Dynamics Help

May 2008
1
0
A mass m is released from a height h and falls to the ground under gravity. It is subject to air resistance which is of magnitude k times its speed. Write down the equation of motion.

Explain what is meant by terminal velocity and determine its value in this case.

Show that the mass hits the ground after a time T given by

T = (kh/mg) + (m/k)(1-exp(-kT/m))

Assumig that k << m and by expanding the right hand side in powers of k/m up to order (k/m)^2, obtain an expression for T which is correct up to and including terms of order k/m.




I've managed to get an equation of motion and a terminal velocity, but I can't derive that middle equation. HELP! :p
 

topsquark

Forum Staff
Apr 2008
3,106
659
On the dance floor, baby!
A mass m is released from a height h and falls to the ground under gravity. It is subject to air resistance which is of magnitude k times its speed. Write down the equation of motion.

Explain what is meant by terminal velocity and determine its value in this case.

Show that the mass hits the ground after a time T given by

T = (kh/mg) + (m/k)(1-exp(-kT/m))

Assumig that k << m and by expanding the right hand side in powers of k/m up to order (k/m)^2, obtain an expression for T which is correct up to and including terms of order k/m.




I've managed to get an equation of motion and a terminal velocity, but I can't derive that middle equation. HELP! :p
Do you mean the T equation?

Let's do it from the start. I'll give only intermediate steps.

\(\displaystyle m \frac{dv}{dt} = -mg + kv\)

\(\displaystyle \int_0^v \frac{dv}{kv - mg} = \int_0^t \frac{dt}{m}\)

\(\displaystyle v(t) = \frac{mg}{k} - \frac{mg}{k} e^{kt/m}\)

\(\displaystyle \int_h^0 dx = \int_0^T \left ( \frac{mg}{k} - \frac{mg}{k} e^{kt/m} \right ) ~dt\)

\(\displaystyle -h = \frac{mg}{k}T - \frac{m^2g}{k^2}e^{kt/g}\)

which leads to the desired equation.

-Dan