physics - serway 11e ch 07: rotational motion and...

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PHYSICS - SERWAY 11E

CH 07: ROTATIONAL MOTION AND GRAVITATION

a,C = v,T 2 / r

v,T = __________ = __________

UNIFORM CIRCULAR MOTION

● In Uniform Circular Motion, an object moves with constant speed in a circular path.

v,T = ______________________________

a,C = ______________________________

r = ______________________________

● When an object completes one lap (__________________ or ___________), it covers a distance of _____ = _________.

- Time for one cycle ____________ (___) in [____]

- Inverse of Period ____________ (___) in [____]

Period is seconds/cycle, frequency is cycles/second. RPM: Revs per Minute: f = RPM / 60

EXAMPLE 1: Calculate the period, frequency, and speed of an object moving in uniform circular motion (radius 10 m) if:

(a) it completes 100 cycles in 60 seconds;

(b) it takes 3 minute to complete 1 cycle.

EXAMPLE 2: The car below takes 10 s to go from A to B, at constant speed. If the semi-circle has radius of 5 m, find its:

(a) period; (b) tangential velocity; (c) centripetal acceleration.

NOTE: Even though the object’s speed is constant, its direction changes, therefore its velocity changes and _________.

Constant speed, but NON-ZERO centripetal acceleration (tangential velocity changes direction).

B A



MORE: UNIFORM CIRCULAR MOTION

PRACTICE 1: A Ping-Pong ball goes in a horizontal circle (radius 5 cm) inside a red cup twice per

second. Find its: (a) period; (b) speed; (c) centripetal acceleration.

EXAMPLE 1: One way to simulate gravity (or “create artificial gravity”) in a space station is to spin it. If a cylindrical space

station (diameter = 500 m) is spun about its central axis, at how many revolutions per minute (rpm) must it turn so that the

outermost points have acceleration equal to the acceleration due to gravity at the surface of the Earth?

PRACTICE 2: A 3kg rock spins horizontally at the end of a 2-m string at 90 rpm. Calculate its: (a) speed; (b) acceleration.

U. CIRCULAR MOTION

a,C = v,T 2 / r

v,T = 2 π r / T = 2 π r f

f = 1 / T = RPM / 60



X _____

ΔX _______

ΔX = ________

( __________ )

________ = _____

ROTATIONAL POSITION & DISPLACEMENT

● Rotational Motion is motion around a ____________ point, that is, in a _______________ path.

- The rotational equivalent of linear POSITION ( ___ ) is Rotational/Angular position ( ___ ).

LINEAR POSITION ROTATIONAL POSITION

- How far you are from the ____________.

- Measured in ______________.

- Origin is where ____________.

- Origin is _______________.

- Direction (+/-) is ______________.

- How far you are from the _____________.

- Measured in ______________.

- Origin is where ____________.

- Origin is _______________:

- Always at the ___________.

- Direction (+/-) is ______________:

CW CCW

● The rotational equivalent of linear DISPLACEMENT ( ____ ) is Rotational Displacement ( ____ ).

- These two quantities are “LINKED” by an equation (and r = radial distance, “radius”):

- This equation “speaks” ____________. Input must be in radians. Output will be in radians.

- One radian is approximately 57o. To convert between radians and degrees, use:

EXAMPLE: An object moves along a circle of radius 10 m from 30o above the positive x-axis to 120o above the +x-axis.

Calculate the object’s (a) angular displacement, and (b) linear displacement.



DISPLACEMENT IN MULTIPLE REVOLUTIONS

● If you make one full revolution around a circle: ΔΘ = ______ = ______ ΔX = ________ = _________.

- If you make N full revolutions: ΔΘ = ______ = ______ ΔX = ________.

- To find out how many revolutions you go through, simply divide ΔΘ by __________ or __________.

- To find out how far from 0o you end up after many revolutions, subtract by 360o until Θ < 360o (or Θ < 2 π).

EXAMPLE: Starting from 0o, you make two 2.2 revolutions around a circular path of radius 20 m. (a) What is your rotational

displacement, in degrees? (b) How many degrees away from 0o are you? (c) What is your linear displacement?



PRACTICE: ROTATIONAL DISPLACEMENT

PRACTICE: While you drive, your tires, all of radius 0.40 m, rotate 10,000 times. How far did you drive, in meters?




PRACTICE: An object moves a total distance of 1,000 m around a circle of radius 30 m.

How many degrees does the object go through? BONUS: How many complete revolutions does it make?




PRACTICE: A car travels a total of 2,000 m and 1140o around a circular path, starting from 0o.

What is the radius of the circular path? BONUS: How far (in degrees) from 0o does the car end up?



