Cylinder on an Inclined Plane

Date: 12/17/97 at 08:18:10
From: Chris Fieldhouse
Subject: Time taken for a uniform cylinder to travel along an inclined 
plane, from rest, accelerated purely by gravity

Please help me.

I am trying to determine if it is possible to arrive at a formula to 
determine the time it will take (in seconds) for a uniform cylinder 
(having outside + possibly inside diameter) to travel a length (in 
meters) along an incline of varying degrees.

Having conducted several tests and using Buckingham's PI theorem, I 
have arrived at:

t = (13.126 L^0.5)
     --------------------
     ( g^0.5 a^0.5)

Where:
L = length of incline in meters
g = gravitational accelaration (9.81)
a = angle of incline (0 taken as horizontal)

This seems to work, but I can't understand why it gives a wrong answer 
for 90 degrees (i.e. dropping the cylinder). Surely, my formula 
doesn't take into account any other factors of energy loss (i.e. 
rotational acceleration, inertia, etc.).

Chris

Date: 12/17/97 at 15:56:31
From: Doctor Anthony
Subject: Re: Time taken for a uniform cylinder to travel along an 
inclined plane, from rest, accelerated purely by gravity

I will give you the conventional treatment for a cylinder rolling down 
a rough plane.  It will include the moment of inertia of the cylinder 
as being an important component in the calculation.

x is the distance measured down the slope, y the outward normal to the 
slope and theta the angle turned through by a radius of the cylinder.  
I is the moment of inertia of the cylinder about its axis, M the mass 
of the cylinder, a its radius, F the force of friction between the 
cylinder and the plane acting up the slope, R the normal reaction of 
the plane on the cylinder.  

Take 'A' as the angle of slope.

The theorems on rigid dynamics that we use are the following:

(1) The centre of mass moves as if all the forces acted at it.

(2) Rotation about the centre of mass is the same as if the centre of 
    mass were a fixed point and the forces acted in their actual 
    positions about this point.

Considering first the motion of the centre of mass:

  M.d^2(x)/dt^2 = Mg.sin(A) - F      (i)

  M.d^2(y)/dt^2 = R - Mg.cos(A)      (ii) 

 For motion about the centre of mass we have:

    I.d^2(theta)/dt^2 = F.a          (iii)

There is no acceleration normal to the plane so d^2(y)/dt^2 = 0  and 

      R = Mg.cos(A)

Since there is no slipping   x = a.theta

                   d^2(x)/dt^2 = a.d^2(theta)/dt^2     (iv)

From (iii) and (iv) we get

          F = (I/a)(1/a)d^2(x)/dt^2

            = (I/a^2)d^2(x)/dt^2

From(i)   (M + I/a^2)d^2(x)/dt^2 = Mg.sin(A)

                                    Mg
                 d^2(x)/dt^2 =   ---------- sin(A)
                                 M + I/a^2

So now we have the acceleration, and can use the constant acceleration 
formula for the motion down the plane.

If the slope is vertical, the term I/a^2 would no longer apply, since 
friction would be zero with zero pressure between cylinder and plane.  
The above analysis has assumed sufficient friction for the cylinder to 
rotate in sympathy with its linear displacement down the plane.  It 
must not be used once the situation no longer applies.

The time to move a distance L can be found from the usual formula

    L = ut + (1/2) x accel x t^2

If initial velocity u = 0 then:

      L = (1/2) x accel x t^2

So              2L
    t = sqrt[---------]
               accel


             2L(M+I/a^2)
    t = sqrt[-----------]
              Mg.sin(A) 


-Doctor Anthony,  The Math Forum
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