Physics Help Forum Average Velocity and Angle of Inclination of a Slope

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Feb 6th 2014, 03:33 AM #1 Junior Member   Join Date: Feb 2014 Posts: 2 Average Velocity and Angle of Inclination of a Slope What is the relationship between a ball rolling down a slope (ramp) and its average velocity. I know we are supposed to use: v^2 = ut + at/2 and F = mgsin(degree). But assume initial velocity is zero. Also if I do an experiment and I get results (make a graph between average velocity and cosine or sine of angle) how can I find the frictional force found within the slope? I'll know there is an error (friction) as the graph is linear but nor propotional. Thanks!
 Feb 6th 2014, 08:03 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 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.
Mar 8th 2014, 09:23 AM   #3
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Join Date: Feb 2013
Location: Greater St. Louis area
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Additonal requirements for block on inclined plane

 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.

 Mar 8th 2014, 09:40 AM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 Given that 'a' is measured in the direction of the plane, the horizontal component of it is: a cos(theta). So the horizontal distance travelled in time t is: x=cos(theta) (mg sin(theta)-mu g cos(theta)) t^2 To maximize x in time t you want to maximize the quantity cos(theta) (mg sin(theta)-mu g cos(theta)) Try taking the derivative with respect to theta and set it to zero - what do you get?
Mar 8th 2014, 10:53 AM   #5
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Join Date: Feb 2013
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 Originally Posted by ChipB Given that 'a' is measured in the direction of the plane, the horizontal component of it is: a cos(theta). So the horizontal distance travelled in time t is: x=cos(theta) (mg sin(theta)-mu g cos(theta)) t^2 To maximize x in time t you want to maximize the quantity cos(theta) (mg sin(theta)-mu g cos(theta)) Try taking the derivative with respect to theta and set it to zero - what do you get?

Thank you so much Chip. That does give me the correct answer!

I just wanted to clarify for anyone else reading this:

I believe that your equation should eliminate the m because with the m gives force not acceleration and the x term should include 1/2 for
x= 1/2 a t^2.

But I got it and must review where I went wrong. Thanks again!

Today's quote by Albert: "There are only two things that are infinite - the universe and human stupidity...and I'm not sure about the former!"

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