ROTATIONAL VELOCITY & ACCELERATION

● The rotational equivalents of linear velocity and acceleration are Rotational Velocity and Rotational Acceleration:

v,AVG = ____________ [ _______ ] _______ = ____________ [ ________ ]

a = ____________ [ _______ ] _______ = ____________ [ ________ ]

● There are 3 additional variables that describe how quickly something rotates (similar to w). They are all related:

w = _________ = _________ = _________ 1 RPM = __________

( ___________ ) 1 Hz = __________

- Often we will convert from any of these three back to _____

● Note that rotational equations work for both:

(1) Points Masses ( ________ ) moving in a circular path; or

(2) Rigid Body/Shape ( ________ ) rotating around themselves.

EXAMPLE 1: A 30-kg disc of radius 2 m rotates at a constant 120 RPM. Calculate its (a) period, (b) angular speed.

EXAMPLE 2: Calculate the rotational velocity for the Earth as it (a) rotates around itself, (b) rotates around the Sun.



PRACTICE: ROTATIONAL VELOCITY & ACCELERATION

PRACTICE: Calculate the rotational velocity (in rad/s) of a clock’s minute hand.

EXTRA: Calculate the rotational velocity (in rad/s) of a clock’s hour hand.



PRACTICE: ROTATIONAL VELOCITY & ACCELERATION

PRACTICE: A wheel of radius 5 m accelerates from 60 RPM to 180 RPM in 2 s. Calculate its angular acceleration.



MOTION EQUATIONS FOR ROTATION

● Just like in linear motion, there are 4* equivalent MOTION equations for Rotation. Same equations, different letters.

- You often use these when given a lot of rotational quantities. Same process: List variables, pick equation, solve.

LINEAR EQUATIONS ROTATIONAL EQUATIONS

vf = vi + a t

vf 2 = vi 2 + 2 a Δx

Δx = vi t + ½ a t 2

Δx = ½ (vi + vf) t *

EXAMPLE 1: A wheel initially at rest accelerates around its central axis, with a constant 4 rad/s2 until it reaches 80 rad/s.

(a) By the time it reaches 80 rad/s, how many degrees will it have rotated through? (b) How long (in s) does this take?

EXAMPLE 2: A very heavy disk, 20 m in radius, takes 1 hour to make a complete revolution, accelerating from rest at a

constant rate. What rotational velocity will the disk have 1 hour after it starts accelerating?



PRACTICE: MOTION EQUATIONS FOR ROTATION

PRACTICE: A tiny object spins with 5 rad/s around a circular path of radius 10 m. The object then accelerates at 3 rad/s2. Calculate its angular speed 8 s after starting to accelerate. BONUS: Calculate its linear displacement in the 8 s.



PRACTICE: MOTION EQUATIONS FOR ROTATION

PRACTICE: The turntable of a DJ set is spinning at a constant rate just before it is turned off. If the turntable decelerates at

3 rad/s2 and goes through an additional 30 rotations before stopping, how fast (in RPM) was the turntable initially spinning?

BONUS: How long (in seconds) does the turntable take to stop?



CONVERTING BETWEEN LINEAR AND ROTATIONAL

● There are tiny equations that “LINK” linear (aka tangential) and rotational (aka angular) variables:

LINEAR ROTATIONAL “LINK”

x Θ --

Δx = x – xo ΔΘ = Θ – Θo Δx = r ΔΘ

v = Δx / Δt w = ΔΘ / Δt v,T =

a = Δv / Δt α = Δw / Δt a,T =

- There are 4 types of acceleration. The equation a,T = ______ refers to ______________ acceleration. More soon!

- When a Shape/Rigid Body rotates around itself, ALL rotational quantities ( ΔΘ, w, α ) are the same at every point.

- Linear speeds (v,T = r w) may be different, since they depend on ________________________________.

EXAMPLE 1: A wheel of radius 8 m spins around its central axis at 10 rad/s. Find the angular AND linear speeds at a point:

(i) at the middle of the wheel (on its central axis);

(ii) at a distance of 4 m from the wheel’s center;

(iii) at the edge of the wheel.

EXAMPLE 2: A small object rotates at the end of a light string. The object reaches 120 RPM from rest in just 4 seconds . If

the object’s tangential acceleration after the 4 seconds is 15 m/s2, calculate the length of the string.



PRACTICE: CONVERTING BETWEEN LINEAR AND ROTATIONAL

PRACTICE: A disc of radius 10 m rotates around itself with a constant 180 RPM. Calculate the linear speed at a point 7 m

from the center of the disc.



PRACTICE: CONVERTING BETWEEN LINEAR AND ROTATIONAL

PRACTICE: A rock rotates around a light, 4-m long string. The rock is initially at rest, but reaches 150 RPM in 3 seconds.

Calculate its tangential acceleration after 3 s. BONUS: Calculate its tangential speed after 3 s.



PRACTICE: ROTATIONAL KINEMATICS

PRACTICE: A 4 m long blade initially at rest begins to spin with 3 rad/s2 around its axis, which is located at the middle of the

blade. It accelerates for 10 s. Find the tangential speed of a point at the tip of the blade 10 s after it starts rotating.



