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Based on Wilson , Buffa, Lou

Chapter 8Rotational Motion

Rotational Motion all points on a body move in a circle around an “axis of rotation”Our bodies will be solid (no fluids) and will have no distortion

Possible distortions: twisting of body (shear)

Measurements of rotational motion are based on angular quantities

Angular Quantities Angle Measuring the angle (measure of rotation) θ

DegreesRevolutionsRadians = arc length/radius = S/r (NO UNITS) radian is the radius length

subtended to the edge of the circle (S)

If S = r then the angle is 1 radian

Conversion of angular Measurements (6 conversions) Degrees to Radians

Xo ( 2 π360o ) = radians

Revolutions to Radians

X revolutions ( 2 π1rev ) = radians

Degrees to Revolutions

Xo ( 1Rev360o ) = revolutions

Radians to Revolutions

X radians ( 1Rev2π ) = revolutions

Revolutions to Degrees

X revolutions ( 360o

1rev ) = degrees

Radians to Degrees

X radians ( 360o

2π ) = degrees

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For any circle the circumference (S) = 2πrComparison of angular measurement

2πr radians = 360o = 1 revolution

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Angular Measurements

Arc Length (S) S = θr θ measured in radians r = length of a radian

Average Angular velocity (ω)

ω = ∆ θ∆ t ω measured in radians/∆t r = length of a

radian

Instantaneous Angular velocity

ω= lim∆ t →0

∆ θ∆ t

Average Angular acceleration (α)

α = ∆ ω∆ t

Instantaneous Angular accleration

α= lim∆ t →0

∆ ω∆t

Conversion of angular motion to linear (tangential motion) Tangential velocity (vt) speed of rotating point

vt = ωr ω measured in radians/∆t r = length of a radian

Tangential acceleration (at)at = αr α measured in radians/∆t2 r = length of a radian

EXAMPLEA wheel with a radius of 0.65 m is rotating

How far will a point on the edge of the wheel move when the wheel turns through 2.5 radians?

S = θr S = 2.5 ( 0.65m) = 1.625m

How far will a point 0.45 m from the center of rotation move in 2.5 radians?S = θr S = 2.5(0.45m) = 1.125 m

What is the angular velocity of this wheel if rotates 5.0 rev in 10.0 seconds

ω = ∆ θ∆ t

ω = 5.0 rev

10.0 sec = 0.50rev/sec

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What is the tangential velocity of a point on the edge of the wheel as the wheel rotates at 0.50rev/sec?

vt = ωr 0.50rev/sec ( 2 π1rev ) = π/sec

Vt = π/sec( 0.65m) = 2.04m/s

Starting from rest the wheel achieves a rotation of 12.0 revolutions/second in 18.0 seconds

a. What is the angular acceleration of this wheel?

α = ∆ ω∆ t =

ωf−ωo

∆ t =

12rev /sec−018 sec = 0.667 rev/sec2

b. What is the tangential acceleration of a point on the edge of the wheel?

at = αr 0.667 rev/sec2 ( 2 πrev ) = 1.334π/s2 or 4.189/s2

at = 1.334π/s2 ( 0.65m) = 0.867π m/s2 or 2.724m/s2

FURTHER EXAMPES Examples

1. A wheel, with a radius of 45 cm is rotated threw the following angles. Determine the arch length a point on the edge of the wheel will move.

a. 2.1 radians (94.5 cm)b. 78o (1.36 radians) (61.5 cm)c. ¾ revolutions (4.712 radians) (212 cm)

2. A merry-go-round is making one complete revolution every 35 seconds. Determine the tangential velocity of different points on this ride; (2π radians)

a. a horse that is 7.80 meters from the “axis of rotation” (1.40 m/s)b. a tiger that is 8.80 meters from the “axis of rotation” (1.58 m/s)

3. Why does the tiger have a faster tangential velocity than the horse?

4. An electric toy is spinning at 10.0rev/sec. Its angular velocity is then increased to 20.0 rev/sec in a time of 4.00 seconds. The diameter of this toy is 9.00cm. (radius 4.50cm). Assume that its center of mass is exactly in the center of its structure.

