1. Circular Motion and Other Applications of Newton’s Laws
Chapter 6
Circular Motion and Other Applications of Newton’s Laws

2. Major topics to be discussed in this chapter
To apply Newton’s 2nd law in:
Uniform circular motion (UCM)
Non-uniform circular motion
Motion in non-inertial (accelerating) frame
Motion in the presence of resistive force

3. Uniform circular motion (UCM)
In UCM, as already discussed, necessitates a centripetal acceleration,
This is a result of the kinematics of the motion.
According to the Newton’s second law, there must exist a force to provide such acceleration. This force, known as centripetal force, denoted as Fr must be pointing also in the direction, as that of the centripetal acceleration’s.
The centripetal force Fr is a “real”, physical force that must be supplied by some physical means (or “agent”), such as a tension in the string, gravitational force, normal force or friction, etc.
Note: In Newton’s 2nd law, Fr = -mar, Fr on the LHS, is the physical force that causes the centripetal acceleration, whereas the RHS, mar is NOT a force but a kinematical requirement of the UCM.

4. The conical pendulum
Two forces act on the conical pendulum: T and mg. The net force is Fnet = T + mg. These forces can be resolved into vertical (j) and horizontal radial ( ) components.
The pendulum’s forces is in equilibrium in the vertical direction:
In the horizontal radial direction, the forces are not in equilibrium. The net force in this component, Fr, is not zero. Fr is the centripetal force that causes the necessary centripetal acceleration in the horizontal plane so that the canonical pendulum executes UCM in the horizontal plane.
The centripetal force here is seen to be supplied by the radial component of the tension force Tsinq. j
Note: the direction of T and radial unit vector are time-varying, but g and j are always constant in time.

5. The conical pendulum (cont.)j
Note:
In the conical pendulum, t0, v, r and q are constants, and independent from the mass m.

6. Example
The mass of the pendulum is kg and L = 1.5 m. If the string can only support a maximum tension of 50.0 N. What is the maximum speed of the pendulum can be supported (before the string breaks)?
In this case, we need to work out how does the speed of the pendulum depend on the tension. j

7. Solution j

8. What is the maximal speed of the car?
A car of mass m = 1500 kg negotiates a curve of radius r = 35.0 m on a horizontal surface. Given the static coefficient ms = between the tires and the road, determines the maximum speed that car can take.

9. Solution
Note:
Although the car is moving, the relevant frictional force for centripetal acceleration is not from kinetic but static friction.
This is because the car’s velocity is in the tangential direction. The car has a tendency to slide away from the circle’s center (but it does not actually slide), hence a static friction is induced in the direction opposing this tendency. The direction of the static friction is in the inwards radial direction and this friction provides the necessary centripetal force for the UCM the car is executing.

10. Solution (cont.)
Note that the maximum speed does not depend on the mass of the car.
Check: We expect that, from command sense, the “sharper” is the curved (i.e., smaller r), the maximum speed allowed should be slower. This expectation is recovered from

11. “Banked Curve”
“Bank curve” allows a care to negotiate a curve despite the absence of friction. In a banked road, the component of the normal force along the horizontal radial direction will provide the required centripetal force.
Note: the direction of n and radial unit vector are time-varying, but g and j are always constant in time.

12. Example
If you were to design a highway with a curve of r = 50m so that any car travelling along that section of the road do not depend on frictional force while negotiating the curve at a speed of km/s without sliding, what is the angle of that section of the road that should be banked?

13. Horizontal circular motion
A pilot is making a stun “loop the loop” at a radius of 2.70 km, with constant speed 225 m/s.
Determine the forces that act on the pilot at (a) the “top”, (b) the “bottom” position.

14. Solution
(a) At the “top” position: j

15. Solution (b) At the “bottom” position: j
Note: the direction of n and radial unit vector are time-varying, but g and j are always constant in time.

16. Non-uniform circular (NUCM) motion vs. uniform circular motion
Comparing UCM and NUCM:
For UCM: v and r are time-independent, and only radial acceleration enters the picture.
In NUCM, apart from the radial acceleration, tangential acceleration will also enter the picture.
For NUCM, v and r may be space- and time-dependent.
Example: The motion of electric fans when electric supply is suddenly switched off: This is a NUCM with constant r but v varies in time.
Example: Roller coaster (r and v time- and position dependent).
Simple harmonic pendulum, swing: (r constant but v time- and position dependent)
A planet’s motion along their orbit: (r and v time- and position dependent)

17. NUCM: Example of a swing (SH pendulum)
In a NUCM, Fnet has both radial and tangential components,
Fnet = Fr + Ft
Fr causes centripetal acceleration,
Ft causes acceleration along the tangential direction.
Active figure 6.8