- Centripetal / Radial (linear) ________

- Tangential (linear) ________

- Total / Acceleration (linear) ________

- Rotational / Angular ________

TYPES OF ACCELERATION IN ROTATION

● There are FOUR types of acceleration in rotation problems

- BUT some exist only if you’re accelerating (spinning faster):

- You always have v T, a C (aka a RAD), and w: - IF accelerating: you also have a T and α:

- a,C __________________________________. - a,T (and α) __________________________________.

- The equation a,T = r α is a way to remember that a,T and α are connected. If one is zero, the other has to be zero.

- Note that IF a,T = 0, then a = _________________________ becomes a = ___________________ = a,C.

EXAMPLE 1: A carousel 10 m in radius completes one cycle every 45 s. A boy stands at the edge of the carousel. Find his:

(a) Tangential velocity

(b) Angular acceleration

(c) Radial acceleration

(d) Tangential acceleration

(e) Total linear acceleration

EXAMPLE 2: A carousel 16 m in radius accelerates from rest with 0.05 rad/s2. A boy stands at the edge of the carousel.

After the carousel has accelerated for 10 s, calculate the boy’s:

(a) Tangential velocity

(b) Tangential acceleration

(c) Radial acceleration

(d) Angular acceleration

(e) Total linear acceleration




PRACTICE: A large disc of radius 10 m initially at rest takes 200 full revolutions to reach 30 RPM. Calculate the total linear

acceleration of a point at half way between the disc’s center and its edge, once the disc reaches 30 RPM.

(You may assume it continues accelerating past that point)




PRACTICE: An object of negligible size moves in a circular path of radius 20 m with 90 RPM. Find its radial acceleration.



ROLLING MOTION (FREE WHEELS)

● So far we have seen Point Masses in a circular path OR Shapes/Rigid Bodies around themselves “FIXED WHEEL”

- In some problems, Shapes/Rigid Bodies are BOTH Rotating ( ___ ) AND MOVING ( ___ ) “FREE WHEEL”

- Cylinder rotating around FIXED axis: - Cylinder rotating around FREE axis:

w _____ 0 BUT v,CM _____ 0 w _____ 0 AND v,CM _____ 0

- If FREE Axis, the total velocity (linear) at the center of mass (usually middle), top, and bottom of the wheel are:

EXAMPLE 1: A wheel of radius 0.30 cm rolls without slipping along a flat surface with 10 m/s. Calculate (a) the angular

speed of the wheel; and (b) the speed of a point at the bottom of the wheel, relative to the floor.

EXAMPLE 2: When a car accelerates from rest for 10 s, its tires experience 8 rad/s2. The tires are 0.40 m in radius.

Calculate: (a) the angular speed of the tires after 10 s; (b) the speeds of points at the top, center, and bottom of the tire.



PRACTICE: ROLLING MOTION

PRACTICE: A long, light rope is wrapped around a cylinder of radius 40 cm, which is at rest on a flat surface, free to move.

You pull horizontally on the rope, so it unwinds at the top of the cylinder, causing it to begin to roll without slipping. You keep

pulling until the cylinder reaches 10 RPM. Calculate the speed of the rope at the instant the cylinder reaches 10 RPM .



CONNECTED WHEELS (STATIC)

● Problems where two wheels (discs, cylinders, etc.) are connected are fairly common in Rotational Kinematics.

- In some cases, the wheels rotate around a FIXED Axis (w BUT no v,CM) fixed pulleys/gears, static bicycles

- In others, the wheels rotate around a FREE Axis (w AND v,CM) bicycle moving on the ground

- Whenever a chain connects two wheels, we have ____________, which yields a set of 4 related equations:

EXAMPLE 1: Two gears (R1 = 2 m, R2 = 3 m) are free to rotate about fixed axes, with a light chain that runs around them,

so they spin together (as above). When you give the smaller pulley 40 rad/s, what angular speed will the larger one have?

EXAMPLE 2: Two pulleys, with radii 0.3 m (left) and 0.4 m (right) attached to a table as shown below. A light cable runs

through the edge of both pulleys, with one end connected to a mass. You pull down on the other end, causing the pulleys to

spin (and the block to move up). When the cable has speed 5 m/s, what will the angular speed of each pulley be?

m



CONNECTED WHEELS: BICYCLES (STATIC)

● Bicycle problems are a bit more complicated than static pulleys/gears, because there are more parts (5 in total):

- The pedals (1) cause the middle sprocket (2) to spin

- The chain connects both sprockets

- The back sprocket (3) is connected to the back wheel (4)

- If the bike is NOT free to move (wheels don’t touch ground), the front wheel (5) doesn’t move or spin (v5 = w5 = 0).

