Chapter 19 Planar Kinetics of a Rigid Body: Impulse and Momentum

19-2

The linear momentum of a rigid body is defined as

L = m vGThis equation states that the linear momentum vector L has a magnitude equal to (mvG ) and a direction defined by vG .

The angular momentum of a rigid body is defined as

HG = IG Remember that the direction of HG is perpendicular to the plane of rotation.

19.1 Linear and Angular Momentum

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Translation

0==

G

GH

mvL

When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because

= 0.

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It is sometimes convenient to compute the angular momentum of the body about the center of rotation O.

HO = ( rG

mvG ) + IG

= IO

When a rigid body is rotating about a fixed axis passing through point O, the bodys linear momentum and angular momentum about G are:

L = m vGHG = IG

Rotation about a fixed axis

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When a rigid body is subjected to general plane motion, both the linear momentum and the angular momentum computed about G are required. L = m vGHG = IG

The angular momentum about point A isHA = IG + (d)mvG

General Plane Motion

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19.2 Principle of Impulse and MomentumAs in the case of particle motion, the principle of impulse and momentum for a rigid body is developed by combining the equation of motion with kinematics. The resulting equations allow a direct solution to problems involving force, velocity, and time.

Linear impulse-linear momentum equation:

L1 + F dt = L2 or (mvG )1 + F dt = (mvG )2

t2

t1

t2

t1

Angular impulse-angular momentum equation:

(HG )1 + MG dt = (HG )2 or IG 1 + MG dt = IG 2

t2

t1

t2

t1

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The previous relations can be represented graphically by drawing the impulse-momentum diagram.

+ =

To summarize, if motion is occurring in the x-y plane, the linear impulse-linear momentum relation can be applied to the x and y directions and the angular momentum-angular impulse relation is applied about a z-axis passing through any point (i.e., G). Therefore, the principle yields three scalar equations describing the planar motion of the body.

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Ex. The space capsule has a mass of 1200 kg and a momentof inertia IG = 900 kg-m2 about an axis passing through G and directed perpendicular to the page. If it is traveling forward with a speed vG = 800 m/s and executes a turn by means of two jets, which provide a constant thrust of 400 N for 0.3 s, determine the capsules angular velocity just after the jets are turned off.

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p.511, 19-7 The space shuttle is located in deep space, where the effects of gravity can be neglected. It has a mass of 120 Mg, a center of mass at G, and a radius of gyration (kG )x = 14 m about the x axis. It is originally traveling forward at v = 3 km/s when the pilot turns on the engine at A, creating a thrust T = 600(1 e-0.3t) kN , where t is in seconds. Determine the shuttles angular velocity 2 s later.

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p. 512, 19-12 The 100-kg flywheel has a radius of gyration about its center of gravity O of kO = 0.225 m. If it rotates counterclockwise with an angular velocity of 1200 rev/min before the brake is applied, determine the time required for the wheel come to rest when a fore P = 1000 N is applied to the handle. The coefficient of kinetic friction between the belt and the wheel rim is k = 0.3. (Hint: Recall from the statics text that the relation of the tension in the belt is given by TB = TC e, where is the angle of contact in radians.)

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p. 514, 19-21For safety reasons, the 20-kg supporting leg of a sign is designed to break away with negligible resistance at B when the leg is subjected to the impact of a car. Assuming that the leg is pin supported at A and approximates a thin rod, determine the impulse the car bumper exerts on it, if after the impact the leg appears to rotate upward to an angle of max = 150o.

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19.3 Conservation of Momentum

Conservation of Linear MomentumIf the sum of all the linear impulses acting on a system of connected rigid bodies is zero (small or nonimpulsive)

21

=

momentumlinearsystem

momentumlinearsystem

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Conservation of Angular MomentumWhen the sum of all the angular impulsescreated by the external forces acting on the system is zero or appreciably small (nonimpulsive)

21

OO momentumgularsystem an

momentumangularsystem

=

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19.4 Eccentric Impact

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19-32

1/

2/

11

22

)()(

)()()()(

AB

AB

BA

AB

vv

vvvv

e

=

=

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The 5-kg slender rod is suspended from the pin at A. If a 1-kg ball B is thrown at the rod and strikes its center with a horizontal velocity of 9 m/s, determine the angular velocity of the rod just after impact. The coefficient of restitution is e = 0.4

Example 19.8

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SolutionConservation of Angular MomentumAngular momentum is conserved about point A since the impulsive force between the rod and the ball is internal.The weights of the ball and rod are non- impulsive.

