Copyright Sautter 2003

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Angular Motion

• Angular motion involves rotation or circular motion. Some elements of circular motion have already been discussed and we will review them here.

• Circular motion (rotation) can be measured using linear units or angular units. Angular units refer to revolutions, degrees or radians.

• The properties of circular motion include displacement, velocity and acceleration. When applied to rotation the values become angular displacement, angular velocity or angular acceleration. Additionally, angular motion can be measured using frequencies and periods or rotation.

• The Greek letters theta (), omega () and alpha () are used to represent angular displacement, angular velocity and angular acceleration

AVERAGE = / t = (2 + 1) / 2

= o t + ½ t2

i = o + t

i = ½ (i2 - o

2) /

s = rVlinear = r

alinear = rf = 1/ T T = 1 / f

1 revolution = 360 degrees = 2 radians = 2 f = 2 / T

Moment of Inertia

• All states of motion are subject to the laws of inertia, that is tend to remain at the same rate and in the same directional orientation.

• In the case of rotational motion, the angular velocity tends to remain unchanged and the plane of rotation persists.

• As you will recall, outside forces can change inertial conditions. In rotation, outside torques must be applied to change an objects rotational inertia.

• Torque, as you remember, is a force applied perpendicularly to the center of rotation.

• τ = F x r

Moment of Inertia

• The tendency of a body to resist changes in its linear state of motion is measured by its mass.

• The tendency of a body to resist changes in its rotational state of motion is measured by its moment of inertia.

• Moment of inertia involves not just the mass of a rotating object but also the distribution of the mass within the object.

• τ = F x r, recall that F = ma, therefore:

• τ = ma x r, since a = r, τ = m r x r =(mr2 )• I = mr2 and τ = I

Moment of Inertia

• τ = I• Note the similarity to F = ma for linear motion.

Instead of a applied force (F) and applied torque (τ) is necessary to provide acceleration.

• Instead of mass (m), the moment of inertia (I) determines the resulting acceleration.

• Instead of linear acceleration (a), angular acceleration () results from the application of a torque.

• Although, the moment of inertia in its simplest form is given as mass times radius squared (mr2), in more complex bodies the value of I must be found by calculus methods (integration) or experimental means.

The Laws of Motion for Rotating Bodies(A summary)

• First Law – A body which is rotating tends to keep rotating at the same rate and in the same plane unless acted on by an outside torque.

• Second Law –

Torque = Moment of Inertia x angular acceleration

(τ = I)• Third Law – for every torque there must be an equal

but opposite torque.

Moment of Inertia

• When the moment of inertia is found by experiment a simplifying technique similar to the center of mass concept is used.

• Remember, the center of mass of an object is a point where all the mass of the body could be concentrated to give the same inertial properties as the actual mass distribution of the body.

• When describing rotational motion, radius of gyration is used in place of the center of mass concept.

• The radius of gyration of a body is the radius of a thin ring of a mass equal to the mass of the body which would give the same rotational characteristics as the actual body.

SphereI = 2/5 mr2

CylinderI = 1/2 mr2

Thin RingI = mr2

Thin RodI = 1/12 mr2

RotationalAxis

RotationalAxis

RotationalAxis

RotationalAxis

I for any object of mass m and radius of gyration rg

I = mrg2

Equations for Rotational Motion

• All bodies in linear motion possess characteristics such as momentum and kinetic energy. All rotating body possess similar properties.

• Linear equations can be converted to rotational analogs by substituting angular values into the linear equations.

• Careful analysis of the units involved will show that the rotational equations yield the appropriate units for each quantity measured (joules for energy, kg m/s for momentum, watts for power, etc.)

These linear values arereplaced by the angular

values shown:

Linear ~ Angular (mass) m ~ I (acceleration) a ~ α (velocity) v ~ ω(displacement) s ~ θ (force) F ~ τ (momentum) p ~ L

Rotational EquationsObtained by Substitution

Linear RotationalF = ma τ = Iα p = mv L = Iω W = Fs Wrot = τθP = Fv Prot = τω

K.E. = 1/2 mv2 K.E.rot = 1/2 Iω2

Rotational Dynamic Problems(a) Find the moment of inertia of a 1 kg disk with radius of 34

cm and rotating at 45 rpm. (b) what is its kinetic energy? © what is its angular momentum ?