EXAMPLE 1: You turn your bicycle upside down for maintenance. The middle and back sprockets have diameters 16 cm

and 10 cm, respectively. You spin the pedals at 8 rad/s. Calculate the resulting angular velocity for the:

(a) middle sprocket

(b) back sprocket

(c) back wheel

(d) front wheel

EXAMPLE 2: You lift your bicycle slightly and begin to spin its back wheel. The middle and back sprockets have diameters

2D and D, respectively. If you spin the back wheel at X RPM, at how many RPM (in terms of X), will the pedals spin?



CONNECTED GEARS: MOVING BICYCLES

● Remember: If the bike doesn’t move when the wheels spin (FIXED Axis) v,front = w,front = ____.

- If the bike IS moving (FREE Axis: w AND v), you also have that v,CM,front = v,CM,back = v,BIKE

- Remember: For Free Axis, we have

- For most bicycles R,front = R,back

EXAMPLE 1: The wheels on your bike have radius of 66 cm. If you ride with 15 m/s, calculate the: (a) linear speeds at the

center of mass of both wheels; and (b) the angular speeds of both wheels.

EXAMPLE 2: The wheels on your bike have radius 70 cm. The middle and back sprockets on your bike have radii 15 cm

and 8 cm, respectively. If you ride with 20 m/s, calculate the angular speed of the:

(a) front wheel

(b) back wheel

(c) back sprocket

(d) middle sprocket



CENTRIPETAL FORCES

● In linear motion, we have forces in the X & Y axes. Now, we’ll have forces in the ___________________ axis.

- Before, we had ΣFX = maX and ΣFY = maY. Now, we have ______________ (remember a,C = v,T 2 / r)

- When writing ΣFC, forces towards the center are ___, forces away are ___, and tangential forces don’t get listed:

EXAMPLE 1: A small 3 kg object on top of a frictionless table is attached to the end of a 2 m string, as shown. If the object

spins once every 2 seconds, calculate the tension on the string.

PRACTICE 1: For the situation above, suppose the string breaks if its tension exceeds 50 N.

Calculate the maximum speed that the object can attain without breaking the string.

EXAMPLE 2: Some crazy fighter pilot (70 kg) in some movie does a nose dive that is nearly circular (radius 300 m). If his

speed at the bottom of the dive is 80 m/s, find his: (a) centripetal acceleration; (b) apparent weight.

NOTE: Centripetal Force is not a force in nature, but simply an indication that a force acts in the centripetal direction.

U. CIRCULAR MOTION

a,C = v,T 2 / r

v,T = 2 π r / T = 2 π r f

f = 1 / T = RPM / 60



CENTRIPETAL FORCES: VERTICAL

EXAMPLE: A Ferris Wheel of radius 50m takes 30 seconds to make a full cycle. An 80 kg guy rides on it. Calculate his:

(a) speed and centripetal acceleration; (b) apparent weight at the bottom; (c)* apparent weight at the top.

EXAMPLE: What is the minimum speed that a rollercoaster cart can have at the top of a vertical loop of radius 10 m so that

the passengers won’t fall, even at the absence of restraints (eg. seat belts)?

Similar question: Find v,MIN for a bucket in a vertical loop so that water doesn’t fall while the bucket is at the top.

PRACTICE: A pendulum is made from a light, 2 m-long rope and a 5-kg small object. When you release the object from rest

as of a certain height, it swings from side to side, attaining a maximum speed of 10 m/s. At the object’s lowest point:

(a) Draw a Free Body Diagram. (b) Find the magnitude of its acceleration. (c) Find the tension on the rope.



CENTRIPETAL FORCES: FLAT & BANKED CURVES

EXAMPLE 1: Find the maximum speed that a 800 kg car can have while going around a flat curve of radius 50 m (without

slipping) if the coefficient of friction between the car and the road is 0.5.

EXAMPLE 2: (a) Find the maximum speed that a 800 kg car can have while going around a banked, frictionless curve of

radius 50 m that makes an angle of 37o with the horizontal. What would happen if the car moves: (b) slower; (c) faster?

PRACTICE 1: You are designing a highway curve to allow

cars to turn, without any banking, at a maximum speed of 50

m/s. The average coefficient of friction between cars and

asphalt, for dry roads, is roughly 0.7. What radius would this

curve have to have, for this to be possible?

PRACTICE 2: For the radius you just found, how much

would you have to bank the same curve, in order to attain

the same maximum speed, but at the absence of friction?



FG = _________

CONCEPT: Universal Law of Gravitation

● Newton’s Universal Law of Gravitation: All objects in the universe _______________________.

- Universal Gravitation Constant (G) = ______________ [ 𝐦𝟑

𝐤𝐠⋅𝐬𝟐 ]

- Not little g! ← local constant

- r is the distance between _______________________.

- Gravitational forces are directed along ______________ connecting 2 objects.

EXAMPLE: Two 30-kg spheres are separated by 5m. What is the gravitational force between them?



