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Author: acroduster1
Time: 04/22/04 07:05 PST
This is a reply to message no. 13707 by rubinho
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Calculating trajectory of ball rolling down slope and off launch ramp
b1ueshift is exactly right.  I'll set up the equation here:

First, you have the Potential Energy of the ball at the top of the first ramp:

PE = m * g * h

Where, as b1ueshift stated, m is the ball mass, g is the acceleration due to gravity, and h is the difference in height between the top of the ramp and the lip of the exit ramp.

Next, you have your kinetic energies at the end of the launch ramp:

KEt = 1/2 * m * v^2

This is the well known average energy equation where m is the ball mass and v is the velocity.

KEr = 1/2 * I * omega^2

This is the energy stored in the rotation of the ball, where I is the moment of inertia of the ball and omega is the angular velocity of the ball.

If you ignore friction and the ball does not slip, the governing equation is PE = KEt + KEr.

Note that you have only one equation and two unknowns (v and omega).  As it turns out, you do have another equation to relate them.  If there is no slipping, the translational speed of the ball is directly related to the rotational speed.  I'll leave that one up to you.

Once you have that relationship, you replace omega in the governing equation and you will have a formula with only v as the unknown.  Solve for it and you have the speed of the ball at the instant it leaves the ramp IN THE DIRECTION OF THE RAMP.  Use that speed and angle in your ballistics equations to see how far the ball will travel.

Luckily, friction (air and surface) will probably be quite low in your case.  Including friction in these equations makes things much tougher.  For the surface friction, the total frictional force depends on the normal force.  As you travel down the ramp and reach the bottom of the valley, your normal force will increase.  You would have to integrate your normal force function to get what the total effect of that friction would be.  Similarly, as your ball slows down in air, the friction that the air provides is less.  To be accurate, more integration would be required.
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