• (a) Idisk = ½ mr2 = ½ (1 kg)(0.34 m)2 = 0.0578 kg m2

• (b) K.E.rot = 1/2 Iω2 = ½ (0.0578)(1.5 )2 = 0.642 joules

• (c) L = Iω = (0.0578)(1.5 ) = 1.28 kg m/s

RotationalAxis

m = 1 kgR = 34 cm = 0.34 m

Frequency = 45 rpm = 45/60 = 0.75 rps

= 2 f = 1.5

Rotational Dynamic ProblemsA disk rolls down an incline1.2 meters high. What is its

speed at the bottom of the hill ?

• P.E. = K.E.rot + K.E.trans

• P.E. = mgh, K.E.rot = ½ Iω2, K.E.trans = ½ mv2

• Idisk = ½ mr2, vtrans = r, = v/r

• K.E.rot = ½ Iω2 = ½ (½ mr2)(v/r)2 = ¼ mv2

• mgh = ¼ mv2 + ½ mv2 , canceling mass from both sides gives:

• gh = ¾ v2 , v = ( 4/3 gh )1/2 = (4/3 x 9.8 x 1.2)1/2 = 3.96 m/s

Potential energy kinetic energyKinetic energy K.E. translation + K.E. rotation

P.E. = K.E. rotation + K.E. translation

RotationalAxis1.2 m

Rotational Dynamic ProblemsA fly wheel with a moment of inertia of 6 kg-m2 is acted on by a

constant torque of 50 N-m. (a) What is the angular acceleration of the wheel (b) How long will it take to reach an angular velocity of

90 radians per second from rest?

• (a) τ = Iα , α = τ/I• α = 50 N-m / 6 kg-m2 = 8.33 rad / sec

• (b) i = o + t, t = (i - o) / • t= ( 90 – 0) / 8.33 = 10.8 seconds

6 kg-m2

50 N-m

τ = Iα

i = o + t

Rotational Dynamic ProblemsA ballerina spins at 1 rps. With outstretched arms her moment of

inertia is 2.4 kg-m2. When she pulls in her arms her moment of inertia becomes 1.2 kg-m2. (a) what is her frequency with her arm

retracted ? (b) How much work is needed to retract her arms?

• (a) Conservation of momentum – Σ Lbefore = Σ Lafter

• Σ(Iω)before = Σ(Iω)after , (2.4 kg-m2 x 1 rps) = (1.2 kg-m2 ω)

• ωafter = 2.4 / 1.2 = 2 rps

• (b) Conservation of energy - Σ energybefore = Σ energyafter

• K.E.rot = ½ Iω2,

• Σ energybefore = ½ (2.4)(2 1)2 = 4.8 2 joules

• Σ energyafter = ½ (1.2)(2 2)2 = 9.6 2 joules

• Energy added = 9.6 2 – 4.8 2 =4.8 2 = 48 joules of work done by the skater

L = IωK.E.rot = 1/2 Iω2

f1 = 1 rpsI = 2.4 Kg-m2

f2 = 2 rps

Rotational Dynamic ProblemsA truck tire of mass 10 kg experiences an angular acceleration of 3 radians per second2 when a torque of 30 N-m is applied.

Find its radius of gyration.

• τ = Iα • 30 N-m = I (5 rad/s2), I = 30/5 = 6 kg-m2

• I = mrg2

• 6 kg-m2 = 10 kg x rg2 , rg = (6 / 10)1/2

• rg = 0.77 m or 77 cm

I = mrg2

τ = Iα

The moment of inertia of a 45 kg grindstone is 5 kg-m2. Find itsradius of gyration ?

(A) 0.13 m (B) 33 cm (C) 2.0 m (D) 96 cm

A flywheel rotating at 120 rad/s has a brake apply 200 N-m of torque toit. It stops in 80 seconds. What is the moment of inertia of the flywheel ?

(A) 113 kg-m2 (B) 36 kg-m2 (C) 200 kg-m2 (D) 300 kg-m2

Find the momentum of inertia of the Earth in kg-m2. Its mass is 6 x 1024 kg and its radius is 6.4 x 106 m.

(A) 9.8 x 1037 (B) 4.0 x 1030 (C) 9.4 x 1018 (D) 2.3 x 1030

A wheel with a moment of inertia of 4.0 kg-m2. It is rotating at 40 rad/s.Find its kinetic energy.

(A) 3.2 x 10-3 j (B) 3.0 x 105 j (C) 7.2 x 102 j (D) 3200 j

Two disks, each with a moment of inertia of 1.0 kg-m2, one of which isrotating at 100 rad/s and the other at rest are press together. Find

new angular velocity of the combination in rad/s.(A) 600 (B) 50 (C) 550 (D) 200

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