CEE 271: Applied Mechanics II, Dynamics– Lecture 25: Ch.17, Sec.4 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Thursday, November 15, 2012

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EQUATIONS OF MOTION: ROTATION ABOUT AFIXED AXIS

Today’s objectives: Studentswill be able to

1 Analyze the planar kineticsof a rigid body undergoingrotational motion.

In-class activities:• Reading Quiz• Applications• Rotation About an Axis• Equations of Motion• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 In rotational motion, the normal component of accelerationat the body’s center of gravity (G) is always .(a) zero(b) tangent to the path of motion of G(c) directed from G toward the center of rotation(d) directed from the center of rotation toward G

ANS: (c)

2 If a rigid body rotates about point O, the sum of themoments of the external forces acting on the body aboutpoint O equals?(a) IGα(b) IOα(c) maG(d) maO

ANS: (b)

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APPLICATIONS

Pin at the center of rotation.

• The crank on the oil-pump rigundergoes rotation about a fixedaxis, caused by the driving torque,M , from a motor.

• As the crank turns, a dynamicreaction is produced at the pin. Thisreaction is a function of angularvelocity, angular acceleration, andthe orientation of the crank.

• If the motor exerts a constant torqueM on the crank, does the crank turnat a constant angular velocity? Isthis desirable for such a machine?

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APPLICATIONS(continued)

• The pendulum of the Charpyimpact machine is released fromrest when θ = 0◦. Its angularvelocity (ω) begins to increase.

• Can we determine the angularvelocity when it is in verticalposition?

• On which property (P ) of thependulum does the angularacceleration (α) depend?

• What is the relationship betweenP and α?

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EQUATIONS OF MOTION: ROTATION ABOUT AFIXED AXIS (Section 17.4)

• When a rigid body rotates about a fixedaxis perpendicular to the plane of thebody at point O, the body’s center ofgravity G moves in a circular path ofradius rG. Thus, the acceleration ofpoint G can be represented by atangential component (aG)t = rGα anda normal component (aG)n = rGω

2.

• Since the body experiences an angular acceleration, itsinertia creates a moment of magnitude, IGα, equal to themoment of the external forces about point G. Thus, thescalar equations of motion can be stated as:

ΣFn = m(aG)n = mrGω2

ΣFt = m(aG)t = mrGα

MG = IGα 6 / 36

EQUATIONS OF MOTION (continued)

• Note that the ΣMG moment equation may be replaced by amoment summation about any arbitrary point. Summingthe moment about the center of rotation O yields

ΣMO = IGα+ rGm(aG)t = [IG +m(rG)2]α

• From the parallel axis theorem, IO = IG +m(rG)2,therefore the term in parentheses represents IO.

• Consequently, we can write the threeequations of motion for the body as:

ΣFn = m(aG)n = mrGω2

ΣFt = m(aG)t = mrGα

MO = IOα

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PROCEDURE FOR ANALYSIS

• Problems involving the kinetics of a rigid body rotatingabout a fixed axis can be solved using the followingprocess.

1 Establish an inertial coordinate system and specify the signand direction of (aG)n and (aG)t.

2 Draw a free body diagram accounting for all external forcesand couples. Show the resulting inertia forces and couple(typically on a separate kinetic diagram).

3 Compute the mass moment of inertia IG or IO.4 Write the three equations of motion and identify the

unknowns. Solve for the unknowns.5 Use kinematics if there are more than three unknowns

(since the equations of motion allow for only threeunknowns).

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DEPENDENT MOTION EXAMPLE

• Given: The uniform slender rodhas a mass of 15 kg.

• Find: The reactions at the pin Oand the angular acceleration ofthe rod just after the cord is cut.

• Plan: Since the mass center, G, moves in a circle of radius0.15 m, it’s acceleration has a normal component toward Oand a tangential component acting downward andperpendicular to rG. Apply the problem solving procedure.

