# Robotics: Linear Velocity of a Piston-Wheel System

As the flywheel rotates at a constant rate of 500 rpm, the piston moves back and forth along a horizontal line. If R = 10 cm and L = 40 cm then what is the linear velocity of the piston, when angle θ = 160°?

## Solution:

In the triangle with sides R and L, name the third side (the base) x.

R and L are constant, but x is a function of time, x(t). The desired velocity is dx/dt.

Apply the Cosine Law:

L2 = R2 + x2 - 2 R x cos θ [equation 1]

Recalling that θ and therefore x are functions of time, perform implicit differentiation:

0 = 2x dx/dt - 2R [ dx/dt cos θ- x sin (θ) dθ/dt ] [equation 2]

To use this equation to find dx/dt, we need R, θ, dθ/dt, and x.

• R = 10
• θ= 160°, cos (θ)= -0.9397, sin (θ)= 0.3420
• To find x when R= 10 and L = 40 we must solve equation [1]:

402 = 102 + x2 - 2 (10) x (-.9397)

This becomes

x2 + 18.79 x - 1500 = 0

The roots are -49.25 and 30.46, and we need the positive root.

Substituting all the values into equation [2] we get:

0 = 2(30.46) (dx/dt) - 2 (10) [ - .9397 dx/dt - 30.46 ( .3420) (52.36)]

(-60.92 - 18.79) dx/dt = 10909

dx/dt = -136.9 cm/sec

The negative sign indicates that x is getting smaller, which is correct when θ < 180°

Written by Val Sawadsky, October 7, 1997