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This isn't really a homework question, I'm just doing it because I'm curious and the result isn't immediately obvious. I'm hoping someone else can verify the solution.

From http://www.phy.duke.edu/~hsg/physics-challenges/challenges.html

A painter is high up on a ladder, painting a house, when unfortunately the ladder starts to fall over from the vertical. Determine which is the less harmful action for the painter: to let go of the ladder right away and fall to the ground, or to hang on to the ladder all the way to the ground.

[tex]W = \int_{\theta_1}^{\theta_2} \tau \, d\theta[/tex]

[tex]K = \frac{1}{2}I \omega^2[/tex]

[tex]I_{rod} = \frac{1}{3}ML^2[/tex]

[tex]v = r\omega[/tex]

and so on

Let the painter have a mass m, the ladder have mass M and length L. [itex]\theta[/itex] is the angle the ladder makes with the wall as it falls.

I'm going to say the criterion for judging the "safety" of the fall is the man's velocity when he hits the ground--he wants a small velocity please!

1: The man stays on the ladder.

I imagined that the ladder was a uniform rod rotating about its fixed end (which is on the ground). Then, I found the work done by the force mg acting on the man at a distance L from the axis of rotation and Mg acting at a distance L/2 (center of mass of the ladder) from the axis of rotation.

[tex]W = \int_{\theta_1}^{\theta_2} \tau \, d\theta = \int_{0}^{\pi/2} mgL\sin{\theta}\,d\theta + \int_{0}^{\pi/2}Mg(L/2)\sin{\theta}\,d\theta = mgL + MgL/2[/tex]

This work goes into changing the rotational kinetic energy of the rod-painter system, whose moment of inertia is [tex]I_{rod} + I_{man} = \frac{1}{3}ML^2 + mL^2[/tex], so

[tex]W = K = \frac{1}{2}I\omega^2[/tex]

[tex]mgL + MgL/2 = \frac{1}{2}(\frac{1}{3}ML^2 + mL^2)\omega^2[/tex]

[tex]2mgL + MgL = (\frac{1}{3}ML^2 + mL^2)\omega^2[/tex]

[tex]g(2m + M) = L(\frac{1}{3}M + m)\omega^2[/tex]

[tex]\omega = \sqrt{\frac{g(2m+M)}{L(M/3+m)}}[/tex]

Now we find the velocity of the man v from the relationship [itex]v = r\omega[/itex]

[tex]v = L\sqrt{\frac{g(2m+M)}{L(M/3+m)}}[/tex]

And finally

[tex]v_{ladder} = \sqrt{\frac{gL(2m+M)}{(M/3+m)}}[/tex]

2: The man jumps off.

This one's easy...

[tex]\frac{1}{2}mv^2 = mgh[/tex]

[tex]v = \sqrt{2gh}[/tex]

So

[tex]v_{freefall} = \sqrt{2gL}[/tex]

Now we can compare these results by the ratio

[tex]\frac{v_{ladder}}{v_{freefall}} = \sqrt{\frac{gL(2m + M)/(M/3+m)}{2gL}} = \sqrt{\frac{2m+M}{2M/3 + 2m}}[/tex]

So if this ratio is greater than unity, [itex]v_{ladder}[/itex] is greater than [itex]v_{freefall}[/itex] and the man should jump off instead of stay on. And actually, it turns out that the ratio will ALWAYS be slightly greater than unity (one of the masses has to be negative to make it otherwise). I thought this result was surprising, since I expected the ladder to help slow him down somehow (don't ask me why though ;P). Does this look right?

From http://www.phy.duke.edu/~hsg/physics-challenges/challenges.html

## Homework Statement

A painter is high up on a ladder, painting a house, when unfortunately the ladder starts to fall over from the vertical. Determine which is the less harmful action for the painter: to let go of the ladder right away and fall to the ground, or to hang on to the ladder all the way to the ground.

## Homework Equations

[tex]W = \int_{\theta_1}^{\theta_2} \tau \, d\theta[/tex]

[tex]K = \frac{1}{2}I \omega^2[/tex]

[tex]I_{rod} = \frac{1}{3}ML^2[/tex]

[tex]v = r\omega[/tex]

and so on

## The Attempt at a Solution

Let the painter have a mass m, the ladder have mass M and length L. [itex]\theta[/itex] is the angle the ladder makes with the wall as it falls.

I'm going to say the criterion for judging the "safety" of the fall is the man's velocity when he hits the ground--he wants a small velocity please!

1: The man stays on the ladder.

I imagined that the ladder was a uniform rod rotating about its fixed end (which is on the ground). Then, I found the work done by the force mg acting on the man at a distance L from the axis of rotation and Mg acting at a distance L/2 (center of mass of the ladder) from the axis of rotation.

[tex]W = \int_{\theta_1}^{\theta_2} \tau \, d\theta = \int_{0}^{\pi/2} mgL\sin{\theta}\,d\theta + \int_{0}^{\pi/2}Mg(L/2)\sin{\theta}\,d\theta = mgL + MgL/2[/tex]

This work goes into changing the rotational kinetic energy of the rod-painter system, whose moment of inertia is [tex]I_{rod} + I_{man} = \frac{1}{3}ML^2 + mL^2[/tex], so

[tex]W = K = \frac{1}{2}I\omega^2[/tex]

[tex]mgL + MgL/2 = \frac{1}{2}(\frac{1}{3}ML^2 + mL^2)\omega^2[/tex]

[tex]2mgL + MgL = (\frac{1}{3}ML^2 + mL^2)\omega^2[/tex]

[tex]g(2m + M) = L(\frac{1}{3}M + m)\omega^2[/tex]

[tex]\omega = \sqrt{\frac{g(2m+M)}{L(M/3+m)}}[/tex]

Now we find the velocity of the man v from the relationship [itex]v = r\omega[/itex]

[tex]v = L\sqrt{\frac{g(2m+M)}{L(M/3+m)}}[/tex]

And finally

[tex]v_{ladder} = \sqrt{\frac{gL(2m+M)}{(M/3+m)}}[/tex]

2: The man jumps off.

This one's easy...

[tex]\frac{1}{2}mv^2 = mgh[/tex]

[tex]v = \sqrt{2gh}[/tex]

So

[tex]v_{freefall} = \sqrt{2gL}[/tex]

Now we can compare these results by the ratio

[tex]\frac{v_{ladder}}{v_{freefall}} = \sqrt{\frac{gL(2m + M)/(M/3+m)}{2gL}} = \sqrt{\frac{2m+M}{2M/3 + 2m}}[/tex]

So if this ratio is greater than unity, [itex]v_{ladder}[/itex] is greater than [itex]v_{freefall}[/itex] and the man should jump off instead of stay on. And actually, it turns out that the ratio will ALWAYS be slightly greater than unity (one of the masses has to be negative to make it otherwise). I thought this result was surprising, since I expected the ladder to help slow him down somehow (don't ask me why though ;P). Does this look right?

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