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EXAMPLE (Solution)

• FBD and Kinetic Diagram

= rG

• Equations of motion:

(+→)ΣFn = man = mrGω

2 ⇒ Ox = 0N

(+ ↓)ΣFt = mat = mrGα ⇒ −Oy + 15(9.81) = 15(0.15)α

("+)ΣMO = IGα+mrGα(rG)

⇒ (0.15)15(9.81) = IGα+m(rG)2

• Using IG = (ml2)/12 and rG = (0.15), we can write:IGα+m(rG)2α = [(15× 0.92)/12 + 15(0.15)2]α = 1.35α

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EXAMPLE (continued)

• After substituting:

22.07 = 1.35α⇒ α = 16.4 rad/s2

• From the second equation in the last page:

−Oy + 15(9.81) = 15(0.15)α

Oy = 15(9.81)− 15(0.15)16.4 = 110N

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CONCEPT QUIZ1 If a rigid bar of length l is released from rest in the

horizontal position (θ = 0), the magnitude of its angularacceleration is at maximum when

(a) θ = 0

(b) θ = 90o

(c) θ = 180o

(d) θ = 0o and θ = 180o

ANS: (d)2 In the above problem, when θ = 90o, the horizontal

component of the reaction at pin A is .(a) zero(b) mg(c) 1

2mω2

(d) None of the aboveANS: (a)

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GROUP PROBLEM SOLVING

• Given: Wsphere = 30 lb,Wrod = 10 lb

• Find: The reaction at the pin Ojust after the cord AB is cut.

• Plan: Draw the free body diagramand kinetic diagram of the rodand sphere as one unit. Thenapply the equations of motion.

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GROUP PROBLEM SOLVING (Solution)

• FBD and kinetic diagram;

30lb

=

Oy

Ox

10lb

mrod(1.0α)

IG,rodαmrod(1.0)(0)

2IG,sphereα

msphere(3)(0)2

msphere(3α)

• Equations of motion for ΣFn = m(aG)n :

Ox = (30/32.2)(3)(0)2 + (10/32.2)(1.0)(0)2

⇒ Ox = 0 lb

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GROUP PROBLEM SOLVING (continued)

30lb

=

Oy

Ox

10lb

mrod(1.0α)

IG,rodαmrod(1.0)(0)

2IG,sphereα

msphere(3)(0)2

msphere(3α)

• ΣFt = m(aG)t: ⇒ Oy = 40− 3.106α from

−Oy + 30 + 10 = (30/32.2)(3α) + (10/32.2)(1.0α)

• MO = Ioα: ⇒ 100 = 9.172α from

30(3.0) + 10(1.0) = [0.4(30/32.2)(1)2 + (30/32.2)(3)2]sphereα

+ [(1/12)(10/32.2)(2)2 + (10/32.2)(1)2]rodα

• Therefore, α = 10.9 rad/s2 and Oy = 6.14 lb

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CONCEPT QUIZ (p. 430)

1 A drum of mass m is set into motion in two ways: (a) by aconstant 40N force (left), and, (b) by a block of weight 40N(right). If αa and αb represent the angular acceleration ofthe drum in each case, select the true statement. (a)

(a) αa > αb

(b) αa < αb

(c) αa = αb

(d) None of the above.

2 In case (b) above (right figure), what is the tension T in thecable? (b)

(a) T = 40 N(b) T < 40 N

(c) T > 40N(d) None of the above.

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EQUATIONS OF MOTION: GENERAL PLANEMOTION

Today’s objectives: Studentswill be able to

1 Analyze the planar kineticsof a rigid body undergoinggeneral plane motion.

In-class activities:• Reading Quiz• Applications• Equations of Motion• Frictional Rolling Problems• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 If a disk rolls on a rough surface without slipping, theacceleration at the center of gravity (G) will andthe friction force will be .(a) not be equal to αr; less than µsN(b) be equal to αr; equal to µkN(c) be equal to αr; less than µsN(d) None of the above

ANS: (c)

2 If a rigid body experiences general plane motion, the sumof the moments of external forces acting on the body aboutany point P is equal to .(a) IPα(b) IPα+maP(c) maG(d) IGα+ rGP ×maP

ANS: (a)

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APPLICATIONS

• As the soil compactor acceleratesforward, the front roller experiencesgeneral plane motion (bothtranslation and rotation).

