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5. Plane Kinetics of Rigid Bodies

5.1 Mass moments of inertia

5.2 General equations of motion

5.3 Translation

5.4 Fixed axis rotation

5.5 General plane motion

5.6 Work and energy relations

5.7 Impulse and momentum

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5.1 Mass moments of inertia

• The moment-of-inertia may be expressed as

𝐼 =𝑟𝑖2𝑚𝑖

• Or about the axis O-O:

𝐼 = න𝑟2𝑑𝑚

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5.1 Mass moments of inertia (contd.)

• If the density is constant throughout the body, the moment of inertia becomes

𝐼 = 𝜌න𝑟2𝑑𝑉

Radius of Gyration

• If all the mass m of a body could be concentrated at a distance k from the axis, the moment of inertia would be unchanged.

• The radius of gyration k of a mass m about an axis for which the moment of inertia is I is defined as

𝑘 =𝐼

𝑚𝑜𝑟 𝐼 = 𝑘2𝑚

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5.1 Mass moments of inertia (contd.)

Transfer of Axes

• If moment of inertia of a body is known about an axis passingthrough mass center, it may be determined easily about anyparallel axis.

𝐼 = ҧ𝐼 + 𝑚𝑑2

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5.2 General Equations of Motion

• The force equation,

𝐅 = 𝑚ത𝐚

tells us that the resultant ΣF of the external forces acting on thebody equals the mass m of the body times the acceleration ത𝐚 ofits mass center G. The moment equation taken about the masscenter,

𝐌𝐺 = ሶ𝐇𝐺

shows that the resultant moment about the mass center of theexternal forces on the body equals the time rate of change ofthe angular momentum of the body about the mass center. 5

5.2 General Equations of Motion (contd.)

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5.2 General Equations of Motion (contd.)Plane Motion Equations

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5.2 General Equations of Motion (contd.)Plane Motion Equations

• The mass center G has an acceleration ത𝐚 and the body has anangular velocity 𝛚 = 𝜔𝐤 and an angular acceleration 𝛂 = 𝛼𝐤,both taken positive in the z-direction. Because the z-directionof both 𝛚 and 𝛂 remains perpendicular to the plane ofmotion, we may use scalar notation 𝜔 and 𝛼 = ሶ𝜔 to representthe angular velocity and angular acceleration.

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𝐅 = 𝑚ത𝐚 and 𝑀𝐺 = ҧ𝐼𝛼

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5.2 General Equationsof Motion (contd.)Alternative Moment Equations

• Alternatively

𝑀𝑃 = ҧ𝐼𝛼 +𝑚ത𝑎𝑑

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5.3 Translation

• For a translating body, then, our general equations for plane motionequation may be written

𝐅 = 𝑚ത𝐚

𝑀𝐺 = ҧ𝐼𝛼 = 010

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5.3 TranslationSample Problem (1)

• The pickup truck weighs 1500kg and reaches a speed of 50km/h from rest in a distance of 60m up the 10-percent inclinewith constant acceleration. Calculate the normal force undereach pair of wheels and the friction force under the reardriving wheels. The effective coefficient of friction betweenthe tires and the road is known to be at least 0.8.

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Solution

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5.3 TranslationSample Problem (2)

• The uniform 30-kg bar OB is secured to the acceleratingframe in the 30 position from the horizontal by thehinge at O and roller at A. If the horizontal accelerationof the frame is a=20 m/s2 compute the force FA on theroller and the x- and y-components of the forcesupported by the pin at O.

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Solution

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5.4 Fixed Axis Rotation• Our general equations for plane motion are directly

applicable and are repeated here.

𝐅 = 𝑚ത𝐚 , 𝑀𝐺 = ҧ𝐼𝛼 , 𝑀𝑂 = 𝐼𝑂𝛼

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5.4 Fixed Axis RotationSample Problem (3)

The 300 kg concrete block is elevated by the hoisting mechanism shown,where the cables are securely wrapped around the respective drums. Thedrums, which are fastened together and turn as a single unit about their masscenter at O, have a combined mass of 150 kg and a radius of gyration about Oof 450 mm. If a constant tension P of 1.8 kN is maintained by the power unit atA, determine the vertical acceleration of the block and the resultant force onthe bearing at O.

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Solution

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5.4 Fixed Axis RotationSample Problem (4)

The uniform slender bar is released from rest in the horizontalposition shown. Determine the value of x for which the angularacceleration is a maximum, and determine the correspondingangular acceleration.

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Solution

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5.5 General Plane Motion

• The dynamics of a rigid body in general plane motioncombines translation and rotation.

𝐅 = 𝑚ത𝐚

𝑀𝐺 = ҧ𝐼𝛼

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5.5 General Plane Motion (contd.)

• Choice of Moment Equation. The application of thealternative relation for moments about any point P,

𝑀𝑃 = ҧ𝐼𝛼 +𝑚ത𝑎𝑑

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5.5 General Plane MotionSample Problem (5)

A metal hoop with a radius r = 150 mm is released from rest onthe 20° incline. If the coefficients of static and kinetic frictionare 𝜇𝑠 = 0.15 and 𝜇𝑘 = 0.12 , determine the angularacceleration of the hoop and the time t for the hoop to move adistance of 3 m down the incline.

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Solution

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Solution (Ctnd.)

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5.5 General Plane MotionSample Problem (6)

A car door is inadvertently left slightly open when the brakes areapplied to give the car a constant rearward acceleration a.Derive expressions for the angular velocity of the door as itswings past the 90° position. The mass of the door is m, its masscenter is a distance ҧ𝑟 from the hinge axis O, and the radius ofgyration about O is 𝑘𝑂 .

