CATAGORY: Other
QUESTION: a solid cylinder of radius 10cm and mass 12kg starts from rest
and rolls without slipping a distance of 6.0 m down a house roof that is
inclined 30degree (see fig.12-31)
a)what is the angular speed of the cylinder about its center as it leaves
the house roof? b)the outside wall of the house is 5.0m high.how far from
the edge of the roof does the cylinder hit the level ground?

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---

          /
        -/-    
     ---/---
        |     (1)    6.0m
       \\\\\\\\*   
       \\\\\\\\\\\*   O 
       \\\\\\\\\\\\\\*     
       \\\\\\30 deg \\\\* (2)
      --------------------*
                 _____  |
         house   |_|_|  |
                 | | |  | 5m
                 -----  |
                        |
        *********************************

We use the principles of conservation of energy but must take into
account the fact that the cylinder is rolling so that rotational
kinetic energy is stored in the cylinder.  The total energy at (1) and
(2) in the figure is given by

(1)  PE(1) + KET(1) + KER(1) = PE(2) + KET(2) + KER(2)

Where KET is the translational kinetic energy given by 

           1
(2)  KET = -mv^2
           2
           
where m is the mass and v is the velocity.

KER is the rotational kinetic energy given by

           1
(3)  KER = -Iw^2 
           2
           
Where I is the rotational moment of inertia about the cylinder axis
and w is the angular rotation rate (radians per unit time).

Since the cylinder rolls without slipping w is related to v by

(4)  v =wr or w = v/r so (3) becomes

           1  v^2
(5)  KER = -I --- 
           2  r^2
           
Initialy KER and KEt are zero.  the initial potential energy
at (1) relative to (2) is given by
 
 (6) PE(1) = mg6.00msin(30degrees) =
 
           =12kg 9.8m/s^2 6.00m sin(30 degrees)
           
           = 352.8 J
           
and 

(7) PE(2) = 0.0 J

Substitution into (1) gives

(8)  PE(1) =  KET(2) + KER(2)

or
               1          1 v(2)^2
(9) 352.8 J =  -mv(2)^2 + -I------
               2          2  r^2 
               
or 

(10) 705.6 J = (m + I/r^2) v(2)^2

Now I for a rotating cylinder is

(11) I = 1/2mr^2

thus 

(12) I/r^2 = 1/2 m

and (10) becomes

(13) 705.6 J = (m + 1/2m) v(2)^2

             = 3/2m v(2)^2
             
             = 1.5(12kg) v(2)^2
             
solving for v(2)

(14) v(2) = sqrt(705.6 J/18kg) = 6.26 m/s

      thus w = v(2)/r = (6.26m/s)/.1m) = 62.6 rad/s

the cylinder leaves the roof at 6.26 m/s at an angle of
30 degrees down from horizontal.  the vertical component
of position is given by

(15) y = 5.00m -(6.26m/s)sin(30degrees)t -1/2(9.8m/s^2)t^2
  
       = 5.00m -3.13m/s t - 4.9 m/s^2 t^2

the horizontal component is given by

(16) x = 6.26m/s cos(30degrees)t

Let y =0.0m and solve (15) for t
                       ______________________________
(17) t = 3.13m/s +/- \/(3.13m/s)-4(- 4.9 m/s^2)5.00m)
         --------------------------------------------
                  2 (-4.9 m/s^2)
                  
                  =  (-1.37s, +0.74s)
                  
The physical answer is +.  The cylinder
lands a distance of

(18)   x = 6.26m/s cos(30 degrees) 0.74s = 4.01m 

*********************
*                   *
* (a) w =62.6 rad/s *
*                   *
* (b) x = 4.01 m    *
*                   *
*********************