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Part 2: 2-D Kinematics and Mechanics
2.1: 2-D Translational Motion
Projectile Motion
Horizontal motion:
Vertical motion:
no acceleration, constant x-component of velocity
free fall
Path is a parabola. The velocity at a point along the path is tangent to the parabola.
Given a launch speed vo and launch angle
o, the launch velocity components are
Example:
A cannon fires a ball with an initial speed of 100 m/s at an angle of 60 with respect to the
horizon. If you can neglect air drag, find (a) the maximum height of the ball, (b) the horizontal
distance traveled by the ball when it reaches this maximum height, and (c) the total horizontal
distance traveled by the ball if it lands at the same height at which it was launched.
Ans. (a) 383 m (b) 442 m (c) 884 m
Example:
Your friend is stranded on a small island. You are cruising at 50 m/s in a small plane in the
horizontal direction at a height of 500 m above the island. You want to drop a care package to
your friend.
(a) If you release the package from rest with respect to you, how far before you reach the
island should you release the package? Neglect air drag in your calculation.
(b) Find the impact speed and impact angle of the package if there is no air drag.
Ans. (a) 505 m (b) 111 m/s at angle of -63
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Uniform Circular Motion
An object travels at a constant speed v along a circular path of radius r.
Quantity
Description
Symbol
mks units
radius
radius of the circular path
r
[m]
period
time for one revolution
T
[s]
frequency
# of revs per time
f
[rev/s]
tangential
velocity
angular speed
the rate that the object travels a distance; the
velocity direction is tangent to the circle
the rate that the object sweeps out an angle
v
[m/s]

[rad/s]
centripetal
acceleration
the size represents the change in the velocity
direction; points towards the center of the circle
ac
[m/s/s]
v
T=1/f
v = 2r / T =2rf = r
r
ac
 = 2 / T = 2f = v / r
ac = v2 / r = 2r = v
Example:
An old “45” vinyl record is meant to spin at 45 rpm. Its diameter is 7 inches (17.78 cm).
Suppose that you glue a pea to the edge of the record and spin the record at this frequency. Find
the pea’s (a) angular speed , (b) tangential speed, and (c) centripetal acceleration .
Now suppose that you glue another pea closer to the center of the record and spin the record.
(a) Is the angular speed of this pea the same as, smaller than, or larger than the angular speed
of the pea at the edge?
(b) Is the tangential speed of this pea the same as, smaller than, or larger than the tangential
speed of the pea at the edge?
(c) Is the centripetal speed of this pea the same as, smaller than, or larger than the centripetal
acceleration of the pea at the edge?
Ans. (a) 4.77 rad/s (b) 41.8 cm/s (c) 196.3 cm/s2 (d) same (e) smaller (f) smaller
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2.2: Translational Mechanics
Force
A force is push or pull exerted by one object on another object.
Force is a vector.
Magnitude: strength of push or pull
Units: mks [Newton = N = kg-m/s2]
U.S. [pound = lb.]
[1 N = 0.2248 lb.]
Direction: direction of push or pull
Newton’s Laws of Motion
1. An object will not change its motion unless acted upon by a nonzero net force.
2. If an object is acted upon by a net force, it will experience an acceleration given by
.
3. If object 1 exerts a force on object 2, then simultaneously object 2 exerts a force on object
1 equal in size and oppositely directed.
Recipe for Using Newton’s Second Law
1. Draw a diagram and identify the object of interest.
2. Draw all of the forces acting on the object as vector arrows.
3. Choose a convenient set of x and y axes. If the object is moving, make the direction of
motion one of the axes.
4. Find expressions for the x and y components of the forces.
5. Apply the law in component form to get two equations:
and
.
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Types of Forces with Examples
1. Weight (near Earth)
Description:
Force of Earth’s gravity pulling down on an object.
Size:
Direction:
W = mg
down (towards Earth’s center)
2. Normal Force
Description:
Pushing force exerted by a surface on an object that is in contact with the surface.
Size:
Direction:
N
perpendicular (normal) to surface
Example:
A 5-kg bock sits atop a table. Find the size of the normal force exerted by the table on the block.
How does the size of the normal force compare to the weight of the block?
Ans. 49 N, equal to the block’s weight
Example:
A 5-kg block slides down a ramp with negligible friction and air drag. The ramp angle, the angle
formed by the ramp surface and the horizon, is 30. Find the acceleration of the block and the
size of the normal force exerted by the ramp on the block. How does the size of the normal force
compare to the weight of the block?
Ans. 4.9 m/s2, down the ramp; 42.4 N, less than the block’s weight
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3. Tension
Description:
Pulling force exerted by a rope or string on an object.
