Non-constant Forces

1. Non-constant Forces • Up until this time, we have mainly dealt with forces that are constant. These produce a uniform, constant acceleration. • Kinematic equations can be used with these forces. • However, not all forces are constant. • Forces can vary with time, velocity, and with position.

2. The Derivative • The derivative yields tangent lines and slopes • We use the derivative to go from • position -> velocity -> acceleration • The Integral • The integral yields the area under a curve • Use the integral to go from • acceleration -> velocity -> position • In these two approaches, F = ma, so methods for determining or using acceleration are also used to fine force Calculus Concepts forForces That Vary With Time

3. Refresher on the Integral • Mathematically, the integral is used to calculate a sum composed of many, many tiny parts. • An example would be with velocity: • v x t gives displacement. If the velocity is changing with time, but a very tiny time change is used to calculate a very tiny displacement, we can assume the velocity was constant during that tiny time change. • If we calculate tiny displacements this way (recalculating our velocity for each time increment), then add the tiny displacements up to get a larger displacement, we have done “integration”.

4. Velocity as a Function ofTime Displacement • Velocity can be represented as follows: • Rearrangement of this expression yields: • What this means is that we can calculate a tiny displacement dx from the velocity v at a given time times a tiny time increment dt.

5. When we sum up these tiny displacements, we use the following notation: • This notation indicates that we are summing up all the little displacements dx starting at position xo at time to until we reach a final position and time, xfand tf. The velocity v may be a function of time, and may be slightly different for one time increment dt and the next time increment. Summing the Displacements

6. Evaluating Integrals Remember that you evaluate polynomial integrals by reversing the process used in taking a polynomial derivative. This is sometimes called an “anti-derivative”. The general method for doing an anti-derivative is:

7. This is what we call a “definite integral” with “limits of integration” • First take the “anti-derivative” • Evaluate the resulting functions using the top limits, and again using the bottom limits • Finally subtract the two values to get the answer • Note: do each side of the equation separately EXAMPLE: v = 3t2 + 2t + 5 (v=dx/dt) to = 1s, tf = 2s, and xo = 4m Evaluating Definite Integrals

8. Sample problem: • Consider a force that is a function of time: • F(t) = (3.0 t – 0.5 t2)N • If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the resulting velocity and position of the particle? vf = 45 m/s xf = 50.6 m

9. Sample problem: • Consider a force that is a function of time: • F(t) = (16 t2 – 8 t + 4)N • If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the resulting change in velocity of the particle? Δv = 1.33 m/s