a. What is its angular acceleration in1. rev/s2

(2.50rev/s2)2. radians/s2 (15.7/s2)

b. What is the tangential acceleration of a point on this toy that is 1. on the outside edge? (70.65 cm/s2)2. 2.00 cm the “axis of rotation” (31.4 cm/s2)

c. Determine the centripetal acceleration at the edge of this toy while rotating at 20rev/sec.(710.6m/s2)d. What is the total acceleration (at) of a point on the edge of this toy? (711.3m/s2)

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A SPECIAL SITUATION Constant Angular Acceleration (α)(alpha) (not to be confused with centripetal acceleration)

Rate of change for angular velocity

Useful rotational kinematic equations that can be applied when angular acceleration is constant

ω = ωo + αt v = v0 + at

θ = ωot + ½ αt2 d = vot + ½ at2

ω2 = ωo2 + 2αθ v2 = vo

2 + 2ad

ω = (ω + ωo)/2 v = (v + vo)/2

EXAMPLES1. A wheel is rolling with a constant velocity of 8.0m/s. The radius of the wheel is 40 cm. It

rolls for a distance of 64 m.a. Find the angular velocity of the wheel.b. What # of revolutions (θ) does the wheel make in 64 m?

2. The same wheel has an initial velocity of 8.0m/s, but it rolls to a stop in 64 meters.a. Find the angular acceleration of this wheel.b. Through what angle (θ) did this wheel turn before it stopped?c. For what amount of time was the wheel rolling?

3. A rotating wheel is propelling a conveyer belt that is moving at a constant 3.60 m/s. The wheel has a radius of 25cm.

a. Determine the angular velocity (ω) of this wheel. (14.4 rad/s)b. How many revolutions will the wheel make if it operates for 26.0 seconds? (59.6

rev)c. The conveyer belt is switched off and it requires 8.0 seconds to stop.

1. Determine the angular acceleration of the wheel. (-1.8 rad/s2)2. What number of radians did the wheel turn threw while stopping?(57.6

rad)a. How many revolutions is this? (9.17 rev)

3. What length of conveyer belt passed over the wheel while it was stopping? (14.4m)

4. What is the average angular velocity of this wheel? (7.2 rad/s)

4. A wheel with a radius of 45.0 cm is rolling at 1450 rev/min. It begins to decelerate at a constant 15.0 rad/s2.

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a. How much time is required for the wheel to reach an angular velocity of 6π rad/s? (8.86sec)

b. Through what angle in radians will the wheel turn while slowing to this angular velocity? (756.3 rad)

c. How far will the wheel roll during this time? (340.3m)ROLLING MOTION without slipping

Wheels are rolling clockwise in this example. The tangential velocity of the bike is 8.40 m/s. Therefore the wheels have a tangential velocity of 8.40 m/s. The radii of the wheels is 0.40m.

A) What is the angular velocity of the wheels? (21/s)

Determine the rolling speed of a point on the wheel relative to surface the wheels are rolling on.

Translational Velocity + Rotational Velocity = Rolling Velocity of a point on the wheel

V + ωr = rolling velocityTORQUE (τ ) a force necessary to produce a change in rotational motion.

This change is dependent on:

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1. Size of the force2. The perpendicular distance from the force line of action to the axis of

rotation ¿) called the lever arm

τ = r⊥ F = r⊥ F sin θ SI unit of torque = m∙N meter-newton

Example 1. A 35N force is applied to nut with a 0.45 m wrench (0.45m lever arm).

a. What amount of torque is applied to the nut? (15.75 m∙N)b. What amount of torque is applied to the nut if the force is applied at an angle

of 25o to the wrench? (6.66 m∙N)2. A 25 kg boy is on a teeter-totter 2.2 m from its fulcrum.

a. How much torque is he applying to the teeter-totter.? (539 m∙N)b. A 21 kg girl is on the opposite side of the teeter-totter. What distance should

shee be from the fulcrum to balance the torque of the boy? (2.62m)3. The handle of a door is .85 m from the hinged edge.

a. What amount of torque would be applied to a the door by a 15N force applied to the handle? (12.75 m∙N)

b. Determine the torque on the door if a 15N force is applied 0.18m from the hinge.

