Circular Motion
Imagine a hammer (athletics variety) being spun in a horizontal circle At a constant speed
Birds-Eye View v ω r
v = rω
Side View T mg ω
We know that the hammer is accelerating….. Because the hammer is constantly changing direction (although the speed is constant)
So from Newtons First and Second Laws, there must be a resultant force Equal to mass x acceleration
For circular motion….. Acceleration = v 2 r or rω 2 (using v = rω )
So the resultant force ….. = mv 2 r or mrω 2 (using v = rω )
Which direction do the resultant force and acceleration act in? Towards the centre of the described circle
So if we look at our original diagram…….
If the circle has a radius,r…. T mg ω
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 = mrω 2
Very important point! The circular force is not an additional force – it is the resultant of the forces present.
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 i)calculate the tension, T in the string ii)calculate the angular velocity, ω in terms of g, r and θ.
Diagram r θ mg T
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
To find the angular velocity, ω… Resolve horizontally Circular motion so we know that there is a resultant force towards the centre Tsinθ = mrω 2 ω = Tsinθ mr
But…… T = mg cosθ so ω = Tsinθ mr Becomes ω = gtanθ r
Easy?