- #1

Hamiltonian

- 296

- 193

- Homework Statement
- A uniform rod of mass M and length is placed against a smooth wall. the coefficient of kinetic friction between the floor and the ladder is cot(phi)/4

where phi the initial angle of inclination of the ladder. Find the time taken by the ladder to slip and become horizontal.

- Relevant Equations
- T = I##\alpha##

F = ma

$$ R - f = m\ddot x$$ $$N - mg = m\ddot y$$

were N and R are the normal reactions from the smooth wall and rough ground.

and f is the friction provided by the ground.

$$ f = \mu N = cot(\phi)N/4$$

i tried to formulate a constraint relation between ##\ddot x## and ##\ddot y## so that I could solve the above equations

since $$x^2 + y^2 = L^2$$

$$ x \frac {dx} {dt} = y \frac {dy} {dt}$$

$$ \frac {dx}{dt} = tan(\theta) \frac {dy} {dt}$$

$$ \ddot x = \ddot y tan(\theta) + \dot y \dot \theta sec^2(\theta)$$

the problem here is the ##\dot y##!

also ##\theta## is the angle made by the ladder at some time during its motion(##\theta## changes with time)

Taking the torques about the center of mass of the rod gives me

$$ (Ncos(\theta) - fsin(\theta) - Rsin(\theta))L/2 = I \frac {d(\omega)} {dt}$$

$$ \frac{L}{2I}\int_0^t t\,dt = \int_\phi^0 \frac {d\theta} {(Ncos(\theta) - fsin(\theta) - Rsin(\theta)}$$

here I end up with an integral that I am unable to solve!

can someone tell me if I am doing something conceptually wrong here or if I have made some major errors in my process?

also would there be an easier way to solve this question as my teacher called it "trivial" and I can't seem to get my head around it.

thanks:)

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