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Non-Uniform Circular Motion


A non-uniform circular motion is a motion which still has circular path but the speed of the object is not constant anymore.

  • For this motion, acceleration is not solely centripetal (radial), but a tangential component shows up too.
  • Tangential acceleration is defined as:
at=dvtdt=d(rω)dt=rdωdt=rα\boxed{a_t=\frac{dv_t}{dt}=\frac{d(r\omega)}{dt}=r\frac{d\omega}{dt}=r\alpha}



where α\alphais known as the angular acceleration which is the rate at which angular velocity is changing.




The magnitude of the net acceleration could be found as: a=ac2+at2\boxed{a=\sqrt{a^2_c+a^2_t}}.

at is tangential acceleration

ac is centripetal (or radial) acceleration



The tangential component is responsible for the change in speed, while the radial component remains responsible for the circular motion.



A mass is traveling in a circular path with radius 1 m. Its speed is changing at a rate of 1.5 m/s2 at an instant when its speed is 4.0 m/s. What is the magnitude of the acceleration of the object? Careful, this isn't straight centripetal acceleration b/c the speed is NOT constant over its path!


Given r = 1m

v = 4 m/s

at = 1.5m/s2

unknown = ac


To find the acceleration magnitude, first we need to find the components of this acceleration. Note that here the speed is changing so, we have a tangential acceleration which is given to us. In addition, the object is moving in a circular path which means that the direction of velocity (remember, velocity is a vector!) is changing as well. So, there should be a centripetal (radial) acceleration as well:


ac = v2/r = (4)2/1 = 16 m/s2​


Now, we can find the magnitude of acceleration vector using its two components:


∣a∣ = ar2+at2


a = sqrt ((1.5)2 +(16)2) = 16.07 m/s2

​

A car travels in a circular track of radius 10 m. Its speed is changing at a rate of 15.0 m/s2 at an instant when its speed is 40.0 m/s. What is the magnitude of the acceleration of the car?


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Extra Practice