so" for that situation, but I don't buy your result.

so lets ignore angular acceleration for 10 degrees as you said

since W is constant and equal to V/R

W = .25(m/s)/15(m)=.01666...units are (rad/s)

Use the kinematic relation

A = Ao + W*t

from this equation we find the time it takes for the man to move through the angle of 10 degrees by solving the above equation for time (t) units in seconds (s)

90 (deg) = pi/2 (rad) = Ao

80(deg) = 4*pi/9 (rad) = A

W= -.01667(rad/s) The sign assumed negative by convention

solve for t

t= 10.5 (s)

from this new starting location in space we will do as you say and analyze the man as if he were in free fall

his equation of motion from this new point must be described as

H = Ho + Vo*t +1/2*g*t^2

So when the man hits the ground his final height is 0

so H=0

Using a little trigonometry his initial height "Ho" is given by

Ho = 15*sin(80 deg) = 14.77 (m)

His initial velocity is given by

Vo = .25(m/s) * cos(80 deg) = .0434 (m/s)

so we solve the following equation for (t)

0 = 14.77 + .0434*t - 4.905*t^2

t=1.74 (s)

That gives him a total elapsed fall time of

10.5(s) + 1.74(s) = 12.24(s)

I hope you can see not to bet the farm on this solution???

Originally posted by joe shmo
well after some thought, the real ( or should I say more precise ) model for the situation of the man riding the radius down to the ground would be modeled by a second order non-linear differential eqyation of the form

-.7357*cos(A) = d^2(A)/dt^2

unluckily for me,... this is not easy to solve for A(t)

So Im going to "buy" your "ignore angular accel ...[text shortened]... 0.5(s) + 1.74(s) = 12.24(s)

I hope you can see not to bet the farm on this solution???

The only thing I see wrong with that solution is the angle would be a lot less than 10 degrees, more like less than 2 degrees of movement and therefore that much less time, however, since he has fast reaction times, he goes into freefall when he sees the pushrod hit the ladder so starts his freefall when the ladder is less than 1 degree off vertical. He wasn't hanging on to the ladder and then falling at a later time. He recognizes what would happen if he just hangs on to the ladder and lets go immediately. I think that will take his fall a lot closer to 1.74 and change seconds.

Originally posted by sonhouse
The only thing I see wrong with that solution is the angle would be a lot less than 10 degrees, more like less than 2 degrees of movement and therefore that much less time, however, since he has fast reaction times, he goes into freefall when he sees the pushrod hit the ladder so starts his freefall when the ladder is less than 1 degree off vertical. He was ...[text shortened]... d lets go immediately. I think that will take his fall a lot closer to 1.74 and change seconds.

If you let that angle be as small as you like, situation1 become indistuingishable from situation 2, i dont like it. however good you may think your intuition is, its not good science, its a guess, nothing more nothing less.

Originally posted by joe shmo
If you let that angle be as small as you like, situation1 become indistuingishable from situation 2, i dont like it. however good you may think your intuition is, its not good science, its a guess, nothing more nothing less.

That wasn't intuition, I calculated the degree of motion, I thought I went through that.
It seems clear when the ladder goes from vertical to horizontal, the first couple of degrees would have almost no contribution from gravity, the ladder would have moved only about 1 foot or about 350 mm, and you can easily see the full circumference would be about 95 meters for a radius of 15 meters, so one degree would be would be about 261 mm so 350 mm would be about 1.4 degrees. Surely you wouldn't argue gravity could possibly be much of a mover at that level.

Originally posted by sonhouse
That wasn't intuition, I calculated the degree of motion, I thought I went through that.
It seems clear when the ladder goes from vertical to horizontal, the first couple of degrees would have almost no contribution from gravity, the ladder would have moved only about 1 foot or about 350 mm, and you can easily see the full circumference would be about 95 m ...[text shortened]... .4 degrees. Surely you wouldn't argue gravity could possibly be much of a mover at that level.

your right, gravity doesnt accelerate the system much at those small angles (although it does accelerate it), but full accleration ( ie 1g) doesn't occurr until just as the system hits the ground. How did you account for that, because it doesnt appear to me that you have?

Originally posted by joe shmo
your right, gravity doesnt accelerate the system much at those small angles (although it does accelerate it), but full accleration ( ie 1g) doesn't occurr until just as the system hits the ground. How did you account for that, because it doesnt appear to me that you have?

All I was interested in was the time it took for the dude to freefall vs the time it took for the ladder to fall. Remember the first post? It is clear after all that work🙂 that the dude hits the ground first. What do you think of the idea the guy would be pushed a bit away from the wall by the amount the ladder actually moves since it would be moving some, so pushing against the guy as he falls.