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Wize High School Grade 12 Physics Textbook > Circular Motion

Uniform Circular Motion

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Uniform Circular Motion


When an object moves uniformly around a circle, speed of the object remains constant but velocity changes. This is called uniform circular motion (UCM).




Examples of uniform circular motion:
  • A vehicle moving around a circular track
  • Amusement park rides, ferris wheel, merry-go-round
  • The moon, or satellites orbiting the Earth in a circular orbit
  • Any object moving in a circle with constant speed!



Wize Tip
We can model any motion that travels along a circular path at a constant speed with UCM. It does not have to travel the complete circle.

Watch Out!
The object is moving at constant speed (magnitude of velocity), but the object has a changing velocity (the direction constantly changes).

  • Velocity vector is always tangent to the path of the motion. So, for a circular motion, it is tangent to the circle.
  • The change in velocity is caused by an acceleration which is pointing into the centre of the circle and it is known as centripetal acceleration:
a=v2r\boxed{a=\frac{v^2}{r}}
  • Centripetal acceleration is perpendicular to velocity vector at any point




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Period and Frequency

  • Period (T)= the time it takes for the object to make one complete revolution or cycle (measured in seconds)
  • Frequency (f)= the number of cycles made in a given time period. Usually it is expressed per second. (measured in Hz where 1 Hz = 1/s).
  • Period and frequency are inversely related as follows:
f=1T\boxed{f=\dfrac{1}{T}}

  • Circumference of a circle of radius rr is 2πr2\pi r.So, time to complete one revolution is:
T=2πrv\boxed{T=\frac{2\pi r}{v}}

  • Angular velocity is defined as the rate at which angles are changing in a circular motion and it could be found as:
ω=2πT\boxed{\omega=\frac{2\pi}{T}}


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Example: A Circulating Ball

A ball attached to a 10 cm long string is having a circular motion with period of 0.59 seconds. If the string makes an angle of 60 degrees with horizontal axis, what is its centripetal acceleration?

Solution:


Note that the radius of circular path in which the ball is moving is different from the length of the string. However, we can find this radius by a little bit of trigonometry knowing the angle the string makes with the horizontal direction.
As length of string is 10 cm, the radius of circular motion is:


r=l⋅Cos60=10⋅12=5 cmr=l\cdot Cos60=10\cdot\frac{1}{2}=5\ cm
Now we can use the period and the radius of this motion to find the speed of the ball:
Period is 2s, hence:


T=2πrvT=\frac{2\pi r}{v}

0.59=2⋅π⋅5⋅10−2v0.59=\frac{2\cdot\pi\cdot5\cdot10^{-2}}{v}

v=0.532 msv=0.532\ \frac{m}{s}
Acceleration is then:

ac=v2r=0.53225⋅10−2=5.66 ms2a_c=\frac{v^2}{r}=\frac{0.532^2}{5\cdot10^{-2}}=5.66\ \frac{m}{s^2}

A 1.6kg ball is attached to a 1.8m string and is swinging in circular motion horizontally at the string's full length. If the string can withstand a tension force of 87N, what is the maximum speed the ball can travel without the string breaking?


A 1000kg car rounds a circular track with radius 10m. If μsmax⁡=0.81,\mu _{s\max }=0.81,what is the maximum speed the car can travel?