PRACTICE: Two spheres are separated by 10m. If the lighter 40kg sphere feels a gravitational force of 1.6×10-9 N, what is the mass of the heavier sphere? EXAMPLE: Two spheres of mass 10kg and 25kg are positioned 5 m apart. Suppose you place a third sphere directly in between the two spheres, so that all objects were on the same line. How far from the 10kg mass would you have to place this third sphere so the net gravitational force on it was zero? PRACTICE: Two spheres of mass 300kg and 500kg are placed in a line 20cm apart. If another sphere of mass 200kg is placed between them, 8cm from the 300kg-sphere, what is the net gravitational force on the 200-kg sphere?



Distance r = ________

R = _________ h = __________

CONCEPT: Center-of-Mass Distance

● From the Universal Law of Gravitation, “r” is center-of-mass distance between 2 objects. What if 1 object is really big?

- Capital letters → __________

- Lowercase letters → __________

EXAMPLE: At what height above Earth is the gravitational force on a 1000-kg satellite equal to 1000N?

● Pro Tip: When looking for R or h, solve for r first, then use r = R + h.

Point Masses

𝐅𝐆 =𝐆𝐦𝟏𝐦𝟐

𝐫𝟐

Planets (Large objects)

𝐅𝐆 =

GRAVITATIONAL CONSTANTS

G = 6.67×10-11 𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m



PRACTICE: A 2,000-kg spacecraft is blasting away from the surface of an unknown planet the same size as the Earth. At 1500km above the surface, an instrument onboard reads the gravitational force to be 18000 N. What is the planet’s mass?

EXAMPLE: Two identical solid spheres of mass 10kg and 60cm in diameter rest on a table, with their surfaces touching. Calculate the gravitational force between them.

GRAV. CONSTANTS

G = 6.67×10-11 𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg RE = 6.37×106 m



CONCEPT: Gravitational Forces in 2D

● To solve for net forces in non-linear arrangements, we must use ________________________.

- Remember that gravity is a force/vector, so we can break it up into its _________________.

1D 2D

EXAMPLE: Calculate the magnitude and direction of the net gravitational force on m1 in the figure. Assume point masses.

Point Masses


𝐫𝟐



CONCEPT: Using Symmetry in 2D Gravitation

● When given equal masses & distances, use symmetry to cancel out vector components.

- Same m’s, r’s → same ________

- Same FG, Θ → same ___________ → cancel if opposite!

EXAMPLE: In the figure below, what is the net gravitational force on mass m if it feels a 5N force from each M on the right?

VECTOR EQUATIONS

Fx = F cos(θ) Fy = F sin(θ)

F = √𝐹𝑥 2 + 𝐹𝑦

2

𝜃 = tan−1 (𝐹𝑦

𝐹𝑥

)



CONCEPT: Finding Net Forces in 2D Gravitation

● To solve 2D Gravitation problems, combine Newton’s Law of Gravity (FG), vector addition, and symmetry. EXAMPLE: Three 50-kg masses are arranged in an equilateral triangle with side length 0.6m. Find the magnitude and

direction of the net gravitational force on the bottom mass. (Equilateral triangles have 60° angles between their sides.)

STEPS FOR 2D GRAV.

1) Label Forces

2) Calculate Forces

3) Decompose & Symmetry

4) Add Components → Fnet



M M

CONCEPT: Acceleration Due to Gravity

● Use Newton’s Law of Gravity to determine the acceleration due to gravity (ag → g) at different distances from a planet.

Any Distance On Surface

m

h m

r

R R

r

- Use g when specifically given/asked for __________. Use gsurf when on the ____________. (“surface gravity”)

- Note that both g’s only depends on (M | m)

- gsurf is a local constant, g decreases as r _____________.

● Your weight at any distance from a planet is the force of gravity → W = FG = GMm

r2 = ________.

- On the surface, W = _______.

EXAMPLE: Compare the exact acceleration due to gravity on the top of Mount Everest, which has a height of 8.85km, with

the surface gravity of the Earth.

GRAV. CONSTANTS

G = 6.67×10-11 𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m

g = _______

gsurf = _______



PRACTICE: You stand on the surface of a mysterious planet with a mass of 6×1024 kg and measure the surface gravity to be 7 m/s2. What must the radius of the planet be? EXAMPLE: An astronaut drops a rock from rest on the surface of an unknown planet. It takes 0.6 seconds to fall 1.5m. If the radius of this unknown planet is 4×106 m, what is the mass?

PRACTICE: How far would you have to be above Earth’s surface for g to be ½ of its surface value?

