Motion of an equalateral triangle, differenciating

printz0r · #1 Oct 24, 2009, 05:01 PM

I am doing a advanced physics problem where I have an equilateral triangle with all of its corners connected to a wheel. The triangle is sitting up right on 2 wheels. It has sides of .2m The wheel to the right is being lifted at a constant rate of .3 m/s. Which causes the wheel on the left to move in the x direction only. The question asks to solve for the angular velocity of the triangle.

The book starts the problem saying that x^2 +y^2 = b^2. They differentiate which leads to x(dx/dt) + y(dy/dt) = 0. How does that differential work, + how do you solve the rest of the problem?

Unknown008 · Oct 24, 2009, 11:40 PM

I don't quite understand what is happening, but I know how they got that differential equation.

$x^2 + y^2 = b^2$

Differentiate with respect to t:

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$

I guess b is a constant here.

Factorise 2:

$2(x \frac{dx}{dt} + y \frac{dy}{dt}) = 0$

That means;

$(x \frac{dx}{dt} + y \frac{dy}{dt}) = \frac02$

$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$

Could you try posting a picture?