Transcript Motion Along a Straight Line at Constant

Book Reference : Pages 24-25
1.
To consider speed & velocity around a circle
2.
To consider acceleration as a change in
velocity
3.
To define an equation for centripetal
acceleration
4.
To define an equation for centripetal force
Velocity v
If an object is moving in a circle
with a constant speed, it’s
velocity is constantly changing....
Because the direction is
constantly changing....
If the velocity is constantly
changing then by definition the
object is accelerating
If the object is accelerating, then
an unbalanced force must exist
Velocity vB
B

C
v
Velocity vB 
Consider an object moving
in circular motion with a
speed v which moves from
Velocity v
point A to point B in t
A
seconds
(From speed=distance / time), the
distance moved along the arc AB,
s is vt
Velocity vA
A
The vector diagram shows
the change in velocity v :
(vB – vA)
The triangles ABC & the
vector diagram are similar
Velocity vB
B

C
Velocity vA
A
Substituting for s = vt
v
Velocity vB 
If  is small, then v / v = s / r
v / v = vt / r
Velocity vA
(a = change in velocity / time)
a = v / t = v2 / r
We can substitute for angular velocity....
a = v2 / r
From the last lesson we saw that:
v = r (substituting for v into above)
a = (r)2 / r
a = r2
In exactly the same way as we can connect force
f and acceleration a using Newton’s 2nd law of
motion, we can arrive at the centripetal force
which is keeping the object moving in a circle
f = mv2 / r
or
f = mr2
Any object moving in a circle is acted upon by a
single resultant force towards the centre of the
circle. We call this the centripetal force
Gravity which keeps satellites in orbit around
Earth and the Earth in orbit around the sun is a
classic example of a centripetal force.
satellite
Gravity
Planet
The wheel of the London Eye has a diameter of
130m and takes 30mins for 1 revolution.
Calculate:
a. The speed of the capsule
b. The centripetal acceleration
c. The centripetal force on a person with a
mass of 65kg
The speed of the capsule :
Using v = r
we know that we do a full revolution (2 rad)
in 30mins (1800s)
v = (130/2) x (2 / 1800)
v = 0.23 ms-1
The centripetal acceleration:
Using a = v2 / r
a = (0.23)2 / (130/2)
a = 7.92 x 10-4 ms-2
The centripetal force:
Using f = ma
F = 65 x 7.92 x 10-4
F = 0.051 N
An object of mass 0.15kg moves around a circular
path which has a radius of 0.42m once every 5s at
a steady rate. Calculate:
a. The speed and acceleration of the object
b. The centripetal force on the object
[.528 ms-1, 0.663ms-2, 0.100N]