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

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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?

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