EQUATIONS GRAV. CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡

𝐠 =𝐆𝐌

𝐫𝟐 𝐠𝐬𝐮𝐫𝐟 =𝐆𝐌

𝐑𝟐

G = 6.67×10-11 𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg RE = 6.37×106 m



𝐫𝟐 𝐫 = 𝐑 + 𝐡

𝐠 =𝐆𝐌


𝐑𝟐

G = 6.67×10-11 𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg RE = 6.37×106 m



𝐫𝟐 𝐫 = 𝐑 + 𝐡

𝐠 =𝐆𝐌


𝐑𝟐

G = 6.67×10-11 𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg RE = 6.37×106 m



CONCEPT: Gravitational Potential Energy

● In Gravitation, we use a different equation for Gravitational Potential Energy.

EXAMPLE: An asteroid at rest falls to Earth from a distance of 6×107 m from Earth’s center. What will be its impact speed?

- There’s still some Grav. Potential Energy at the surface!

−𝐆𝐌𝐦

𝐫→ −

𝐆𝐌𝐦

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m

𝐠𝐬𝐮𝐫𝐟 =𝐆𝐌

𝐑𝟐 𝐠 =

𝐆𝐌

𝐫𝟐

𝐔𝐆 = −𝐆𝐌𝐦

𝐫

𝐊𝐢 + 𝐔𝐢 + 𝐖𝐍𝐂 = 𝐊𝐟 + 𝐔𝐟

Grav. Potential Energy (Simple)

𝐔𝐆 = 𝐦𝐠𝐡

Grav. Potential Energy (Gravitation)

𝐔𝐆 =

● Alternative for solving kinematics problems

- g is [ constant | changing ]

● ____________ use kinematics!

- As r changes, g is [ constant | changing ]



PRACTICE: How much energy is required to move a 1000-kg object from Earth’s surface to a height twice Earth’s radius?

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m


𝐑𝟐 𝐠 =

𝐆𝐌

𝐫𝟐


𝐫




EXAMPLE: Two identical small planets of mass 7×1022 kg and radius 2×106 m are initially at rest 5×1010 m apart. What is

the speed of each planet when the two eventually collide?

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m


𝐑𝟐 𝐠 =

𝐆𝐌

𝐫𝟐


𝐫




PRACTICE: You launch a rocket with an initial speed of 5×103 m/s from Earth’s surface. At what height above the Earth will

it have ¼ of its initial launch speed? Assume the rocket’s engines shut off after launch.

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m


𝐑𝟐 𝐠 =

𝐆𝐌

𝐫𝟐


𝐫




CONCEPT: Escape Velocity

● Escape velocity is the minimum launch speed for an object to escape → object stops very far away & never returns.

- When an object goes really far away, FG → _____

- When an object stops, 𝐯𝐟 = ____

- If 𝐯𝐟 > 0, then it wasn’t the minimum launch speed.

● Escape Velocity comes from Conservation of Energy:

- vesc only depends on big mass M and initial distance r.

EXAMPLE: How fast must a 5kg rock be thrown to escape the Sun, if thrown directly outward near Earth's orbital distance?

Final (farrrr) KF = _______

UF = _______

Initial

Ki = 1

2mvi

2

Ui = −𝐆𝐌𝐦

𝐫𝐢

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

rSun-Earth = 1.5×1011 m

MSun = 2×1030 kg

RSun = 6.96×108 m


𝐯𝐞𝐬𝐜 = √𝟐𝐆𝐌

𝐫

vesc = ________



PRACTICE: a) How fast does a spaceship have to go to escape Earth, starting from the launch pad on the surface?

Assume it burns all its fuel very fast and then shuts off the engines. b) One idea to make getting to space easier is to build a

space elevator, a large platform high above Earth’s surface where spaceships can land and take off. How high above Earth

would this platform have to be for the escape velocity to be 1/5 of its surface value?

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m

𝐊𝐢 + 𝐔𝐢 = 𝐊𝐟 + 𝐔𝐟


𝐫



EXAMPLE: You land on the surface of a large, spherical asteroid. You set off walking in one direction and you realize

you’ve arrived back to your spaceship after walking 25km. You grab a tool from your toolbox and drop it, noting it takes

about 30 seconds to hit the ground from 1.4 m high. How fast would you have to jump to escape this asteroid?

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m


𝐑𝟐 𝐠 =

𝐆𝐌

𝐫𝟐

𝐊𝐢 + 𝐔𝐢 = 𝐊𝐟 + 𝐔𝐟


𝐫



CONCEPT: Intro to Satellite Motion

● A satellite is any object that orbits another. Examples: (1) Moon around Earth (2) Earth around Sun.

- The shape of an orbit depends on its _____________ and _____________.

1) ____________ 2) ____________ 4) ______________

3) ____________

- These values change as __________ changes!

- Assume circular orbits (simpler), unless told otherwise.

EXAMPLE: You stand on a large tower while exploring a mysterious planet. From that height, the minimum speed to go

around the planet without crashing is 2000m/s, the circular orbit speed is 5000m/s, and the escape speed is 10,000m/s.

