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Kinematic Equations for Uniform Acceleration

Table of Contents
Introduction
Problems
Answers

      
Introduction

Linear Angular

Constant
v = vo  +  at =  o  +  t a and
x = vot  +  1/2 at2   =  o t  +  1/2 t2   a and
v2 = vo2  +  2ax 2 =  o  +  2 a and

Observe that with constant acceleration, the angular equations are the same as the linear equations with angular terms replacing linear terms.

Table of Contents

     

Problems

1.     An automobile engine slows down from 5500. rpm to 900. rpm in 4.50 seconds.  Calculate (a) its angular acceleration, assumed uniform, and (b) the total number of revolutions the engine makes in this time.
    
2. Astronauts are tested for the stress of takeoff in a centrifuge which takes 1.00 min to turn through 30.0 complete revolutions before reaching its final speed.  Calculate (a) its angular acceleration, and (b) final speed in rpm.
   
3. A wheel with a diameter of 50.0 cm accelerates uniformly from 225 rpm to 411 rpm in 5.43 s.  How far will a point on the edge of the wheel travel in this time period?
   
4. The tires of a car make exactly 75 revolutions as the car reduces its speed from 130.0 km/h to 60.0 km/h.  The tires have a diameter of 80.0 cm.  Calculate (a) angular acceleration, and (b) time it takes car to come to a stop assuming uniform deceleration.

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Answers

1.     An automobile engine slows down from 5500. rpm to 900. rpm in 4.50 seconds.  Calculate (a) its angular acceleration, assumed uniform, and (b) the total number of revolutions the engine makes in this time.
    
   
(a)   /\ =  900. rpm - 5500. rpm  =  - 4600. rpm

5500 rev        1 min           2r
------------  x  -----------  x  ---------- =  576 rad s -1  =  o
    min              60 s           1 rev
    

- 4600 rev        1 min           2r
--------------  x  -----------  x  ---------- =  - 482 rad s -1  =  /\ 
    min                 60 s           1 rev
   

        /\         - 482 rad s -1 
= -------  =  -----------------  =  - 107 rad s -2   
          /\t            4.50 s
    

(b)    =  o t  +  1/2 t2  =  576 rad s -1 x 4.50 s  +  1/2 x ( - 107 rad s -2 ) x ( 4.50 s )2 = 1510 rad
   

                     1 rev
1510 rad  x -----------  =  240. revolutions
                     2rad
    
   

   
2. Astronauts are tested for the stress of takeoff in a centrifuge which takes 1.00 min to turn through 30.0 complete revolutions before reaching its final speed.  Calculate (a) its angular acceleration, and (b) final speed in rpm.
   
   
(a)    o  =  0   thus        =  1/2 t2

   
      
                      2rad
30.0 rev   x  ------------  =  188 rad  = 
                       1 rev      
   
   

            2           2 x 188 rad
  =  --------  =  ------------------ =  0.104 rad s -2   
            t2              ( 60.0 s )2     
  
    

(b)        =  t  =  0.104 rad s -2  x  60.0 s  =  6.24 rad s -1  
   
   

   
3. A wheel with a diameter of 50.0 cm accelerates uniformly from 225 rpm to 411 rpm in 5.43 s.  How far will a point on the edge of the wheel travel in this time period?
   
   
225 rev          1 min           2r
------------  x  -----------  x  ---------- =  23.6 rad s -1  =  o
    min              60 s           1 rev
   
    

 411 rev        1 min           2r
------------  x  -----------  x  ---------- =  43.0 rad s -1  =  f
    min              60 s           1 rev
   
    

          /\        43.0 rad/s  -  23.6 rad/s 
  =  ------  =  ---------------------------------  =  3.57 rad s -2  
           /\t                          5.43 s
   
    

=  o t  +  1/2 t2  =  23.6 rad s -1 x  5.43 s  + 1/2 x 3.57 rad s -2  x  ( 5.43 s )2  =  181 rad
   
   

   
4. The tires of a car make exactly 75 revolutions as the car reduces its speed from 130.0 km/h to 60.0 km/h.  The tires have a diameter of 80.0 cm.  Calculate (a) angular acceleration, and (b) time it takes car to come to a stop assuming uniform deceleration.
   
   
(a)          130.0 km            1 h            1000 m                                                      vo         36.1 m s -1 
               -------------  x  -----------  x  --------------  =  36.1 m/s  =  vo         o = ----- =  ----------------  =  90.3 rad s -1 
                   1 h               3600 s           1 km                                                         r            0.400 m
  
    

60.0 km            1 h            1000 m                                                                      vf         16.7 m s -1 
-------------  x  -----------  x  --------------  =  16.7 m/s  =  vf                         = ----- =  ----------------  =  41.8 rad s -1
    1 h               3600 s           1 km                                                                        r           0.400 m
   
    

                2r
75 rev x ---------   =  471 rad  = 
               1 rev
  
    

2  =  o2  +  2                    Solve for alpha.
   

   

            2  -  o2       ( 41.8 rad s -1 )2  -  ( 90.3 rad s -1 )2 
  =  ------------------  =  -----------------------------------------------  =  - 6.80 rad s -2  
               2                                2 x 471 rad       
   
   

(b)                /\                       /\          0  -  90.3 rad s -1  
            =  -------   =>    /\t  =  --------  =  ------------------------  =  13.3 seconds
                      /\t                                         - 6.80 rad s -2 
     

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