• What are the loads experienced bythe roller shaft or bearings?

• The forces shown on the roller’sFBD cause the accelerations shownon the kinetic diagram. Is the point Athe IC?

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APPLICATIONS (continued)

• The lawn roller is pushed forward with a force of 200 Nwhen the handle is at 45◦.

• How can we determine its translation acceleration andangular acceleration?

• Does the acceleration depend on the coefficients of staticand kinetic friction?

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APPLICATIONS(continued)

• During an impact, the center ofgravity of this crash dummy willdecelerate with the vehicle, butalso experience anotheracceleration due to its rotationabout point A.

• Why?

• How can engineers use this information to determine theforces exerted by the seat belt on a passenger during acrash?

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GENERAL PLANE MOTION (Sec. 17.5)

• When a rigid body is subjected toexternal forces and couple-moments, it can undergo bothtranslational motion and rotationalmotion. This combination is calledgeneral plane motion.

• Using an x− y inertial coordinatesystem, the equations of motionsabout the center of mass, G, may bewritten as:

ΣFx = m(aG)x (1)ΣFy = m(aG)y (2)MG = IGα (3)

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GENERAL PLANE MOTION (continued)

• Sometimes, it may be convenient towrite the moment equation about apoint P other than G. Then theequations of motion are written asfollows:

ΣFx = m(aG)x (4)ΣFy = m(aG)y (5)MP = Σ(Mk)P (6)

• In this case, Σ(Mk)P represents thesum of the moments of IGα andrG/P ×maG about point P .

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FRICTIONAL ROLLING PROBLEMS

• When analyzing the rolling motion of wheels, cylinders, ordisks, it may not be known if the body rolls without slippingor if it slides as it rolls.

• For example, consider a disk withmass m and radius r, subjected to aknown force P .

• The equations of motion will be:ΣFx = m(aG)x ⇒ P − F = maGΣFy = m(aG)y ⇒ N −mg = 0ΣMG = IGα⇒ Fr = IGα

• There are 4 unknowns (F,N, a, andaG) in these three equations.

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FRICTIONAL ROLLING PROBLEMS (continued)

• Hence, we have to make anassumption to provide anotherequation. Then, we can solve forthe unknowns.

• The 4th equation can be obtainedfrom the slip or non-slip conditionof the disk.

• Case 1: Assume no slipping and use aG = αr as the 4th

equation and DO NOT use Ff = µsN . After solving, youwill need to verify that the assumption was correct bychecking if Ff ≤ µsN .

• Case 2: Assume slipping and use Ff = µkN as the 4th

equation. In this case, aG 6= αr.25 / 36

PROCEDURE FOR ANALYSIS

Problems involving the kinetics of a rigid body undergoinggeneral plane motion can be solved using the followingprocedure.

1. Establish the x− y inertial coordinate system. Draw boththe free body diagram and kinetic diagram for the body.

2. Specify the direction and sense of the acceleration of themass center, aG, and the angular acceleration, α, of thebody. If necessary, compute the body’s mass moment ofinertia IG.

3. If the moment equation MP = Σ(Mk)P is used, use thekinetic diagram to help visualize the moments developedby the components m(aG)x, m(aG)y, and IGα.

4. Apply the three equations of motion.

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PROCEDURE FOR ANALYSIS(continued)

5. Identify the unknowns. If necessary (i.e., there are fourunknowns), make your slip or no-slip assumption (typicallyno slipping, or the use of aG = αr, is assumed first).

6. Use kinematic equations as necessary to complete thesolution.

7. If a slip-no slip assumption was made, check its validity!!!

Key points to consider1 Be consistent in using the assumed directions. The

direction of aG must be consistent with α.2 If Ff = µkN is used, Ff must oppose the motion. As a test,

assume no friction and observe the resulting motion. Thismay help visualize the correct direction of Ff .

