Grafana in space: Monitoring Japan's SLIM moon lander in real time
circular.pdf
1. Uniform & Non-Uniform Circular Motion
Uniform & Non-Uniform Circular Motion
I P S I T A M A N D A L
2. References
An Introduction To Mechanics
by Daniel Kleppner & Robert Kolenkow,
Chapter 1 (sections 1.8, 1.12, 1.13, 1.14)
Fundamentals of Physics
by Jearl Walker, David Halliday, & Robert Resnick,
Chapter 4 (section 4-5)
Introduction to Classical Mechanics
by David Morin, Chapter 3 (section 3.5)
3. Circular Motion
Speed of a particle moving in a circular path can be
constant variable
Uniform Non-uniform
Motion is in a plane described by a two-dimensional (2d)
coordinate system
sun
Ferris
wheel
4. Polar Coordinates ...
Recall ( flipped class lecture at home ):
Directions of unit vectors vary with position
We derived
1
.
ˆ
i
ˆ
j
θ
θ̂
ˆ
r
8. Change in Velocity
Velocity is a vector ⤇ has (1) magnitude (speed), (2) direction
A change in velocity results in acceleration ⤇ a net force acts on the
body
The change may involve magnitude and / or direction ⤇ a body in
circular motion ( ) is accelerating
= change in velocity
9. Clicker Question
According to Newton’s laws, in an inertial frame, what is the net force
acting on a body in a circular motion ?
A: Zero
B: Non-zero
C: Can be zero or non-zero depending on whether speed is constant or
changing
13. Uniform Circular Motion
Constant (uniform) speed:
No tangential acceleration
Velocity changes in direction
⤇ points tangentially to the circle:
Radially inward acceleration
with a uniform magnitude
Centripetal
Acceleration
14. Centripetal acceleration arises from a centripetal force:
Centripetal force accelerates a body by changing the direction of
its velocity without changing its speed
Centripetal Force
15. Period T of uniform circular motion = time for one revolution
(one complete trip around the circle)
Period of Motion
Distance = Speed x Time
⇒ Circumference of the circle = v T
⇒ 2 π r = v T
⇒ T = 2 π r / v
16. Examples
Swinging a ball on the end of a string ⤇ tension provides the
centripetal force
Satellite in a circular orbit around Earth
⤇ gravity provides the centripetal force
17. Examples ...
Car moving in a horizontal circle on a level surface ⤇ friction provides
the centripetal force
Death spiral in figure skating ⤇ the man is the center of rotation (one
toe dug into the ice in a pivot position), exerting centripetal force to
keep his partner moving in a circle
18. Problem 1
A bob of mass m hangs from a string of length L. Conditions have been
set up so that the mass swings around in a horizontal circle, with the
string making an angle of ϕ with the vertical. What is the angular
speed ω of the bob?
R = L sin ϕ
ϕ
L
R
m
Weight = m g
Tension = T
ϕ
19. Problem 1 ...
ϕ
T
T cos ϕ
T sin ϕ
a = v2 / R
m g
Bob
0 = T cos ϕ - m g ( y-direction )
- m a = - T sin ϕ ( r-direction )
gives
a / g = tan ϕ
⇒ v2 = R g tan ϕ
⇒ R2 ω2 = R g tan ϕ
⇒ ω2 = g tan ϕ / ( L sin ϕ)
20. Problem 2
A car of mass m moves at a constant speed v around a banked circular
track of radius R . If the friction is negligible (slippery conditions like ice
on a highway or oil on a racetrack), what bank angle φ prevents sliding?
φ
v
φ
φ
Weight = m g
Normal reaction = N
φ
N
N cos φ
N sin φ
a = v2 / R
m g
0 = N cos φ - m g ( y-direction )
- m a = - N sin φ ( r-direction )
gives
tan φ = v2 / ( g R )
22. Non-Uniform Circular Motion
Speed varies:
Velocity changes in direction + magnitude ⤇ points tangentially to the
circle
Both radial & tangential components of acceleration are nonzero
Tangential component of a = rate of change of speed
Tangential component of a is in the same (opposite) direction as the
velocity if the particle is speeding up (slowing down)
Exercise / Tutorial: Prove that
23. Examples
Roller coaster cars ⤇ slow down and speed up as they move around a
vertical loop
David swinging sling in a vertical circle
24. Problem 3
Analyze the forces as the roller coaster goes through the top of a hill,
the bottom of a valley, top of a loop, down the side of a loop
25. Problem 3: Solution
(1) Top of the hill
Centripetal force is supplied by gravity
& possibly even the safety harness
Normal reaction = N ≥ 0
Weight = m g
Fnet = N – m g (in vertically upward direction)
⇒ - centripetal force = N – m g
⇒ - m v1
2 / R1 = N – m g
How fast can the coaster can go until the rider just (barely) loses contact with
the seat ?
N = 0
⇒ m v1
2 / R1 = m g
⇒ v1
2 = g R1
At higher speeds, N = m ( g - v1
2 / R1 ) says that the normal force will be negative!
This just means that for v1
2 / R1 > g the rider will fly off the coaster ( N=0 ) unless
a safety harness supplies an extra downward force ( Fsafety ) pulling the rider
downward, providing the remaining centripetal force : m v1
2 / R1 = m g + Fsafety
velocity v1
N
mg
R1
26. Problem 3: Solution ...
(2) Bottom of the valley
Normal reaction = N
Weight = m g
Fnet = N – m g (in vertically upward direction)
⇒ centripetal force = N – m g
⇒ m v2
2 / R2 = N – m g
velocity v2
N
R2
mg
27. Problem 3: Solution ...
(3) Top of the loop
Normal reaction = N ≥ 0
Weight = m g
Fnet = - N – m g (in vertically upward direction)
⇒ - centripetal force = - N – m g
⇒ m v3
2 / R2 = N + m g
If the speed is too low, N = m ( v3
2 / R2 – g ) says that the normal force will be
negative ! This just means that for v3
2 / R2 < g, the car would fall off the track.
To prevent this, roller coasters have wheels on both sides of the track.
velocity v3
N
R2
mg
28. Problem 3: Solution ...
(4) Down the side of the loop
Normal reaction = N ≥ 0
Weight = m g
Fnet,x = N ( in radially inward direction )
⇒ centripetal force = N
⇒ m v4
2 / R2 = N
Fnet,y = - m g ( in vertically upward direction )
⇒ tangential force = - m g
⇒ - m aθ = - m g
⇒ aθ = g
N
R2
velocity v4
mg
aθ
aR
aθ
aR
a
29. Points to Remember
An object moving in a circle:
Always has a tangentially directed velocity
Always has a radially inward component of acceleration
Always has a net force acting on it
Has a tangential component of acceleration if its speed changes with
time
30. Problem 4
An object of mass m is constrained to move in a circle of radius r. Its
tangential acceleration is given by at = b + c t2
, where b and c are
constants. If v = v0
at t = 0, determine the radial component of the
acceleration
Radial acceleration is: