Ch. 7 (Section 7.3)

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Ch. 7 (Section 7.3) Circular Motion

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Ch. 7 (Section 7.3). Circular Motion. Describing Circular Motion. An object undergoes acceleration moving at a constant speed. Explain!. Circular Motion Velocity. Recall v = Δd/Δt so in circular motion v = Δr/Δt Velocity vector of object moving in a circle is tangent to the circle . - PowerPoint PPT Presentation

Transcript of Ch. 7 (Section 7.3)

Page 1: Ch. 7 (Section 7.3)

Ch. 7 (Section 7.3)

Circular Motion

Page 2: Ch. 7 (Section 7.3)

Describing Circular Motion

• An object undergoes acceleration moving at a constant speed. Explain!

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Circular Motion Velocity

• Recall v = Δd/Δt so in circular motion v= Δr/Δt

• Velocity vector of object moving in a circle is tangent to the circle

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Circular Motion Acceleration

• Acceleration of object in uniform circular motion ALWAYS points towards middle of circle = centripetal acceleration

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Centripetal Acceleration• Centripetal acceleration = (velocity)2 divided by the

radius of the circle; ac = v2/r• ALWAYS points to center or circle!• To measure speed: measure the period, time (T) it

takes for object to complete one revolution.• In one revolution distance = circumference of circle

= 2πr so v = 2πr/T • If ac = v2/r then ac= (2πr/T)2÷r 4π2r/T2

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Centripetal Acceleration Practice 1. A test car moves at a constant speed around a

circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2, what is the car’s tangential speed?

a = v2/r

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2. A dog sits 1.5 m from the center of a merry-go-round. The merry-go-round is set in motion, and the dog’s tangential speed is 1.5 m/s. what is the dog’s centripetal acceleration?

a = v2/r

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Centripetal Force

• Constant force applied causing circular motion

• Recall 2nd Law: F = ma, if in uniform circular motion a = ac

• So: Centripetal Force = Fc = mv2/r

• Fc = m (4π2r/T2) • Points to center of circle, same

direction as acceleration

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Example Problem pg. 165

• A 13-g rubber stopper is attached to a 0.93-m string and swung in a circle, making one revolution in 1.18 sec. Find the tension force exerted by the string on the stopper.

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Nonexistent Force• Imagine you’re riding in a car and it

suddenly stops, what happens to you? • Is there a force on you? Why/ why not?• Imagine riding in a car again, and it

makes a sharp left turn, what happens to you?

• Is there a force on you? Why/ why not? • Hence there is NOT an outward “force”

on you, it is a fictitious force we call the centrifugal force, explained by Newton’s 1st Law

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FUN DEMO

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Changing Circular Motion: Torque

• Point mass: object moving in circular path (ex. rubber stopper swung in circle, person on merry go round etc…)

• Rigid rotating object: entire mass that rotates around its on axis (ex: a cd, the entire merry go round, a revolving door or ordinary door)

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Torque

• How do you open a door?• Should you push next to the hinges?• Should you push in, towards the hinges?• To open the door most easily, you push at a

distance far from the hinges (axis of rotation), and in a direction perpendicular to the door.

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Torque

• Distance from the axis of rotation = lever arm• The product of the perpendicular force

applied AND the lever arm = TORQUE• Greater torque = greater change in rotational

motion

Sesaw will balance when magda = mbgdb

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Chapter 8

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Newton’s Law of Universal Gravitation• Everything pulls on everything else!• Greater the mass = greater the force• Greater the distance = weaker the force

Force = (mass1 x mass2) (distance)2

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The Universal Gravitation Constant

M m

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1. What is the force of gravity between a 100 kg astronaut and a 50 kg astronaut floating 2 m apart in deep space?

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Calculate the force of gravity on you, at Earth’s Surface

• 1 pound = ~4.45 Newtons• Mass of Earth = 5.97×1024 kg• Radius of Earth = 6.38×106 m

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Section 8.2: Using the Law of Universal Gravitation

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Gravity can be a Centripetal Force• Centripetal force:

– Center seeking– Any force that is directed at right angels to the path of

a moving object that tends to produce circular motion

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Centripetal Force Keeps the Moon and Planets Where they Are!

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Satellites and Gravity• The force of the Earth’s gravity keeps satellites in

orbit– Satellites have horizontal velocities large enough that

their falling distance is equal to the curvature of the Earth

• 29,000km/hr

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What would happen to Earth if…1. The sun’s gravity

increased?2. The Earth’s inertia

decreased?3. The Earth’s inertia

increased?

Balance is key for and orbit to happen!

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Combining Equations to solve for Orbital speed and Period

• Recall Fc = mv2/r & F = GMEm/r2

• GMEm/r2 = mv2/r • getting V by itself:• v = √GME/r

• Period: T = 2π√r3/GME

• Both speed, v & period, T are independent of mass

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Example Problem Pg. 187

• A satellite orbits Earth 225 km above its surgace. What is its speed in orbit and its period?

• Radius of Earth = 6.36 × 106 m• Mass of Earth = 5.97 × 1024 kg• G = 6.67 × 10-11 N*m2/kg2

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Gravitational Field

• gravity is always in the direction of the force• To find strength of the field you can place a

small body of mass (m) in the field and measure the forceon the bodyg= F/mon Earth’s surface the gravitation field = 9.80 N/kg

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Einstein’s Theory of Gravity

• Einstein proposed that gravity is not a force but an affect of space itself

• Mass causes space to be curved and other bodies are accelerated because of the way they follow this curved space