Chapter 5: Circular Motion and Gravity
Circular Motion and Gravity
Uniform Circular Motion
Definition: The motion of an object traveling at a constant (uniform) speed in a circular path.
Acceleration: Despite constant speed, an object in uniform circular motion is always accelerating because its velocity (a vector with magnitude and direction) is continuously changing direction.
If there were no acceleration, the object would move in a straight line at a constant speed.
Direction of Acceleration for Uniform Circular Motion
Coordinate System:
Tangential (t) axis: Always aligned with the instantaneous velocity.
Radial (r) or Centripetal (c) axis: Always perpendicular to the tangential axis, pointing toward the center of the circular path.
Role of Tangential Acceleration: A tangential acceleration component changes the speed of an object. In uniform circular motion, the speed is constant, so there is no tangential acceleration.
Role of Radial/Centripetal Acceleration: Since speed is constant, any acceleration must be radial, changing only the direction of velocity.
Derivation of Direction: By considering the change in velocity vector (\Delta \vec{v} = \vec{v2} - \vec{v1}) over a small time interval, it can be shown that the direction of the average acceleration points toward the center of the circle at a point midway between the two velocity vectors.
Centripetal Acceleration: The acceleration in uniform circular motion is always directed toward the center of the circle. This is called centripetal acceleration (meaning "center-seeking").
Example (Interactive Question, Page 5): A rock twirled on a string in a clockwise direction. At any point P, its acceleration is directed toward the center of the circle.
Magnitude of Acceleration for Uniform Circular Motion
The magnitude of centripetal acceleration ( a_r ) is given by:
a_r = \frac{v^2}{r}
Where:
v is the constant speed of the object.
r is the radius of the circular path.
Derivation Sketch (Page 6): Using similar triangles formed by position vectors ( \Delta \vec{r} ) and velocity vectors ( \Delta \vec{v} ), and the definitions of speed and acceleration, the formula can be derived:
\frac{\Delta v}{v} = \frac{\Delta r}{r} \implies \Delta v = \frac{v \Delta r}{r}
a_r = \frac{\Delta v}{\Delta t} = \frac{v}{r} \frac{\Delta r}{\Delta t} = \frac{v}{r} v = \frac{v^2}{r}
Circular Motion Terminology and Definitions
Period (T): The time it takes for an object to complete one full revolution around the circumference of the circle.
For constant speed v , the speed can also be expressed in terms of the period and radius (R):
v = \frac{\Delta r}{\Delta t} = \frac{2 \pi R}{T}
Conversely, the period can be found as:
T = \frac{2 \pi R}{v}
Frequency (f): The rate of rotation, defined as the number of revolutions per unit time.
f = \frac{1}{T}
Centripetal Force
Newton's Second Law in Circular Motion: For an object undergoing uniform circular motion, a net force must be acting in the radial direction to produce the centripetal acceleration.
\sum Fr = m ar = \frac{m v^2}{r}
Nature of Centripetal Force: This "net radial force" is called centripetal force (center-seeking force).
It is NOT a new fundamental force. Instead, it is the sum of all existing forces (e.g., tension, gravity, friction, normal force) that happen to have a net component directed toward the center of the circle.
Example (Interactive Question, Page 13): If a string twirling a rock breaks, the centripetal force (tension) is removed, and the rock flies off in a straight line tangent to the circle at the point of release, due to its inertia.
Forces in Action (Problem Examples):
Car Turning on a Flat Road (Page 14): The static friction force between the tires and the road provides the necessary centripetal force.
f_s = m \frac{v^2}{r}
Since N = mg (on a flat road), and fs \leq \mus N , the maximum speed is given by \mus mg = m \frac{v{max}^2}{r} , leading to v{max} = \sqrt{\mus g r} .
For r=50.0 \ m and \mus=0.90 , v{max} = \sqrt{0.90 \times 9.8 \ m/s^2 \times 50.0 \ m} \approx 21 \ m/s .
Whirling Stone (Page 15): The tension in the string provides the centripetal force. If the rotation rate doubles (period halves), the tension increases by a factor of four.
F_T = \frac{m v^2}{r} = \frac{m (2 \pi r / T)^2}{r} = \frac{4 \pi^2 m r}{T^2}
If T \rightarrow T/2 , then FT \rightarrow \frac{4 \pi^2 m r}{(T/2)^2} = \frac{4 \pi^2 m r}{T^2/4} = 4 \frac{4 \pi^2 m r}{T^2} = 4 FT .
Barrel of Fun Ride (Page 16): The normal force from the wall provides the centripetal force, while static friction prevents the rider from sliding down when the floor drops.
Banked Curve (No Friction, Page 17): The horizontal component of the normal force provides the centripetal force.
N \sin \theta = \frac{m v^2}{r}
N \cos \theta = mg
Dividing these equations yields \tan \theta = \frac{v^2}{gr} . This formula determines the ideal banking angle for a given speed and curve radius.
For r=50 \ m and v=15 \ m/s , \tan \theta = (15^2) / (9.8 \times 50) \approx 0.459 , so \theta \approx 24.6^\circ .
Banked Curve (With Friction, Page 19): Friction can act up or down the incline, adding to or subtracting from the radial component of the normal force, allowing for a range of safe speeds.
