Site hosted by Angelfire.com: Build your free website today!
Main Desktop Lessons Desktop Review Desktop Lab Desktop Chart Desktop

      
Rotational Kinetic Energy

Table of Contents
Introduction
Problems
Answers

      
Introduction

A body rotating about an axis has Rotational Kinetic Energy.

KE =  ( mv2 )                        Each particle making up the body has Kinetic Energy. 

    
v = r   
   

KE =  ( mr2 2 )   =   ( mr2 ) 2       is the same for all the particles. 
   

I =  ( mr2 )    
   

KE  =  I 2                      Observe the similarity to KE = mv2  for Translational Motion.  

   
Rolling objects have both Translational Kinetic Energy and Rotational Kinetic Energy.

KETotal  =   mv2  +  I 2             Translational and rotation velocity about center of mass 

Table of Contents

     

Problems

1.      A 7.27 kg bowling ball with a radius of 9.00 cm rolls without slipping down a lane at 4.55 m/s.  Calculate the total kinetic energy.    Moments of Inertia
   
2. Calculate the Kinetic Energy of the Earth with respect to the Sun.  massEarth = 6.0 x 10 24 kg   
radiusEarth = 6.4 x 10 6 m         Earth is 1.5 x 10 8 km from the Sun.
   
3. A 1640 kg merry-go-round has a radius of gyration of 8.20 m.  How much net work is required to accelerate from rest to a period of 8.00 s?
   
4. A 1100. kg has four tires each with a mass of 30.0 kg that includes the mass of the wheel.  The radius of the tire is 80.0 cm.  The radius of gyration of the tire and wheel is 29.8 cm.  Calculate (a) the total kinetic energy of the car when traveling 115 km/h, and (b) the fraction of kinetic energy in the tires and wheels.  (c) If the car initially at rest is pulled by a tow truck with a force of 2500. N, what is the acceleration of the car ignoring frictional losses?  (d) What percent error would you make in (c) if you ignored the rotational inertia of the tires and wheels?

Table of Contents

   

Answers

1.      A 7.27 kg bowling ball with a radius of 9.00 cm rolls without slipping down a lane at 4.55 m/s.  Calculate the total kinetic energy.    Moments of Inertia
   
   
I = 2/5 MR2  =  2/5 x 7.27 kg x ( 0.0900 m )2  =  0.0236 kg m2  

    

          v          4.55 m s -1  
= ----  =  -----------------  =  50.6 rad s -1  
          r          0.0900 m

   

KE Total  =  KE Translational  +  KE Rotational  =  mv2  +  I2  

   
KE Total  =  x 7.27 kg x ( 4.55 m s -1 )2  + x 0.0236 kg m2 x ( 50.6 rad s -1 )2 

  
KE Total  =  105 J
   
2. Calculate the Kinetic Energy of the Earth with respect to the Sun.  massEarth = 6.0 x 10 24 kg   
radiusEarth = 6.4 x 10 6 m         Earth is 1.5 x 10 8 km from the Sun.
   
   
KE Total  =  KE Rotation Earth  +  KE Revolution Earth  

  
                            2 rad
Rotation  =  ----------------------  =  7.27 x 10 -5  rad s -1 
                       24 h x 3600 s/h

   

                                   2 rad
Revolution  =  ----------------------------------------  =  2.00 x 10 -7  rad s -1 
                       364.25 d x 24 h/d x 3600 s/h

  

IRotation  =  2/5 mR2  =  2/5 x 6.0 x 10 24 kg x ( 6.4 x 10 6 m )2  =  1.0 x 10 38 kg m2 

   
I Revolution  =  mR2  =  6.0 x 10 24 kg x ( 1.5 x 10 11 m )2  =  1.4 x 10 47 kg m2 

  
KETotal  =  IRotationRotation2  + IRevolutionRevolution2  

   
KETotal  =  x 1.0 x 10 38 kg m2 x ( 7.27 x 10 -5  rad s -1 ) 2   +  x 1.47 x 10 47 kg m2 ( 2.00 x 10 -7  rad s -1 ) 2 

  
KETotal  =  2.9 x 10 33 J
3. A 1640 kg merry-go-round has a radius of gyration of 8.20 m.  How much net work is required to accelerate from rest to a period of 8.00 s?
   
   
The net work is equal to the gain in Kinetic Energy.
   

I = Mk2  =  1640 kg x ( 8.20 m )2  =  1.10 x 10 5 kg m2  

   
          2 rad
= ------------  =  0.785 rad s -1 
          8.00 s 

   

KE = I 2  =  x 1.10 x 10 5 kg m2  x ( 0.785 rad s -1 )2  

   
KE = Work = 3.39 x 10 4 J
   
4. A 1100. kg has four tires each with a mass of 30.0 kg that includes the mass of the wheel.  The radius of the tire is 80.0 cm.  The radius of gyration of the tire and wheel is 29.8 cm.  Calculate (a) the total kinetic energy of the car when traveling 115 km/h, and (b) the fraction of kinetic energy in the tires and wheels.  (c) If the car initially at rest is pulled by a tow truck with a force of 2500. N, what is the acceleration of the car ignoring frictional losses?  (d) What percent error would you make in (c) if you ignored the rotational inertia of the tires and wheels?
    
    
(a)   I 4 tires = 4 x Mk2  =  4 x 30.0 kg x ( 0.298 m )2  =  10.7 kg m2  
   

115 km          1 h            1000 m
-----------  x  -----------  x  -------------  =  31.9 m/s = v
  1 h             3600 s          1 km 

   

          v          31.9 m s -1  
= ----  =  -----------------  =  39.9 rad s -1  
          r          0.800 m

   

KE Total  =  KE Translational  +  KE Rotational  =  mv2  +  I2  

  
KE Total  =  x 1100 kg x ( 31.9 m s -1 ) 2  + x 10.7 kg m2 x ( 39.9 rad s -1 ) 2 

   
KE Total  =  5.68 x 10 6 J

   

(b)    KE wheels =    I2  =  x 10.7 kg m2 x ( 39.9 rad s -1 ) 2  =  4.26 x 10 3 J

   

                                                 KE wheels             4.26 x 10 3 J
Fraction of KE in Wheels  =  -----------------  =  --------------------  = 7.50 x 10 -4  
                                                 KE Total             5.68 x 10 6 J

   

                                                 a
(c)   F = ma  +  I  =  ma  +  I ----                          Solve for a.    a = r
                                                 r    

   

              F                                2500. N
a  =  ---------------  =  ----------------------------------------------  =  2.25 m s -2  
          m  +  I/r           1100. kg + 10.7 kg m2 / 0.800 m

   

                 2500. N
(d)   a  =  --------------  =  2.27 m s -2                  a without tires
                 1100. kg

   

                             a without  -  a with                  2.27 m s -2  -  2.25 m s -2 
Percent Error = ------------------------ x 100  =  ----------------------------------- x 100 = 0.9 %
                                  a with                                             2.25 m s -2  
   

Table of Contents