Originally Posted by **ChipB** You asked about a ball rolling down a ramp, but to simplify let's consider a block sliding down the ramp. The reason is that a rolling ball introduces issues having to do with angular momentum that complicates the analysis. if you truly mean a rolling ball then the analysis requires that you provide the ball's diameter and mass, and whetger it's solid or hollow.
The sum of forces that act in the direction of the ramp is:
F = mg sin(theta) - umg cos(theta)
where u = coefficient of friction. Hence the acceleration of the block is:
a = mg sin(theta) - ug cos(theta)
The time 'T' it takes to slide to the end of the ramp, starting with zero velocity, can be found from:
D = (1/2)a T^2.
The average velocity will then be the distance travelled divided by time:
V_avg = D/T
Hope this helps. |

This was a very helpful post but I have a problem which expands on this one.

A block stands at rest and slides down a plane inclined at an angle theta with a coefficient of kinetic friction u (mu) . What should theta be so that the block travels a given horizontal distance in the minimum amount of time?

I got the above with the addition of x = D cos theta.

Therefore t = ( 2 D cos theta/ (g sin theta - u g cos theta) )^(1/2)

Simplifying

t = (2 D/ g)^(1/2) (cos theta)^(1/2) (sin theta - u cos theta)^(-1/2)

Now take dt/d theta and set to zero

(2 D/ g)^(1/2) *

[ (1/2) (-sin theta)^(-1/2) * (sin theta - u cos theta)^(-1/2)]

-[ (cos theta)^(1/2) * (cos theta + u sin theta)^(-3/2)]

setting this to zero eliminates the constant term but I am left with an imaginary part. Can the imaginary part just be eliminated? Did I over-complicate this?

Thanks for any help.