Alta High Physics

Rotational Motion

Chapter 8

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Important Variable θ (theta) – angular displacement – radians ω (omega) – angular velocity – radians/sec α (alpha) = angular acceleration – radians/sec2

t – still time and still in seconds

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Measuring Angular Displacement

θ = L/r 360° = 2π radians

= 1 revolution v= ωr =

Linear velocity =

angular velocity x radius a = αr

Linear acceleration =

angular acceleration x radius

θ

r

L

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Sample Problems

If an exploding fireworks shell makes a 10° angle in the sky and you know it is 2000 meters above your head, how many meters wide is the arc of the explosion?

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Solution

θ = L/rθ = 10° x π rad/180° = 0.17 rad

0.17 rad = L/2000 meters

L = 2000 meters x 0.17 rad = 349 meters

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Sample Problems

Convert the following measures from Radians to degrees: 3.14 rad 150 rad 24 rad

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Solution 3.14 rad x 180°/π rad = 180° 150 rad x 180°/π rad = 8594° 24 rad x 180°/π rad = 1375°

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Sample Problems

What is the linear speed of a child seated 1.2 meters from the center of a merry-go-round if the ride makes one revolution in 4 seconds?

What is the child’s acceleration?

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Solutiona)v = ωr and ω=2π rad/4 sec = 1.6 rad/s

since r = 1.2 m then v = 1.58 rad/s x 1.2 m = 1.9 m/s

b) Since the linear velocity is not changing there is no linear or tangential acceleration, but the child is moving in a circle so there is centripetal or radial acceleration which you will recall fits the equation:a c = v2/r and since v = ωr ac = ω2r so a c = (1.9 m/s)2/1.2 m = 3.0 m/s2 orac = ω2r = (1.6 rad/s)2(1.2 m) = 3.0 m/s2

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Angular Kinematic Equations

Linear Angular

V = v0+at ω=ω0+ αt

x = v0t+½at2 θ=ω0t+½ αt2

V2 = v02+2ax ω2=ω0

2+2αθ

Vavg = (v0+vf)/2 ωavg=(ω0+ωf)/2

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Sample Problem

A angular velocity of wheel changes from 10 rad/s to 30 rad/s in 5 seconds.

a) What is its angular acceleration?b) What is its angular displacement

while it is accelerating?

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Solution

α = (ωf – ω0)/t =

=(30 rad/s – 10 rad/s)/5 sec = 4 rad/s2

θ = ω0t = ½αt2

= 10 rad/s + ½(4 rad/s2)(5 sec)2

= 100 rad

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Torque

Torque = Fr = force x radius Torque is measured in

newton•meters which means that it has the same units as work in a linear system.

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Rotational Inertia

The rotational inertia of an object depends upon its shape and its mass.

The equations for each shape are given on page 223 in the text

Torque = Iα = rotational inertia x angular

acceleration

Hence Fr = Iα

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Sample Problem

A force of 10 newtons is applied to the edge of a bicycle wheel (a thin ring mass 1 kg and radius of 0.5 meters). What is the resulting angular acceleration of the wheel?

If the wheel was at rest when the force was applied and the force is applied for 0.4 seconds what is the angular velocity of the wheel immediately after it is applied?

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Solution

Since Fr = Iα then α = Fr/I and since the wheel is a thin ring:

I = mr2 = (1 kg)(0.5 m)2 = 0.25 kg m2

So α = (10 N)(0.5 m)/0.25 kg m2

= 20 rad/s2

ωf = ω0 + αt

= 0 rad/s + (20 rad/s2)(0.4 sec)

= 8 rad/s

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Problems Types Finding angles when distance and

length of arc are known Converting from revolutions to radians

and radians to degrees Finding angular velocity and angular

acceleration if radius and their linear counterparts are known

Using angular kinematic equations Calculating Torque