Direction of Torque Clockwise ------ negative Counter clockwise --------- positive

Rotational Static Equilibrium

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An object with an axis of rotation is Balanced τnet = 0.0 m∙N and ω = 0.0/s Not rotating

Conditions for Static Equilibrium 1. ∑ F=0

2. ∑ τ=0

3. The object cannot be moving

Example Rotational Static Equilibrium

----- 1.0m ------ ---------------- 2.2m---------------

F1 F21. A diver (arrow) with a mass of 45kg is standing on the end of the diving board. The mass

of the diving board is 11kga. Find the force being applied to supports F1 and F2

1. Find the pivot point a. Determine the torque of the known forces. Include the direction (positive or negative)

( remember the Net torque is 0.0 mN)b. Solve for unknown torque and its associated force or lever arm

2. Using linear forces determine remaining forces

τ1 + τ2 + τboard + τdiver = 0.0 mN

F1 r1 + F2 r2 +FBrB FD rD = 0.0 mNF1 1.0m + F2 0.0 m +107.8N(0.6m) + 441N ( 2.2m) = 0.0 mNF1 1.0m + 0.0 mN +64.68mN + 970.2 mN = 0.0 mNF1 1.0m = -1034.9 mNF1 = -1034.9 N

F1 + F2 + FB + FD = 0.0 N -907.2N + F2 +(-64.68N) + (-441N) = 0.0N F2 = 1412.88 N = 1412N ANSWER F1 = -1034.9 N F2 = 1412 N

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ROTATAIONAL INERTIA (I) or moment of inertia (I)Inertia is defined as a property of matter that cause that matter to resist any change in its velocity.

This was applied to translational motion only and was directly proportional to the mass of the matter

rotational inertia (I) is defined as a property of matter that resists any change in its angular velocity.

I is dependent on the mass of the object and the distribution of the mass within the object

To find the moment of inertia for various objects we will refer to a chart, or a hand out.

You are responsible for the applications of I = mr2

This is proportional to the product of the mass of the matter and the displacement of the mass from its axis of rotation

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τ = r⊥ Fnet = rF⊥ =r ma⊥ = rm αr = r2mαo τnet

= ∑ τ= τ1 + τ2 + τ3 …+ τn o τnet = ¿ r2 )α

I = Icm + Md2 for rotating object with axis not located at the center of mass

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Torque required for angular acceleration (α) τ = Iα

α≈ τI

angular acceleration is directly proportional to the net torque applied to a rotating mass and inversely proportional to the moment of inertia of that mass

Example: 1. Find the angular acceleration of a disc that has a moment of

inertia of 12.3 kg m2 when a net torque of 43.1 m∙N is applied to the disc. (3.5/s2)

a. If the frictional torque acting on this disc is 7.3 m∙N, what must the applied torque be? (50.4 m∙N)

2. A solid uniform sphere with a mass of 15.0 kg and a radius of 25.0 cm has an axis of rotation directly through its center.

a. What amount of net torque is required give this sphere an angular acceleration of 12.0/s2? (4.50 m∙N)

3. A solid disc with a mass of 25 kg and a radius of 0.35m is rotating on an axis 0.060 m from its center of mass.

a. What is the moment of inertia for this mass? (1.62 kgm2)b. What amount of net torque is required to give this mass an

angular acceleration of 18/s2? (29.18 m∙N)

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ROTATIONAL WORK & ENERGY

Rotational Work W = Fs = F (r⊥θ) = (F r⊥)θ = τθW = τθ for a single force

θ is in radians

Rotational PowerP = W/t = (τθ)/t = τ (θ/t) = τω

Rotational Kinetic EnergyKE = ½ Iω2

Example 1. 25 mN of torque are required to rotate a wheel through 15

revolutions.a. How much work is required? (2,356 J)

b. What amount of power is required to perform this work in 8.0 seconds? (295 watts)

2. A torque of 18 mN causes a wheel to rotate at 63 rad/s. a. What amount of power is required achieve this angular

velocity? (1,100 watts)

b. How many revolutions will this wheel make in 5.0 seconds? (50 rev)

3. A wheel with a moment of inertia of 37 kg m/s is rotating at 45 RPM.

a. What is the rotational KE of this wheel? (410J)

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Total Kinetic Energy

KE total =KE rotational + KE translational

KE Total = ½ Iω2 + ½ mv2

Example1. A uniform solid 2.5 kg cylinder with a radius of 0.15m rolls

without slipping at a speed of 2.2 m/s. KE total =KE rotational + KE translational

KE Total = ½ Iω2 + ½ mv2

= ½ (1/2 Mr2) ω2 + ½ mv2 = ½ (1/2∙ 2.5 kg∙(0.15m)2 (14.66/s)2 + ½ 2.5 kg ( 2.2m/s)2 = ½(0.028125kgm2)(215.1/s2) + 6.05J= 3.025J + 6.05J= 9.075J