18. Vertical circular motion with non-uniform speed
A ball is attached to the end of a cord with length r, and set into a vertical circular motion.
Determine the tension T as a function of the instantaneous speed v, and the angle q.
Only tension force and gravity is acting on the ball, hence Fnet = T + mg
The trick is to resolve Fnet into their radial and tangential components, Fnet = Fr + Ft = Fr + Ft

19. Solution

20. Tension at some specific positions
What is the smallest possible value of the tension, and when will this happen? What is the value of the instantaneous speed of the ball at this instance?
Answer: Tmin = 0. It cold occur at the top position if v2top=gr.

21. Motion can be observed/analysed in both inertial as well as non-inertial frames
So far all phenomena discussed are analysed in inertia frames.
Neverthress, as a matter of principle, these phenomena can also be analysed in non-inertial frames.
The figure below illustrates that the rotational motion of a block can be analysed by the man in the frame S (inertial frame), or by the girl in the rotating, non-inertial frame S’.

22. Listen to the description of her observation by the girl in frame S’.
S’ says: “As I observe it, the block is not experiencing any acceleration at all because it is stationary. Obviously the net force on the block in the plane I am sitting on is zero. However, I do agree that there is a tension T from the cord acting on the block. Since, according to me, there is not net force experienced by the block in the plane I am sitting on, the tension force T must be cancelled by some other force I call fictitious force, Ffic, whose magnitude must be equal to that of T, but with a direction opposite to that of T, so that the net force on the block in the plane Fnet = T + Ffic = 0. In other words, the magnitude of fictitious force is Ffic= T= mv2/r. This Ffic can only be felt by me (as well as the block) who are in the non-inertial frame. S’ S

23. Listen to the description of his observation by the man in frame S.
S says: “Hey, my frame is an inertial frame, but yours is not. The block, according to me, is executing a UCM, hence it is experiencing an centripetal acceleration, which is provided by the tension T from the cord. From my point of view, it is then obvious that the net force on the block is not zero but equal to a value of T = mv2/r. I don’t see any force acting on the block in the plane of rotation other than the tension T. In particular, I don’t see there is any ‘fictitious force’ at all, as claimed by you. So I conclude that the fictitious force is an effect seen only by observer in the non-inertial frame” S’ S

24. Fictitious force
All objects in an non-inertial frame will feel the effect from the fictitious force
The fictitious force act on the object in the non-inertial frame as though they are real force, but it could not be associated with any “source” .
As a comparison, examples of real forces, which has a physical origin, are electromagnetic forces, gravity, tension from a cord, friction, contact force etc.
The fictitious force does not exist in inertial frame. The fictitious force arise only in non-inertial frame.

25. Centrifugal force: An example of fictitious force
When a car negotiates a curve, the drive feels a force pushing him/her towards the radially outward direction.
Such a force, known as centrifugal force, is an example of fictitious force. It arises in the car executing an accelerating motion. During the acceleration interval, the car is an inertial frame, with respect to the Earth.
From the view point of an observer at rest wrp to the Earth, the driver in the car appears to experience a non-zero net force pointing in the radially inward direction (this force is just the centripetal force).

26. More on fictitious force
Fictitious force is also known as inertial force.
Despite being “fictitious” the fictitious force has real effect on the objects in the non-inertial frame, e.g.:
The object in the cars slides outward; you feel a force pushing you towards the wall in an rotating platform; the coriolis force is responsible for the circulation in ocean currents that regulate the Earth’s climate.

27. Application of centrifugal force

28. Fictitious force in translational systems
Consider an accelerating train, in which a pendulum is hung to the ceiling.
The accelerating train is an non-inertial frame; the Earth is an inertial frame.
For an inertial observer, the forces on the pendulum are:
The non-inertial observer sees ax j i

29. Pendulum as an “acceleration meter”
How would the non-inertial observer inside the train know what the acceleration of the train ax is?
ANS
The non-inertial observer can know what the acceleration is by measuring the angle q made by the pendulum with respect to the vertical.
q and T measured by both observers in the inertial and non-inertial frames are the same.
The relation between q and ax: ax

30. Motion with resistive forces
Motion can also happen in a fluid medium (e.g., air, water)
Such medium will present a resistive force R to the object while it moves through the medium.
The magnitude of R depends on
(i) the type of the medium,
(ii) the shape of the object,
(iii) the type of surface of the object,
(iv) the speed of the object.
The direction of R is always in that to oppose the motion of the object, i.e.
Usually, R increases with v
The detailed relations between R and v could be complicated, and computer simulation may be required for such purpose.