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SolutionConservation of Angular MomentumWe have

Since (vG )2 = 0.52 , then

22

22

2221

21

)1)(5(121)5.0())(5()5.0())(1()5.0)(9)(1(

)5.0()()5.0()()5.0()()()()(

++=

++=

=+

GB

GGRBBBB

AA

vv

IvmvmvmHH

22 6667.1)(5.05.4 += Bv

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SolutionCoefficient of RestitutionWe have

Solving

22

22

11

22

)(5.06.309

)()5.0(4.0)()()()(

B

B

GB

BG

v

vvvvve

=

=

=

+

rad/s 287.3m/s 96.196.1)(

2

2

===

Bv

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p. 526, 19-38 The satellites body C has a mass of 200 kg and a radius of gyration about the z axis of kZ = 0.2 m. If the satellite rotates about the z axis with an angular velocity of 5 rev/s, when the solar panels are in a position of

= 0, determine the angular velocity of the satellite when the solar panels are rotated to a position of = 90. Consider each solar panel to be a thin plate having a mass of 30 kg. Neglect the mass of the rods.

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p. 527, 19-44 The 15-kg thin ring strikes the 20-mm-high step. Determine the smallest angular velocity the ring can have so that it will just roll over the step at A without slipping.

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p. 529, 19-52 The mass center of the 1.5-kg ball has a velocity of (vG)= 1.8 m>s when it strikes the end of the smooth 2.5-kg slender bar which is at rest. Determine the angular velocity of the bar about the z axis just after the impact if e = 0.8.

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Chapter 19Planar Kinetics of a Rigid Body:Impulse and Momentum 2Translation 4 5 6 719.2 Principle of Impulse and Momentum 9 10 11 12 13 14 15 16p.511, 19-7The space shuttle is located in deep space, where the effects of gravity can be neglected. It has a mass of 120 Mg, a center of mass at G, and a radius of gyration (kG)x = 14 m about the x axis. It is originally traveling forward at v = 3 km/s when the pilot turns on the engine at A, creating a thrust T = 600(1 e-0.3t) kN , where t is in seconds. Determine the shuttles angular velocity 2 s later. 18p. 512, 19-12The 100-kg flywheel has a radius of gyration about its center of gravity O of kO = 0.225 m. If it rotates counterclockwise with an angular velocity of 1200 rev/min before the brake is applied, determine the time required for the wheel come to rest when a fore P = 1000 N is applied to the handle. The coefficient of kinetic friction between the belt and the wheel rim is k = 0.3. (Hint: Recall from the statics text that the relation of the tension in the belt is given by TB = TC e, where is the angle of contact in radians.) 20 21 22 23 24 25 26 27 28 2919.4 Eccentric Impact 31 32Example 19.8 34 35 36p. 526, 19-38The satellites body C has a mass of 200 kg and a radius of gyration about the z axis of kZ = 0.2 m. If the satellite rotates about the z axis with an angular velocity of 5 rev/s, when the solar panels are in a position of = 0, determine the angular velocity of the satellite when the solar panels are rotated to a position of = 90. Consider each solar panel to be a thin plate having a mass of 30 kg. Neglect the mass of the rods. 38p. 527, 19-44The 15-kg thin ring strikes the 20-mm-high step. Determine the smallest angular velocity the ring can have so that it will just roll over the step at A without slipping. 40 41p. 529, 19-52The mass center of the 1.5-kg ball has a velocity of (vG)= 1.8 m>s when it strikes the end of the smooth 2.5-kg slender bar which is at rest. Determine the angular velocity of the bar about the z axis just after the impact if e = 0.8. 43 44