Size:
Direction:
T
a rope can only pull in a direction parallel to the rope
Example:
A 5-kg block is suspended from the ceiling by a single rope. Find the tension in the rope. How
does the size of the tension compare to the weight of the block?
Ans. 49 N, equal to the block’s weight
Example:
A 5-kg block is suspended from the ceiling by two ropes. Rope 1 makes an angle of 30 with
respect to the ceiling. Rope 2 makes an angle of 45 with respect to the ceiling. Find the
tensions in the two ropes. How does the sum of the tensions compare to the weight of the block?
Ans. 36 N in Rope 1 and 44 N in Rope 2; the sum of the tensions is greater than the block’s
weight.
Example:
Consider the sketch below. The blocks are released from rest. Friction and air drag are
negligible. The motion of the pulley is also negligible. (a) Find the size of the acceleration of the
blocks and the tension in the string if m1 = 2 kg and m2 = 5 kg. (b) Recalculate the acceleration
and tension if the block’s switch places.
m1
Ans. (a) 5g/7, 14 N (b) 2g/7, 14 N
m2
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4. Friction

Static Friction
Description:
Force exerted between objects whose surfaces are in contact when the
objects are trying to slide across each other.
Size:
fs
The maximum static friction can reach is sN.
s
N
Direction:

coefficient of static friction [unitless]
normal force
parallel to the surfaces and opposite the intended direction of motion
Kinetic Friction
Description:
Force exerted between objects whose surfaces are in contact when the
objects are sliding across each other.
Size:
fk
fk = kN always
k
N
Direction:
coefficient of kinetic friction [unitless]
normal force
parallel to the surfaces and opposite the direction of motion
NOTE 1: The coefficients of friction are independent of the area of contact and the speed of the
object. They only depend on the types of surfaces in contact.
NOTE 2: When an object is on the verge of slipping, its acceleration is still zero and the static
friction force is at its maximum value so can be placed by sN.
NOTE 3: When an object is sliding, the kinetic friction force can be placed by kN.
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Example:
Consider the situation described in the last example of the Tension section. (a) If m1 = 2 kg and
m2 = 5 kg, what is the smallest that the coefficient of static friction can be if the blocks are to
remain motionless when released? (b) What if the blocks switch places?
Ans. (a) 2.5 (b) 0.4
Example:
A child is trying to pull a 20-kg sled across level ground by means of a rope that makes an angle
of 30 with respect to the ground. The coefficient of static friction between the snow and the
sled runners is 0.3 and the coefficient of kinetic friction is 0.1. (a) How hard does the child have
to pull the rope to get the sled moving? (b) How hard does the child have to pull the rope to keep
the sled moving at a constant speed?
Ans. (a) 57.9 N (b) 21.3 N
Example:
A box is kicked so that it has an initial speed of 3 m/s. It slides 90 cm along the floor before
coming to rest. Find the coefficient of kinetic friction between the box and the floor.
Ans. 0.51
2.3: Universal Gravity
Description:
Size:
A fundamental attractive force exerted between two objects.
(Newton’s Law of Universal Gravitation)
G = 6.672x10-11 N-m2/kg2
(Universal Gravitation Constant)
m = gravitational mass of object
r = distance between gravitational centers of objects
Direction:
attractive along the line joining the gravitational centers
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Acceleration due to Earth’s Gravity
The acceleration due to Earth’s gravity depends on the altitude h of the object. This follows
from the Law of Universal Gravitation. Make m1=ME = 5.98x1024kg (mass of Earth), m2 =m
(mass of object), and r = RE + h where RE = 6380 km (radius of Earth). Then use Newton’s
Second Law to write
so
At the surface of Earth, h = 0 and we know this acceleration is g. Thus, it follows that
Example:
How much weight do you lose if you fly in airplane at an altitude of 20, 000 feet?
Ans. About 0.2%
Example:
At what altitude would you weigh half as much as what you weigh on Earth?
Ans. 2644 km = 1640 miles (Earth’s atmosphere is roughly 75 km thick.)
Weight on Other Planets and Moons
W = mgother where
Example:
How does your weight on the Moon compare to your weight on Earth? The mass of the Moon is
7.35x1012 km and its radius is 1738 km.
Ans. Moon weight is 16.5% of Earth weight.
Example:
How close to the Moon do you have to be so that its gravitational pull on you is cancelled by the
gravitational pull of the Earth? The gravitational centers of the Earth and Moon are about
3.92x105 km apart.
Ans. 39122 km from its center or 37384 km from its surface
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2.4: Circular Motion Mechanics
Centripetal Force
If an object moves along a circular arc, there must be a centripetal force directed towards the
center of the circle that provides the centripetal acceleration.
