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Wize University Physics Textbook (Master) > Circular Motion

Non-Uniform Circular Motion

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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.

Wize Tip
In a uniform circular motion, the angular velocity is constant but it is changing in a non-uniform circular motion.


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.

Watch Out!
The net acceleration does not point towards the center of the circle in non-uniform 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?

Extra Practice
Vertical circular motion
Practice: Vertical String
A box of mass 2.50 kg is attached to a string and whirled in a vertical circle of radius 1.0 m. At the exact top of the path, the tension in the string is 3 times the ball’s weight.
Ferris wheel
Practice: Ferris Wheel
A 65 kg man is riding a Ferris wheel with radius of 15 m, that moves at speed 2.355 m/s. Determine the normal force that the man experiences at the top of the ferris wheel and at the bottom.
Vertical circular motion
A box of mass 2.50 kg is attached to a string and whirled in a vertical circle of radius 1.0 m. At the exact top of the path, the tension in the string is 3 times the ball’s weight.
Ferris wheel
A 65 kg man is riding a Ferris wheel with radius of 15 m, that moves at speed 2.355 m/s. Determine the normal force that the man experiences at the top of the ferris wheel and at the bottom.
Doug is on a ferris wheel, which is spinning counterclockwise relative to you. The ride is over and the wheel is starting to slow down. What direction is his acceleration at the top of the ferris wheel?
Circular motion
A race car starts from rest on a circular track. The car increases its speed at a constant rate at as it goes once around the track in the counterclockwise direction. Find the angle that the total acceleration of the car makes with the radius of the circle, at the moment when the car completes first full circle.
A runner is running at a speed of 10m/s along a circular track of radius 25 m.
(a) Find the time it takes to complete one lap when he is running at constant speed.
(b) Find the radial acceleration during this time.