Vertical Circular Motion (Water Can, Page 20):
Minimum speed at the top: For the water not to fall out, the normal force (or tension) at the top must be at least zero. The gravitational force alone provides the centripetal force: mg = m \frac{v{min}^2}{r} \implies v{min} = \sqrt{gr} .
For r=1.0 \ m , v_{min} = \sqrt{9.8 \ m/s^2 \times 1.0 \ m} \approx 3.13 \ m/s .
Normal force at the bottom: Using energy conservation, the speed at the bottom would be v_{bottom} = \sqrt{5gr} . The forces are normal force (up) and gravity (down):
N - mg = m \frac{v_{bottom}^2}{r} = m \frac{5gr}{r} = 5mg
N = 6mg .
Conical Pendulum (Ball on String, Page 23):
The horizontal component of tension provides the centripetal force, and the vertical component balances gravity.
Centripetal acceleration: a_r = g \tan \theta .
Speed of rotation: v = (\sin \theta) \sqrt{\frac{gL}{\cos \theta}} , where L is the string length and r = L \sin \theta . For L=0.8 \ m and \theta=20^\circ , v \approx 0.988 \ m/s .
Non-Uniform Circular Motion
Definition: Motion in a circle where the object's speed is varying.
Acceleration Components: For this to occur, there must be two components of acceleration:
Centripetal (radial) acceleration ( ar ): Perpendicular to the direction of motion, responsible for changing the object's direction. ( ar = v^2/r )
Tangential acceleration ( a_T ): Parallel (or anti-parallel) to the direction of motion, responsible for changing the object's speed.
Total Acceleration: The total acceleration (
a) is the vector sum of these two components:\vec{a} = \vec{ar} + \vec{aT}
Its magnitude is given by: a = \sqrt{ar^2 + aT^2}
Forces for Non-Uniform Circular Motion: The net force on the object points in the same direction as the total acceleration. Therefore, there will be components of force in both the radial and tangential directions.
Example (Interactive Question, Page 26): A car speeding up on a circular ramp will have acceleration pointing both inward (centripetal) and forward (tangential).
Example (Interactive Question, Page 28): A car slowing down on a circular exit ramp will have a net force pointing both inward (centripetal) and backward (anti-tangential).
Problem (Page 29): Car accelerating from rest to 22 \ m/s on a 90^\circ circular path of r = 87 \ m .
(a) Tangential acceleration ( aT ): Using kinematics, vf^2 = v0^2 + 2 aT \Delta s , where \Delta s = (1/4) 2 \pi r = \pi r / 2 . This gives a_T \approx 1.77 \ m/s^2 .
(b) Radial acceleration ( ar ) when v = 15 \ m/s : ar = v^2/r = (15 \ m/s)^2 / (87 \ m) \approx 2.59 \ m/s^2 .
Newton’s Law of Universal Gravitation
Statement: Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Formula:
F = G \frac{m1 m2}{r^2}
Where:
F is the gravitational force between the two objects.
m1 and m2 are their masses.
r is the distance between their centers (for spherical objects).
G is Newton’s universal gravitational constant: G = 6.67 \times 10^{-11} \ N \cdot m^2/kg^2 .
Key Observations:
Infinite Range: The force acts over all space; every object exerts a gravitational pull on every other object.
Proportional to Mass: The force is directly proportional to each mass.
Action-Reaction Pairs: The force one object feels from another is equal in magnitude and opposite in direction (Newton's Third Law). Their accelerations, however, will differ if their masses differ.
Spherical Objects: For spherical objects, the distance r is measured from their geometric centers (a concept Newton proved using calculus).
Interactive Question (Page 32): A spaceship is never beyond the pull of Earth's gravity because gravitational force has an infinite range.
Problem (Page 33): The magnitude of the gravitational force on the Moon from the Earth is equal to the magnitude of the gravitational force on the Earth from the Moon (Newton's 3rd Law).
Gravity at the Surface of the Earth
The weight of an object ( F_g ) on Earth's surface is due to gravity:
Fg = mg = G \frac{ME m}{r_E^2}
Where ME is Earth's mass and rE is Earth's radius.
Acceleration due to gravity ( g ):
g = \frac{G ME}{rE^2}
Calculation: g = (6.67 \times 10^{-11} \ N \cdot m^2/kg^2) \frac{(5.97 \times 10^{24} \ kg)}{(6.38 \times 10^6 \ m)^2} \approx 9.78 \ m/s^2 .
Gravitational and Inertial Mass
Gravitational Mass ( m_{gravitational} ): The mass that determines the strength of the gravitational force an object experiences or exerts.
Inertial Mass ( m{inertial} ): The mass that quantifies an object's resistance to acceleration (as in Newton's Second Law, F = m{inertial} a ).
The Equivalence Principle: Experiments have consistently shown that gravitational mass and inertial mass are equivalent ( m{gravitational} = m{inertial} ).
This equivalence explains why all objects near Earth's surface fall with the same acceleration ( a = g ), regardless of their mass (assuming negligible air resistance):
G \frac{m{gravitational} ME}{rE^2} = m{inertial} a
If m{gravitational} = m{inertial} , then a = G \frac{ME}{rE^2} = g .
Problem (Page 36): To reduce the gravitational force to 1/2 of its surface value, one must go a distance of h = (\sqrt{2} - 1)rE \approx 0.414 rE \approx 2.64 \times 10^6 \ m above Earth's surface.