The centripetal force can be a single force, a component of a single force, or a combination of
forces.
Recipe for Centripetal Force Problems
1. Draw a diagram and identify the object of interest.
2. Draw all of the forces acting on the object as vector arrows.
3. Draw a set of axes. Make one of the axes along the radius of the circle. Call the
direction along this axis towards the circle’s center the positive direction.
4. Find expressions for the x and y components of the forces.
5. Apply Newton’s Second Law Law in component form to get two equations:
and
.
6. Replace the acceleration component along the circle radius with the centripetal
acceleration.
ac = v2 / r = 2r = v .
Example: Gravity as a centripetal force.
Find the period of the Moon’s orbit around the Earth assuming a circular orbit. The gravitational
centers of the Earth and Moon are about 3.92x105 km apart. The masses and radii of the Earth
and Moon appear in earlier examples in the Gravity section.
Ans. about 28 days
Example: Normal force as a centripetal force.
An amusement park ride consists of a large hollow cylinder, oriented
vertically, that spins on a vertical axis as shown. Find an expression for the
minimum tangential speed the person must have so that she remains “glued”
to the wall if the coefficient of static friction is s.
Ans.
r
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Example: Normal force component as a centripetal force.
A variation of the amusement park ride in the previous
example is to have tilted walls instead of vertical walls.
This eliminates the need for static friction between the
person and wall. Verify this. Find the minimum
tangential speed of the rider if she stays in place as shown
without the need of static friction.
r

Ans.
Example: Normal force and gravity as centripetal force
A rope is tied to the handle of a bucket of water. You take hold of the end of the rope and swing
the bucket in a vertical circle. Find the minimum speed of the bucket if no water is to spill out at
the top of the circle.
Ans.
Example: Normal force and gravity as centripetal force
A pilot in a jet plane traveling at 200 m/s (450 mph) performs a loop-the-loop where the radius
of the circle is 2 km. How big is the normal force exerted on the pilot by the seat compared to
the pilot’s weight (a) at the top of the loop when the pilot is upside down, and (b) at the bottom
of the loop?
Ans. (a) equal to the pilot’s weight (b) three times the pilot’s weight
Example: Tension as centripetal force
Consider the situation below. A 2-kg block (M) is
hanging by a string that goes through a hole in a
table top. The string makes a 90 angle and is tied
to a 20-gram ball (m). If the ball is set in motion
on the table top, it will travel in a circle and keep
the block stationary. How fast must the ball travel
if the diameter of the circular path is 30 cm?
Ans. 12.12 m/s = 27 mph
m
M
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2.5: Drag Forces
Description: A force exerted by the particles of a fluid (gas or liquid) on an object moves
through the fluid. Symbolized by R.
Size:
R = bvn
v = speed of object
n = number, typically between 1 and 2 (lower viscosity makes n larger)
b = coefficient that depends on object and fluid
Direction:
opposite the direction of motion
Air Drag
R = bv2
b = (1/2)AD
units [kg/m]
 = density of air [kg/m3]
(~ 1.3 kg/m3)
A = largest cross-sectional area of object perpendicular to the motion [m2]
D = drag coefficient of object [unitless]
Terminal Speed
If an object is falling down, its weight and the drag force are directed oppositely. As the speed
increases, the drag force increases in size. When the drag force equals the weight, there is no
more acceleration and the object reaches its terminal speed, vT.
At terminal speed, bvT2 = mg, so
Example:
A skydiver with a mass of 60 kg steps out of a plane at an altitude of 2500 m.
(a) Assuming that the skydiver maintains a b-coefficient of 0.2 kg/m, find his terminal speed.
(b) Suppose that the parachute doesn’t open. What is the impact speed of the skydiver?
(c) What would be the impact speed if the chute doesn’t open and there was no air drag?
Ans. (a) 54 m/s (120 mph) (b) 54 m/s (120 mph) (c) 221 m/s (495 mph)
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2.6 Non-constant Forces
If one or more forces acting on an object are not constant, there are two options one can use to
analyze the motion.
Option 1 Differential Equation
Apply Newton’s Second Law and write it as a differential equation. Try to solve the differential
equation.
Option 2 Iteration Method
Apply Newton’s Second Law and find the equation for the acceleration.
Break the time up into small intervals of length t.
Assume that the acceleration is constant over a time interval.
Knowing the position and velocity at the beginning of a time interval (time t), use the equations
of motion for constant acceleration to find the position and velocity at the end of the time
interval (time t+ t).
x(t+t) = x(t) + vx(t)t + (1/2) ax(t) t 2
vx(t+t) = vx(t) + ax(t) t