1

Part 2: 2-D Kinematics and Mechanics

2.1: Projectile Motion

Horizontal motion: no acceleration, constant x-component of velocity

Vertical motion: free fall

Path is a parabola. The velocity at a point along the path is tangent to the parabola.

𝑥 = 𝑥𝑜 + 𝑣𝑥𝑜𝑡

𝑣𝑥 = 𝑣𝑥𝑜

𝑦 = 𝑦𝑜 + 𝑣𝑦𝑜𝑡 −1

2𝑔𝑡2

𝑣𝑦 = 𝑣𝑦𝑜 − 𝑔𝑡

𝑣𝑦2 = 𝑣𝑦𝑜

2 − 2𝑔(𝑦 − 𝑦𝑜)

𝑦 = 𝑦𝑜 +1

2(𝑣𝑦𝑜 + 𝑣𝑦)𝑡

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

𝑣𝑥𝑜 = 𝑣𝑜 cos 𝜃𝑜

𝑣𝑦𝑜 = 𝑣𝑜 sin 𝜃𝑜

Given a launch speed vo and launch angle

o, the launch velocity components are

2

2.2: 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

the rate that the object travels a distance; the

velocity direction is tangent to the circle

v [m/s]

angular speed the rate that the object sweeps out an angle [rad/s]

centripetal

acceleration

the size represents the change in the velocity

direction; points towards the center of the circle

ac [m/s/s]

T = 1 / f

v = 2r / T =2rf = r

= 2 / T = 2f = 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

ac

v

r

3

2.3: Force and the Laws of Motion

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

Σ�⃗� = 𝑚�⃗� .

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: Σ𝐹𝑥 = 𝑚𝑎𝑥 and Σ𝐹𝑦 = 𝑚𝑎𝑦.

4

Types of Forces with Examples

1. Weight (near Earth)

Description: Force of Earth’s gravity pulling down on an object.

Size: W = mg

Direction: 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: N

Direction: 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

5

3. Tension

Description: Pulling force exerted by a rope or string on an object.

Size: T

Direction: 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.

Ans. (a) 5g/7, 14 N (b) 2g/7, 14 N

m1

m2

6

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 coefficient of static friction [unitless]

N normal force

Direction: 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 coefficient of kinetic friction [unitless]

N normal force

Direction: 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.

7

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

2.4: Universal Gravitation

Description: A fundamental attractive force exerted between two objects.

Size: 𝐹 = 𝐺𝑚1𝑚2 𝑟2⁄ (Newton’s Law of Universal Gravitation)

G = 6.672x10-11

N-m2/kg

2 (Universal Gravitation Constant)

m = gravitational mass of object

r = distance between gravitational centers of objects

Direction: attractive along the line joining the gravitational centers

8

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 = 5.98x1024

kg (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

𝐹 = 𝐺𝑀𝐸𝑚 (𝑅𝐸 + ℎ)2 = 𝑚𝑎⁄ so

𝑎 = 𝐺𝑀𝐸 (𝑅𝐸 + ℎ)2⁄

At the surface of Earth, h = 0 and we know this acceleration is g. Thus, it follows that

𝑔 = 𝐺𝑀𝐸 𝑅𝐸2⁄

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 𝑔𝑜𝑡ℎ𝑒𝑟 = 𝐺𝑀𝑜𝑡ℎ𝑒𝑟 𝑅𝑜𝑡ℎ𝑒𝑟2⁄

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

9

2.5: 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:

Σ𝐹𝑥 = 𝑚𝑎𝑥 and Σ𝐹𝑦 = 𝑚𝑎𝑦.

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. 𝑣 = √𝑟𝑔/𝜇𝑠

r

10

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.

Ans. 𝑣 = √𝑟𝑔 tan𝜃

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. 𝑣 = √𝑟𝑔

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

r

M

m

11

2.6: 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/m

3)

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

𝑣𝑇 = √𝑚𝑔 𝑏⁄

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)

12

2.7 Torque

A torque can be exerted by one object upon another by application of a force. A net torque

causes a change in the rotational motion of the object.

The size of the force, the point where the force is applied, and the direction of the force are all

important in determining the torque .

Size: = r F sin mks units [N-m]

Direction: Right Hand Rule – Fingers trace r into F, thumb gives direction.

(The size and direction information can be expresses using the vector cross-product, written as

𝜏 = 𝑟 × �⃑� .)