physics - clutch ch 06: centripetal forces &...

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PHYSICS - CLUTCH

CH 06: CENTRIPETAL FORCES & 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

PHYSICS - CLUTCH


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

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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

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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.

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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?

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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?

PHYSICS - CLUTCH


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?

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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

PHYSICS - CLUTCH


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

PHYSICS - CLUTCH


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

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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

)

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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

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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 = _______

PHYSICS - CLUTCH


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

PHYSICS - CLUTCH


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

PHYSICS - CLUTCH


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

PHYSICS - CLUTCH


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

𝐫𝟐 𝐠𝐬𝐮𝐫𝐟 =

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

PHYSICS - CLUTCH


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

𝐓𝐬𝐚𝐭𝟐 =

𝟒𝛑𝟐𝐫𝟑

𝐆𝐌

G = 6.67×10-11

ME = 5.97×1024 kg

RE = 6.37×106 m

r

v

Fg

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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

PHYSICS - CLUTCH


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 = __________

PHYSICS - CLUTCH


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 - CLUTCH


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.

PHYSICS - CLUTCH


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 = ____________

PHYSICS - CLUTCH


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



𝐆𝐌

T2 = _________

PHYSICS - CLUTCH


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


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

PHYSICS - CLUTCH


physics - clutch ch 06: centripetal forces &...

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