# Circular Motion Problem

• Apr 13th 2008, 07:51 PM
shinn
Circular Motion Problem
A train is moving around a circular track of radius 800m at an uniform speed of 80km/hr. A light inelastic wire is attached to the roof of the carriage and has a small package attached at the other end.

Qns: Find the angle of inclination of the wire with the vertical. (Hint: the radius of the circle described by the package is approx. the same as that of the circular track.)

The answer at the back of the book says 3.60 degrees, but I'm having trouble to approach this question after resolving the forces on the package...

Anyone helps will be greatly appreciated,
• Apr 14th 2008, 04:08 AM
topsquark
Quote:

Originally Posted by shinn
A train is moving around a circular track of radius 800m at an uniform speed of 80km/hr. A light inelastic wire is attached to the roof of the carriage and has a small package attached at the other end.

Qns: Find the angle of inclination of the wire with the vertical. (Hint: the radius of the circle described by the package is approx. the same as that of the circular track.)

The answer at the back of the book says 3.60 degrees, but I'm having trouble to approach this question after resolving the forces on the package...

Anyone helps will be greatly appreciated,

The package is moving in a circle at constant speed, so the net force in the horizontal (radial, actually) direction will be equal to the centripetal force:
$\displaystyle \sum F_r = \frac{mv^2}{r}$

The only downward force is the weight of the package, so the vertical force is w = mg.

So the angle this makes will be
$\displaystyle \theta = tan^{-1} \left ( \frac{\frac{mv^2}{r}}{mg} \right ) = tan^{-1} \left ( \frac{v^2}{rg} \right )$

Convert v into m/s then do the calculation. I get
$\displaystyle \theta = tan^{-1} \left ( \frac{(22.222222~m/s)^2}{(800~m)(9.8~m/s^2)} \right ) = 3.604^o$

(Make sure your calculator is in "degree" mode.)

-Dan