Predict the shape of the orbit for the following launch velocities: a) 1,500 m/s ; b) 4000 m/s ; c) 6,000 m/s ; d) 15,000 m/s



CONCEPT: Kepler’s Laws

● Kepler noticed all orbits (stars, planets, satellites) obeyed three laws:

1) Kepler’s First Law: All orbits (even circular) are _____________ with the Sun at one focus.

- No planet or physical object at center or other focus.

- Eccentricity → # between 0 and 1, how _____________ the orbit is.

- Near 0 is very (circular/elliptical)

- Near 1 is very (circular/elliptical)

2) Kepler’s Second Law: In an orbit, equal areas of the orbit are swept out in equal times.

3) Kepler’s Third Law: The square of the period “T” is proportional to the cube of the radius “r” of the orbit.

- For 2 satellites orbiting mass M, the ratio ______ is constant.



CONCEPT: Kepler’s First Law

● All orbits are ______________ (even circular ones) with the Sun at one focus. Nothing physical at other focus.

● Eccentricity of orbit (e) → # between 0 and 1, measures how _________________ the orbit is.

- Lower eccentricities (near 0) are nearly (circular / elliptical).

- Higher eccentricities (near 1) are very (circular / elliptical).

- Eccentricity relates the aphelion & perihelion with the semi-major axis:

EXAMPLE: Earth’s closest distance to the Sun is 1.471×1011m, while its farthest distance is 1.521×1011m.

Calculate a) Earth’s semi-major axis and b) orbital eccentricity.

ELLIPTICAL ORBITS

𝐚 =𝐑𝐚+𝐑𝐩

𝟐

𝐑𝐚 = 𝐚(𝟏 + 𝐞)

𝐑𝐩 = 𝐚(𝟏 − 𝐞)

● Major axis → ________ axis (Length: )

- Closest distance: Perihelion/periapsis

- Farthest distance: Aphelion/apoapsis

- Semi-major axis: 𝐚 = _______

● Minor axis → ________ axis (Length: )

Rp = ____________

Ra = ____________



CONCEPT: Kepler’s Third Law

● For any circular orbit, the orbital period (T) squared is proportional to the orbital radius (r) cubed.

- The relationship between r and T depends only on (M | m).

- If you’re given 2 out of 3 (M, r or h, T) use Kepler’s 3rd Law!

● For any 2 objects orbiting the same mass M, the ratio r3

T2 = ______________.

- Units can be non-SI when comparing, as long as they are consistent.

EXAMPLE: The Earth orbits the Sun once a year at a distance of 150 million kilometers, while Mars orbits every 687 days at 228 million kilometers. Calculate the mass of the Sun using a) Earth and b) Mars. Do you get the same number?

KEPLER’S 3RD LAW

𝐓𝐬𝐚𝐭𝟐 =

𝟒𝛑𝟐𝐫𝟑

𝐆𝐌

T2 = _________



EXAMPLE: Jupiter orbits once every 11.86 years and orbits at a distance of 5.2 AU, where 1 AU is the average distance

between Earth and the Sun. If Neptune’s orbital distance from the Sun is 30.11 AU, how long does it take to complete its

orbit (in years)? Calculate without using MSUN.

PRACTICE: Io and Ganymede are two of Jupiter’s four Galilean moons. Io orbits at an average distance of 422,000km in 1.77 days. What is Ganymede’s average orbital distance (in km), if it takes 4 times longer to orbit Jupiter?



𝐫𝟐 𝐫 = 𝐑 + 𝐡

𝐠 =𝐆𝐌

𝐫𝟐 𝐠𝐬𝐮𝐫𝐟 =

𝐆𝐌

𝐑𝟐



𝐆𝐌

G = 6.67×10-11 𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m



𝐫𝟐 𝐫 = 𝐑 + 𝐡

𝐠 =𝐆𝐌


𝐑𝟐



𝐆𝐌

𝐫𝟏 𝟑

𝐓𝟏 𝟐

= 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 =𝐫𝟐

𝟑

𝐓𝟐 𝟐

G = 6.67×10-11 𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m



vsat = __________

CONCEPT: Velocity of a Satellite

● For a satellite in circular orbit, the gravitational force keeps it in _____________________________.

- The orbital speed & distance are related by:

- M = mass of planet

- r = orbital distance (not height!)

- For every value of r, there is an exact speed required to maintain circular orbit.

EXAMPLE: Calculate the height of the International Space Station, which orbits at 7,670 m/s in a nearly circular orbit.

