Transcript Circular Motion

Circular Motion
Imagine a hammer (athletics variety)
being spun in a horizontal circle
At a constant speed
Birds-Eye View
ω
v
r
v = rω
Side View
ω
T
mg
We know that the hammer is
accelerating…..
Because the hammer is constantly
changing direction (although the
speed is constant)
So from Newton’s First and
Second Laws, there must be a
resultant force
Equal to
mass x acceleration
For circular motion…..
Acceleration = v2
r
or rω2 (using v = rω )
So the resultant force …..
= mv2
r
or mrω2 (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….
ω
T
mg
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
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
θ
T
mg
r
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 = mrω2
ω=
Tsinθ
mr
But……
T = mg
cosθ
so ω =
Tsinθ
mr
Becomes ω = gtanθ
r
Easy?