Title: Circular Motion
1Circular Motion
2Imagine a hammer (athletics variety) being spun
in a horizontal circle
3Birds-Eye View
?
v
r
4v r?
5Side View
?
T
mg
6We know that the hammer is accelerating..
- Because the hammer is constantly changing
direction (although the speed is constant)
7So from Newtons First and Second Laws, there
must be a resultant force
- Equal to
- mass x acceleration
8For circular motion..
- Acceleration v2
- r
- or r?2 (using v r? )
9So the resultant force ..
- mv2
- r
- or mr?2 (using v r? )
10Which direction do the resultant force and
acceleration act in?
- Towards the centre of the described circle
11So if we look at our original diagram.
12If the circle has a radius ,r.
?
T
mg
13We 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 mr?2
14Very important point!
- The circular force is not an additional force
it is the resultant of the forces present.
15Typical exam style question
- Ball hangs from a light piece of inextensible
string and describes a horizontal circle of
radius,r and makes an angle ? with the vertical . - If the mass of the ball is m kg
- calculate the tension, T in the string
- calculate the angular velocity, ? in terms of g,
r and ?.
16Diagram
?
T
mg
r
17To find the tension.
- Resolve vertically
- Ball is not moving up or down so vertical
components must be equal -
Tcos? mg so T mg cos?
18To find the angular velocity, ?
- Resolve horizontally
- Circular motion so we know that there is a
resultant force towards the centre
Tsin? mr?2 ? Tsin? mr
19But
so ? Tsin? mr Becomes ?
gtan? r
20Easy?