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m 𝐠𝐧𝐞𝐚𝐫 =𝐆𝐌

𝐑𝟐 𝐠𝐟𝐚𝐫 =

𝐆𝐌

𝐫𝟐

𝐯𝐬𝐚𝐭 = √𝐆𝐌

𝐫



PRACTICE: Suppose that you used some geometry and kinematics to estimate that the Earth goes around the Sun with an orbital speed of approximately 30,000 m/s (60,000 mph), and that the Sun is approximately 150 million kilometers away from the Earth. Use this information to estimate the mass of the Sun. EXAMPLE: Two satellites are in circular orbits around a mysterious planet that is 8×107 m in diameter. The first satellite has mass 68kg, orbital radius 6×108 m, and orbital speed 3000 m/s, while the other has mass 84 kg, orbital radius 9×108 m/s. What is the orbital speed of this second satellite? PRACTICE: You throw a baseball horizontally while on the surface of a small, spherical asteroid of mass 7×1016 kg and diameter of 22km. What is the minimum speed so that it just barely goes around the asteroid without hitting anything?

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m 𝐠 =𝐆𝐌


𝐆𝐌

𝐑𝟐


𝐫

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg RE = 6.37×106 m 𝐠 =

𝐆𝐌


𝐆𝐌

𝐑𝟐


𝐫

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg RE = 6.37×106 m 𝐠 =

𝐆𝐌


𝐆𝐌

𝐑𝟐


𝐫



vsat = _______

T2 = _______

CONCEPT: Orbital Period of a Satellite

● We can also relate the orbital speed and orbital period T, the time it takes to complete 1 orbit, using circular motion.

- This equation is also called ___________________.

- Orbital speed, period, distance all interdependent. As distance increases, v _____________, T _____________.

EXAMPLE: What is the orbital period and speed of the International Space Station orbiting 400km above Earth’s surface?

SATELLITE MOTION


𝐫

SATELLITE MOTION GRAV. CONSTANTS


𝐫 𝐯 =

𝟐𝛑𝐫

𝐓



𝐆𝐌

G = 6.67×10-11

ME = 5.97×1024 kg

RE = 6.37×106 m

r

v

Fg



PRACTICE: A satellite orbits at an orbital period of 2 hours around the Moon. What is the satellite’s orbital altitude?

EXAMPLE: A small moon orbits its planet with a speed of 7500 m/s. It takes 28 hours to complete 1 full orbit. What is the

mass of this unknown planet?

PRACTICE: A distant planet orbits a star 3 times the mass of our Sun. This planet of mass 8x1026 kg feels a gravitational

force of 2x1026 N. What is this planet’s orbital speed and how long does it take to orbit once?

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m

MMoon = 7.35×1022 kg

RMoon = 1.74×106 m

𝐠𝐧𝐞𝐚𝐫 =𝐆𝐌


𝐆𝐌

𝐫𝟐


𝐫 𝐯𝐬𝐚𝐭 =

𝟐𝛑𝐫

𝐓



𝐆𝐌

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m

MMoon = 7.35×1022 kg

RMoon = 1.74×106 m



𝐆𝐌

𝐫𝟐



𝟐𝛑𝐫

𝐓



𝐆𝐌

EQUATIONS CONSTANTS


𝐫𝟐 𝐫 = 𝐑 + 𝐡 G = 6.67×10-11

𝐦𝟑

𝐤𝐠⋅𝐬𝟐

ME = 5.97×1024 kg

RE = 6.37×106 m

MMoon = 7.35×1022 kg

RMoon = 1.74×106 m



𝐆𝐌

𝐫𝟐



𝟐𝛑𝐫

𝐓



𝐆𝐌



CONCEPT: Geosynchronous Orbit

● Geosynchronous orbit → satellite’s orbital period synchronizes with Earth’s rotation: _______ = ________

- The satellite stays above the same place on the surface!

- This is the only distance where a circular geosynchronous orbit is possible.

EXAMPLE: What is the height of Earth’s geosynchronous orbit?

SATELLITE MOTION



𝟐𝛑𝐫

𝐓



𝐆𝐌 𝐓𝐬𝐚𝐭 =

𝟐𝛑𝐫

𝐯𝐬𝐚𝐭



𝐫 𝐯 =

𝟐𝛑𝐫

𝐓




𝟐𝛑𝐫

𝐯𝐬𝐚𝐭

𝐫𝐬𝐲𝐧𝐜 = √𝐆𝐌𝐓𝐩

𝟐

𝟒𝛑𝟐

𝟑

G = 6.67×10-11

ME = 5.97×1024 kg

RE = 6.37×106 m

rsync = __________



EXAMPLE: Calculate the period of Mars’ rotation if a satellite in synchronous orbit around Mars travels at 1450 m/s.



𝐫 𝐯 =

𝟐𝛑𝐫

𝐓




𝟐𝛑𝐫

𝐯𝐬𝐚𝐭

𝐫𝐬𝐲𝐧𝐜 = √𝐆𝐌𝐓𝐩

𝟐

𝟒𝛑𝟐

𝟑

G = 6.67×10-11

ME = 5.97×1024 kg

RE = 6.37×106 m

MMars = 6.42×1023 kg

RMars = 3.4×106 m



physics - serway 11e ch 07: rotational motion and...

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