Curvlinear motion- ladder problem

**heatly** · September 30th 2010, 03:45 PM

A ladder is resting against a vertical wall.The top slides down in the -y direction and the bottom in the -x direction.
What is dtheta/dt (where theta is the angle btw x axis and ladder)when the top of the ladder has -2.6m/s j and the angle btw the x axis and ladder is 26deg.

I have worked out that the i component of vel is 0.634039m/s,but dont know how to calculate dtheta/dt.
Length of ladder is 0.155meters

ans=-18.663 rad/s

**Ackbeet** · September 30th 2010, 04:24 PM

This is a related rates problem. You need a general relationship between theta and y. After you get that, you can differentiate it (possibly using implicit differentiation) to get a relationship between dtheta/dt and dy/dt.

**skeeter** · September 30th 2010, 04:35 PM

Originally Posted by heatly

A ladder is resting against a vertical wall.The top slides down in the -y direction and the bottom in the -x direction.
What is dtheta/dt (where theta is the angle btw x axis and ladder)when the top of the ladder has -2.6m/s j and the angle btw the x axis and ladder is 26deg.

I have worked out that the i component of vel is 0.634039m/s,but dont know how to calculate dtheta/dt.
Length of ladder is 0.155meters

a ladder $L = 0.155 \, m$ ... for a flea circus?

$\theta = \arctan\left(\frac{y}{x}\right)$

$\displaystyle \frac{d\theta}{dt} = \frac{x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}}{x^2} \cdot \frac{x^2}{x^2+y^2} = \frac{x \cdot \frac{dy}{dt} - y \cdot \frac{dx}{dt}}{L^2}$

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