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DEPENDENT MOTION EXAMPLE

• Given: A spool has a mass of 200 kg and a radius ofgyration (kG) of 0.3 m. The coefficient of kinetic frictionbetween the spool and the ground is µk = 0.1.

• Find: The angular acceleration (α) of the spool and thetension in the cable.

• Plan: Focus on the spool. Follow the solution procedure(draw a FBD, etc.) and identify the unknowns.

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EXAMPLE (Solution)

• The free body diagram and kinetic diagram for the bodyare:

IG a maG =

1962 N

• Equations of motion:

ΣFy = m(aG)y : NB − (200kg)(9.8 m/s2) = 0

⇒ NB = 1962N (7)

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EXAMPLE (continued)

• Note that aG = (0.4)α. Why ?• ΣFx = m(aG)x:

T − 0.1NB = 200aG = 200(0.4)α

T − 196.2 = 80α

• ΣMG = IGa:

450− T (0.4)− 0.1NB(0.6) = 20(0.3)2α

450− T (0.4)− 196.2(0.6) = 1.8α

• Solving these two equations, we getα = 7.50 rad/s2 and T = 797N 30 / 36

CONCEPT QUIZ1 An 80 kg spool (kG = 0.3 meter) is on a rough surface and

a cable exerts a 30N load to the right. The friction force atA acts to the and the aG should be directed to the

.

(a) right, left

(b) left, right

(c) right, right

(d) left, left

ANS: (c)

0.2 m

A

30N

α

0.75m

G

2 For the situation above, the moment equation about G is?(a) 0.75(FfA)− 0.2(30) = −(80)(0.32)α(b) −0.2(30) = −(80)(0.32)α(c) 0.75(FfA)− 0.2(30) = −(80)(0.32)α+ 80aG(d) None of the above

ANS: (a)31 / 36

GROUP PROBLEM SOLVING

• Given: A 80 kg lawn roller has a radiusof gyration of kG = 0.175 meter. It ispushed forward with a force of 200N .

• Find: The angular acceleration ifµs = 0.12 and µk = 0.1.

• Plan: Follow the problem solvingprocedure.

• Solution: The moment of inertia of the roller about G isIG = m(kG)2 = (80)(0.175)2 = 2.45kg ·m2

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GROUP PROBLEM SOLVING (continued)

• FBD:

• Equations of motion:

ΣFx = m(aG)x

FA − 200 cos 45 = 80aG (8)ΣFy = m(aG)y

NA − 784.8− 200 sin 45 = 0 (9)ΣMG = IGα

−0.2FA = 2.45α (10)• We have 4 unknowns: NA, FA, aG and α.• Another equation is needed to allow solving for the

unknowns.

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GROUP PROBLEM SOLVING (continued)

• The three equations we have now are:

FA − 200 cos 45◦ = 80aG

NA − 784.8− 200 sin 45◦ = 0

−0.2FA = 2.45α

• First, assume the wheel is not slipping.Thus, we can write aG = rα = 0.2α

• Now solving the four equations yields: NA = 926.2N ,FA = 61.4N , α = −5.01 rad/s2, and aG = −1.0 m/s2

• The no-slip assumption must be checked. IsFA (= 61.4N) ≤ µsNA (= 111.1N) ?

• Yes, therefore, the wheel rolls without slip.

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CONCEPT QUIZ

1 A slender 100 kg beam is suspended by a cable. Themoment equation about point A is?

(a) 3(10) = 1/12(100)(42)α

(b) 3(10) = 1/3(100)(42)α

(c) 3(10) = 1/12(100)(42)α+(100aGx)(2)

(d) None of the above.

ANS: (c)

����������������������������

��������

4m3m

10N

A

2 Select the equation that best represents the ‘no-slip’assumption.(a) Ff = µsN(b) Ff = µkN(c) aG = rα(d) None of the above.

ANS: (c)35 / 36

note

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