OR

KE total =KE rotational + KE translational

KE Total = ½ Iω2 + ½ mv2 v = ωr, ω = vr

= ½ (___ Mr2)ω2 + ½ mv2 ____ refers to

= ½ (___ Mr2)(vr )2 + ½ mv2 multiplier for I

= ½ (___ Mr2)v2

r2 + ½ mv2

KE Total = ½ (___) mv2 + ½ mv2

KE Total = ½ (___) mv2 + ½ mv2 = ½ (1/2) mv2 + ½ mv2

= ¼ mv2 + ½ mv2

= ¾ mv2 KE total

= ¾ (2.5kg)(2.2m/s)2 = 9.075J

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WORK-ENERGY THEOREM Wnet = ∆KE = KEF – KEO Wnet = ½ Iωf

2 – ½ Iωo2 = ½ I(ωf

2 – ωo2)

Example 1. A grinding disc with a mass of 2.5 kg and a radius of 32 cm has

an angular velocity of 3.2/s. a. How much work is required to stop the rotation of this

grinding wheel? (0.66J)

b. Determine the work required to increase its angular velocity from 3.2/s to 8.7/s. (4.2J)

2. A uniform cylindrical hoop is released from rest at the top of an inclined plane. The height of the inclined plane is 0.25m. The hoop is allowed to roll down the incline without slipping and no loss of energy due to friction. What is the linear velocity of the hoop at the bottom of the incline?

h = 0.25m I = mr2 linear velocity at base of incline?

PE = KErotational + KEtranslational mgh = ½ Iω2 + ½ mv2

mgh = ½ (1mr2)(v2

r2 ) + ½ mv2

gh = ½v2 + ½ v2 = 1 v2

v = √ gh v = √( 9.8 ms2 ) ∙(0.25 m) = 1.56 m/s

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3. A uniform solid sphere is released from rest at the top of an inclined plane. The height of the inclined plane is 0.25m. The sphere is allowed to roll down the incline without slipping and no loss of energy due to friction. What is the linear velocity of the sphere at the bottom of the incline?

h = 0.25m I = 2/5mr2 linear velocity at base of incline?

PE = KErotational + KEtranslational

mgh = ½ Iω2 + ½ mv2

mgh = ½ (2/5mr2)(v2

r2 ) + ½ mv2

gh = 1/5 v2 + ½ v2 = 710v2

v2 = 107 gh

v = √ 107

gh v = √ 107

(9.8 m/ s2(0.25 m)) =1.87m/s

ANGULAR MOMENTUM (L)

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Momentum = m∙v ρ = mv v = ωr

Angular momentum (L)(L) = r⊥ρ = r⊥∙mv = m∙ r⊥ v = m∙ r⊥ωr = (mrr )ω

L = mrrω = Iω

L = Iω

Momentum is changed by force ∆ρ= F∆t

Angular momentum is changed by torque

τ = Iα = I ∆ ω

∆ t = ∆( Iω)

∆t = ∆ L∆ t

τ∆t = ∆L change in angular momentum

Units for angular momentum mr2(rad/s) = kg∙m2/s

Angular Momentum is a vector; It is in the direction of the angular velocity (ω)

L⃗= Iω⃗

Examples1. A 0.75 kg particle is rotating clockwise as viewed from above at

4.7rad/s. The radius of the rotation is 20.0 cm. What is the angular momentum of this particle? Give magnitude and direction. (0.141kg m2/s away from you)

2. A 35 kg solid sphere with a radius of 0.45m has an angular velocity of 150 rev/min. Determine the angular momentum of this sphere. (55.7kgm2/s)

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τ = Iα = I∆ ω∆ t = ∆(Iω)

∆t = ∆ L∆ t

τ ∙∆t = ∆L F∆t = ρ

Conservation of Angular Momentum

Lo = Lf

Iωo = Iωf

Example 1. A skater has a moment of inertia of 95kgm2 with his arms

outstretched, and a moment of inertia of 68 kgn2 with his arms pulled close to his chest. With his arms outstretched he is spinning at he is spinning at 1.25 rev/s. Determine his rotation when he pulls his arms close to his chest. (1.75 rev/sec)