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0.29 Phy1260: circular motion -- the mathematics of circular motion  (Page 4/7)

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Assume that the arc length is equal to one-half the circumference of the circle. This arc represents a subtended angle of 180 degrees. Then,

s = circumference/2 = pi * radius

angle = (pi * radius)/radius = pi

From this we can see that pi radians is equal to 180 degrees.

Earlier we saw that 2*pi radians equal 360 degrees.

Facts worth remembering

  • One radian is equal to approximately 57.3 degrees
  • Pi radians is equal to 180 degrees
  • 2*Pi radians are equal to 360 degrees

Tangential displacement versus angular displacement

Consider the case of a 1.5-radian angular displacement of a wheel in a given time interval. What is the corresponding displacement of apoint on the circumference of the wheel? Assume that the radius of the wheel is 0.5 meters.

angle = s/r, or

s = angle * r, or

s = 1.5 * 0.5m = 0.75m

A simple solution

Thus, we see that the tangential displacement of a point on the circumference of a wheel due to a given angular displacement of the wheel in radians is simplythe product of the displacement and the radius of the wheel.

Solving for the same result using angular displacement in degrees would be somewhat morecomplicated.

Tangential speed versus angular velocity

A similar simplification occurs when dealing with the angular velocity of a wheel and the tangential speed of a point on the circumference of the wheel.

As in linear measurements, the average angular velocity of a wheel is equal to the angular displacement of the wheel divided by the time interval duringwhich the displacement takes place.

Measurement of angular velocity in radians

w = dA/dT

By substitution,

w = (s/r)/dT = s/(r*dT)

where

  • s is the length of an arc along the circumference of a circle
  • dA is an angular displacement in a given time interval
  • dT is the time interval
  • r is the radius of the circle
  • w is the angular velocity

Another example

Consider the case of a 1.5-radian/second angular velocity of a wheel. What is the corresponding tangential speed of a point on the circumference ofthe wheel? Assume that the radius of the wheel is 0.5 meters.

The tangential speed is equal to the tangential displacement, s, divided by the time interval over which the displacement occurs. Given the above information,we can write:

w = s/(r*dT)

Given that

v = s/dT

Substitution yields

v; = w*r, or

v = (1.5/s)*(0.5m) = 0.75 m/s

Another simple relationship

Once again, if you keep your units straight, the tangential speed of a point on the circumference of the wheel is simply equal to the angular velocity inradians per second multiplied by the radius of the wheel.

Facts worth remembering

tangential displacement = dA * r

tangential speed = w * r

where

  • tangential displacement is the distance that a given point travels around the circumference of a circle as a function of anangular displacement in radians.
  • tangential speed is the speed at which a point travels around the circumference of a circle as a function of an angularvelocity in radians.
  • w is the angular velocity in radians/second
  • dA represents angular displacement
  • r represents radius
;

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Source:  OpenStax, Accessible physics concepts for blind students. OpenStax CNX. Oct 02, 2015 Download for free at https://legacy.cnx.org/content/col11294/1.36
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