1. Circular Motion

2. Imagine a hammer (athletics variety) being spun in a horizontal circle
At a constant speed

3. Birds-Eye View ω v r

4. v = rω

5. Side View ω T mg

6. We know that the hammer is accelerating…..
Because the hammer is constantly changing direction (although the speed is constant)

7. Equal to mass x acceleration
So from Newton’s First and Second Laws, there must be a resultant force
Equal to
mass x acceleration

8. For circular motion…..
Acceleration = v2 r
or rω2 (using v = rω )

9. So the resultant force …..
= mv2 r
or mrω2 (using v = rω )

10. Which direction do the resultant force and acceleration act in?
Towards the centre of the described circle

11. So if we look at our original diagram…….

12. If the circle has a radius ,r….ω T mg

13. We can find the resultant force by resolving in the plane of the circle.
The only force acting in the horizontal plane is the tension
So by resolving T = mrω2

14. Very important point!
The ‘circular force’ is not an additional force – it is the resultant of the forces present.

15. Typical exam style question
Ball hangs from a light piece of inextensible string and describes a horizontal circle of radius,r and makes an angle θ with the vertical .
If the mass of the ball is m kg
calculate the tension, T in the string
calculate the angular velocity, ω in terms of g, r and θ.

16. Diagram θ T mg r

17. To find the tension…. Resolve vertically
Ball is not moving up or down so vertical components must be equal
Tcosθ = mg
so T = mg
cosθ

18. To find the angular velocity, ω…
Resolve horizontally
Circular motion so we know that there is a resultant force towards the centre
Tsinθ = mrω2
ω = Tsinθ mr

19. But……
T = mg
cosθ
so ω = Tsinθ mr
Becomes ω = gtanθ r

20. Easy?