# Dynamics, ball in motion

1. Apr 30, 2006

### danielI

Hey!

In a test of resistance to motion in an oil bath, a small steel ball of mass m is released from rest at the surface. If the resistance to motion is given by R = kv where k is a constant, derive an expression for the depth h required for the ball to reach a velocity v.

$$mg - kv = ma = mv\frac{dv}{ds}$$

$$\int_0^hds = \int_0^{v_t}\frac{mv}{mg-kv}dv$$

$$h = -\frac mk\int_0^{v_t}\frac{-kv}{mg-kv}dv$$

$$h = -\frac mk\int_0^{v_t}\frac{mg-kv-mg}{mg-kv}dv$$

$$h = -\frac mk\left[v_t - \int_0^{v_t}\frac{mg}{mg-kv}dv\right]$$

$$h = -\frac mk\left[v + \frac{mg}{k}\int_0^{v_t}\frac{-k}{mg-kv}dv\right]$$

$$h = -\frac mk\left[v + \frac{mg}{k}\ln(mg-kv_t)\right]$$

$$h = -\frac {m^2g}{k^2}}\ln(mg-kv_t)-\frac{mv_t}{k}$$

$$h = \frac {m^2g}{k^2}}\ln\left(\frac{1}{mg-kv_t}\right)-\frac{mv_t}{k}$$

$$h = \frac {m^2g}{k^2}}\ln\left(\frac{1}{1-(kv_t)/(mg)}\right)-\frac{mv_t}{k}$$

I've calculated this two more times and got the same result.

Why is this so?

Last edited: Apr 30, 2006
2. May 1, 2006

What happened to your constant of integration?
Once you have substituted the initial conditions and solved for the constant of integration, you should get the correct answer.