31. Simplified models for resistive motion
We will only discuss simplest model of resistive motion in a fluid medium:
Model 1: R ~ - v
This is an approximate model for objects that move with a relatively slow speed, or for object with relatively small cross sectional area.
Model 2: R ~ - v2
This is an approximate model for objects that move with a relatively larger speed, or for object with relatively large cross sectional area.

32. Our purpose
We wish to analyse the motion of these resistive model. Specifically, we wish to know how the speed of the object varies with time and position.
We will do so by solving the Newton’s second law for these particles, which is basically a second order differential equation.
Knowing the forces of the problem (the “dynamics”) allows us to obtain how the object’s velocity and position changes with time (the “kinematics”).

33. Model 1: R proportional to -v
The resistive force, or otherwise known as the drag force, is given by
R=-bv
b is a constant that depends on the characters of the medium, the geometry and surface type of the object.
A good system that is well described by this model is shown in the next slide.

34. Example for the Model 1.
First, write down the Newton’s second law: j

35. The qualitative behavior of the particle’s kinematics
The velocity obeys the first order differential equation. v as a function of t and y can be obtained by solving this equation.
Nevertheless, useful insight of the kinematical bevaviour can be obtained without explicitly solving the equation by examining the limits of the equation.

36. The t limit The initial conditions at t=0 v = 0 and dv/dt = g
When t increases, v start to increase, so will the magnitude of R (initially 0 at v=0, t=0), since R =-bv.
As t, R ® mg, the net acceleration anet = g + R/m will finally reduced to |anet|=0. At this stage, the velocity reaches its maximal speed, and the particle will keep on moving at this constant speed, known as the terminal velocity, vT.

37. Terminal velocity
Active figure 6.15

38. Solving the velocity equation quantitatively
To know the detail behavior of v between 0 and vT, we need to solve the first order velocity equation
t = m/b is the time constant that characterise the time scale the particle to achieve terminal velocity.

39. Quantitative behavior of v as a function of time
The terminal velocity can be recovered by setting t>>t
v(t = 0) = 0
v at t = t is:
v(t = t) = vT(1-e-1)=0.632vT

40. A few terminal velocities

41. Model 2: R proportional to -v2
This model describes objects moving through fluid medium at relatively high speed.
R = -½ DrAv2
D is an empirical dimensionless quantity called drag coefficient
r medium’s density
A effective cross sectional area of the object

42. Terminal velocity
Newton’s 2nd law: j

43. Solving the kinematic equation quantitativelyj

44. Example
A baseball kg, radius 3.7 cm, effective cross sectional area A=4.2 x 10-3 m2, is to the air. It is known that the air density is 1.20 kg/m3, and the terminal velocity of the base ball is measured to be vT = 43 m/s. What is the drag force on the base ball when its instantaneous speed is 40.2 m/s? Assume model No. 2.

45. Solution

46. Quick Quiz basketball baseball vT,baseball vT,basketball
A baseball and a basketball, having the same mass, are dropped through air from rest such that their bottoms are initially at the same height above the ground, on the order of 10 m or more. Which one strikes the ground first?
(a) the baseball
(b) the basketball
(c) both strike the ground at the same time
basketball
baseball
vT,baseball
vT,basketball

47. Quick Quiz 6.7
Answer: (a). The basketball, having a larger cross-sectional area, will have a larger force due to air resistance than the baseball. This will result in a smaller net force in the downward direction and a smaller downward acceleration.

48. A conceptual question yi =1000m yi =1000m vT vmoon Moon Earth
On Earth, where there is an atmosphere, a baseball is dropped from a helicopter hovering 1000 m above the ground. On the moon, where there is no atmosphere and the acceleration due to gravity is 1/6 the value on Earth, a similar experiment is performed where the same baseball is dropped from a space shuttle hovering 1000 m above the surface. The speed of the baseball when it hits the surface of the moon will be a) about 1/36, b) about 1/6, c) about equal to, or d) significantly greater than the speed of the baseball when it hits the ground on Earth?
yi =1000m
yi =1000m vT
vmoon
Moon
Earth

49. ANSWER
(d). From table 6.1, the approximate terminal speed of the baseball on Earth will be vT = 43 m/s. On the moon, the final speed is determined completely by the constant acceleration due to gravity and the distance of the fall. Using equation 2.13 and considering yi and vmoon to be zero,
the terminal speed is found to be 57 m/s which is 33 % greater than the terminal speed on the Earth. This example illustrates the large effect air resistance can have on reducing the speed of falling objects. The absence of air on the moon accounts for the large quantity of impact craters on its surface.