Angular Quantities

      
Introduction

Every point in a body rotating about a fixed axis moves in a circle whose center is on the axis has a radius, r.  Any point on the radius will sweep out the same angle in the same time.  

One radian, rad, is defined as the angle subtended by an arc whose length is equal to the radius.  In the drawing to the left, the length, l, is equal to the radius, r.  = 1 rad.

                                  l
By definition      = --- = Angular Displacement
                                  r

360^o = 2rad  

         
= --- = Angular Velocity
           t

Omega, , is the average when measured over long periods of time and instantaneous when the time approaches zero.  Units are rad/s.

As shown in the drawing to the left, the instantaneous velocity is larger as you move farther from the axis, but , is constant.  Thus the   is independent of the radius.

        
= ------ = Angular Acceleration   
          t

is the average angular acceleration when measured over long periods of time and instantaneous when the time approaches zero.

There is a simple relationship between Angular and Linear measurements.

x = r
  
v = r

a_T = r

Observe that the relationship only gives the tangential acceleration.  In the section on circular motion, we saw that an object also has centripetal acceleration, a_C.  The total linear acceleration of a particle is

a  =  a_T  +  a_C 

Remember that a_T  and  a_C are perpendicular to each other.

In the section on circular motion, we related centripetal acceleration to the linear velocity and radius.  We can also relate them to angular velocity.
      

           v²        ( r )²  
_C = -----  =  ---------- =  ²r
           r              r
   

At times we relate the angular velocity to the frequency of rotation, f.  The frequency is the number of complete rotations in one seconds.  One revolution is equal to an angular displacement of 2rad.  

     
1 rev      2rad  
------- =  ----------      thus   = 2f
  1 s           1s 

   
Period, T, is the time for 1 revolution.  
   

          1                          1
T  =  ----     and    f  =  -----
          f                           T

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Problems

1.	The Sun subtends an angle of about 0.5o when viewed from Earth.  The Sun is 150 million km from Earth.  What is the radius of the Sun?
2.	A 0.25 m grinding wheel rotates at 3500 rpm.  Calculate its angular velocity in rad/s.
3.	A laser beam is directed at the Moon, 380,000 km from Earth.  The beam diverges at a of 1.8 x 10 -5 rad.  What is the diameter of the spot on the Moon?
4.	A bicycle with 68.0 cm diameter tires travels 8.50 km.  (a) How many revolutions do the tires make in this trip?  (b) If it makes the trip in 55.0 minutes, what is the angular velocity in rad/s?  (c)  What is the period of rotation of a tire?
5.	(a) Calculate the angular velocity of the Earth as it revolves around the Sun.  (b) Calculate the angular velocity of the Earth as it rotates.
6.	A 70.0 cm diameter wheel accelerates uniformly from 160. rpm to 350. rpm in 3.50 s.  Calculate (a) angular acceleration, (b) tangential acceleration, and (c) centripetal acceleration.

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Answers

1.	The Sun subtends an angle of about 0.5o when viewed from Earth.  The Sun is 150 million km from Earth.  What is the radius of the Sun?
	2rad 0.5o x ----------- = 0.009 rad               360o      x = diameter of Sun = r = 1.50 x 10 8 km x 0.009 rad = 1 x 10 6 km                         diameter       1 x 10 6 km  radius =  ------------- = ------------------ = 5 x 10 5 km                       2                     2       Rad is not a true unit and does not have to be canceled.
2.	A 0.25 m grinding wheel rotates at 3500 rpm.  Calculate its angular velocity in rad/s.
	3500 rev       60 s         2rad ------------ x ---------- x ------------ = 3700 rad/s  1 min          1 min          360o
3.	A laser beam is directed at the Moon, 380,000 km from Earth.  The beam diverges at a of 1.8 x 10 -5 rad.  What is the diameter of the spot on the Moon?
	x = diameter of spot = r = 380,000 km x 1.8 x 10 -5 rad = 6.8 meters
4.	A bicycle with 68.0 cm diameter tires travels 8.50 km.  (a) How many revolutions do the tires make in this trip?  (b) If it makes the trip in 55.0 minutes, what is the angular velocity in rad/s?  (c)  What is the period of rotation of a tire?
	(a) Circumference of tire = d = x 0.340 m = 1.07 m         trip                      8.50 x 103 m -------------------  =  --------------------- = 7940 revolutions circumference           1.07 m     (b)  7940 rev        2rad       1 min       -------------- x ----------- x ----------- = 15.1 rad/s        55.0 min        1 rev          60 s     (c)  55.0 min       60 s       ------------- x ---------  =  0.416 s       7940 rev       1 min
5.	(a) Calculate the angular velocity of the Earth as it revolves around the Sun.  (b) Calculate the angular velocity of the Earth as it rotates.
	(a) 1 rev               1 d            1 h            2rad  ------------  x  --------  x  -----------  x -----------  =  1.996 x 10 -7 rad/s 364.25 d        24  h        3600 s         1 rev      (b) 1 rev         1 h            2rad  -------- x ----------- x ------------  =  7.27 x 10 -5 rad/s 24 h        3600 s        1  rev
6.	A 70.0 cm diameter wheel accelerates uniformly from 160. rpm to 350. rpm in 3.50 s.  Calculate (a) angular acceleration, (b) tangential acceleration, and (c) centripetal acceleration.
	(a)           /\ = 350. rpm  -  160. rpm  =  190. rpm     190. rev       2rad           1 min -----------  x  -----------  x  -----------  =  0.332 rad/s  =    1 min          1 rev           3600 s                  /\         0.332 r s -1    =  --------   =  ----------------  = 0.0949 rad s -2              /\t              3.50 s     (b)         aT  =  r  =  0.700 m x 0.0949 rad s -2  =  0.0664 m s -2   (c)          aC  =  2r  =  ( 0.332 r s -1 )2  x  0.700 m  =  0.0772 m s -2

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