Transcript Centripetal Motion

Centripetal Motion
Motion towards the center of a circle
Terms
• Uniform circular motion is the motion of an
object in a circle with constant or uniform
speed
• Circumference is the distance of one complete
cycle around the perimeter of a circle
• Period is the time needed to make one cycle
around a circle
• Centripetal means center seeking
• Centrifugal means away from the center
Calculating average speed in a circle
• Average speed = circumference/time
• Circumference = 2 * pi * radius
so average speed = 2 * π * R
t
where R = radius
π = 3.14
t = time in s
Tangential
• The direction of the velocity vector of a circle is
constantly changing since the direction is
constantly changing
• The direction of the velocity vector at one instant
is the direction of a tangent line
• A tangent line is a line that touches a circle at one
point but does not intersect it
circle
tangent
tangent
Why an object moving in a circle is
accelerating
• An object moving in a circle is accelerating
since the velocity is changing due to direction
• Reminder that velocity is speed and direction
and acceleration is a change in velocity
• The speed of the object may be constant but
the velocity is changing due to a change in
direction
Centripetal Force
• Any object moving in a circle is experiencing
centripetal force
• There is a physical force pushing or pulling the
object toward the center of the circle
• Centripetal and centrifugal have very different
meanings
Calculating Acceleration in Circular
Motion
• Acceleration = 4 * π2 * R or
T
• Where v= speed
R = radius
T = period in s
v2
R
Net Force
• Reminder F = ma
Fnet = m * v2 or Fnet = m* 4*π2 * R
R
T
• Where m = mass
v = speed
R = radius
T = period in s
Sample Problem 1
• A 900kg car moving at 10m/s takes a turn with
a radius of 25m. Determine the acceleration
and net force acting on the car.
m = 900kg
v = 10m/s
R = 25m
• a = v2/R so a = (10m/s)2/25m = 4m/s2
• Fnet = m * a = 900kg * 4m/s2 = 3600N
Sample Problem 2
• A 95kg halfback makes a turn on the football
field. The halfback sweeps out a path that is a
portion of a circle with a radius of 12m. The
halfback makes a quarter of a turn around the
circle in 2.1s. Determine the speed, acceleration
and net force acting on the halfback.
• m = 95kg
R = 12m
traveled .25 of the circumference of 2.1s
Solution
• v = d/t = 2 * π * R/T
v = .25 * 2 * π * 12m = 8.97m/s
2.1s
• a = v2
R
• a = (8.97m/s)2 = 6.71m/s2
12m
• Fnet = m * a
• Fnet = (95kg) * (6.71m/s2)
• Fnet = 637N