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Solution

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Solution (Ctnd.)

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5.5 General Plane MotionSample Problem (7)

The drum A is given a constant angular acceleration 𝛼0 of3 rad/s2 and causes the 70-kg spool B to roll on the horizontalsurface by means of the connecting cable, which wraps aroundthe inner hub of the spool. The radius of gyration ത𝑘 of the spoolabout its mass center G is 250 mm, and the coefficient of staticfriction between the spool and the horizontal surface is 0.1.Determine the tension T in the cable and the friction force Fexerted by the horizontal surface on the spool.

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Solution

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Solution

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5.5 General Plane MotionSample Problem (8)

The uniform steel beam of mass m and length l is suspended bythe two cables at A and B. If the cable at B suddenly breaks,determine the tension T in the cable at A immediately after thebreak occurs. Treat the beam as a slender rod and show that theresult is independent of the length of the beam.

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Solution

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5.6 Work and EnergyWork of Forces and Couples

• The work done by a force F is given by

𝑈 = න𝐅. 𝑑𝐫 𝑜𝑟 𝑈 = න(𝐹 cos 𝛼) 𝑑𝑠

𝑈 = න𝑀 𝑑𝜃

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5.6 Work and EnergyKinetic Energy

(a) Translation. The translating rigid body of Figure (a) has a

mass m and all of its particles have a common velocity v.

𝑇 =1

2𝑚𝑣2

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5.6 Work and EnergyKinetic Energy(contd.)

(b) Fixed-axis rotation.

𝑇 =1

2𝐼𝑂𝜔

2

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5.6 Work and EnergyKinetic Energy(contd.)

(c) General plane motion.

𝑇 =1

2𝑚 ҧ𝑣2 +

1

2ҧ𝐼𝜔2

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5.6 Work and EnergyKinetic Energy(contd.)

• The kinetic energy of plane motion may also be expressed interms of the rotational velocity about the instantaneouscenter C of zero velocity. Because C momentarily has zero

velocity, 𝑇 =1

2𝐼𝑂𝜔

2 for the fixed point O holds equally well

for point C, so that, alternatively, we may write the kineticenergy of a rigid body in plane motion as

𝑇 =1

2𝐼𝐶𝜔

2

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5.6 Work and EnergyPotential Energy and the Work-Energy Equation

• Gravitational potential energy 𝑉𝑔 and elastic potential energy𝑉𝑒 were covered in detail before. Recall that the symbol 𝑈′ (ratherthan 𝑈) is used to denote the work done by all forces except theweight and elastic forces, which are accounted for in the potential-energy terms.

• The work-energy relation was introduced before for particle motionand was generalized to include the motion of a general system ofparticles. This equation

𝑈1−2′ = ∆𝑇 + ∆𝑉𝑔 + ∆𝑉𝑒

Power

• If the force F and the couple M act simultaneously, the total instantaneous power is

𝑃 = 𝐅. 𝐯 + 𝑀𝜔38

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5.6 Work and EnergySample Problem (10)

The wheel rolls up the incline on its hubs without slipping and is pulled by the100-N force applied to the cord wrapped around its outer rim. If the wheelstarts from rest, compute its angular velocity 𝜔 after its center has moved adistance of 3 m up the incline. The wheel has a mass of 40 kg with center ofmass at O and has a centroidal radius of gyration of 150 mm. Determine thepower input from the 100-N force at the end of the 3-m motion interval.

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Solution

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Solution (Cntd.)

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5.6 Work and EnergySample Problem (9)

In the mechanism shown, each of the two wheels has a mass of 30 kg and acentroidal radius of gyration of 100 mm. Each link OB has a mass of 10 kg andmay be treated as a slender bar. The 7-kg collar at B slides on the fixed verticalshaft with negligible friction. The spring has a stiffness k=30 kN/m and iscontacted by the bottom of the collar when the links reach the horizontalposition. If the collar is released from rest at the position 𝜃 = 45° and iffriction is sufficient to prevent the wheels from slipping, determine

(a) the velocity𝑣𝐵 of the collar as it first strikes the spring and

(b) the maximum deformation x of the spring.

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Solution

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Solution (Cntd.)

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5.7 Impulse and MomentumLinear Momentum

• The linear momentum of any mass system,𝐆 = 𝑚ത𝐯

𝐅 = ሶ𝐆 𝑎𝑛𝑑 𝐆1 +න𝑡1

𝑡2

𝐅𝑑𝑡 = 𝐆2

• In planar applications:

𝐹𝑥 = ሶ𝐺𝑥 𝐹𝑦 = ሶ𝐺𝑦

and

(𝐺𝑥)1+න𝑡1

𝑡2

𝐹𝑥 𝑑𝑡 = (𝐺𝑥)2

(𝐺𝑦)1 +න𝑡1

𝑡2

𝐹𝑦 𝑑𝑡 = (𝐺𝑦)2

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5.7 Impulse and MomentumAngularMomentum

• Angular momentum about the mass center:

𝐻𝐺 = ҧ𝐼𝜔

𝑀𝐺 = ሶ𝐻𝐺 𝑎𝑛𝑑 𝐻𝐺 1 +න𝑡1

𝑡2

𝑀𝐺𝑑𝑡 = 𝐻𝐺 2

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5.7 Impulse and MomentumAngularMomentum (contd.)

• The angular momentum 𝐻𝑂 about any point 𝑂 is easily written as

𝐻𝑂 = ҧ𝐼𝜔 + 𝑚 ҧ𝑣𝑑

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5.7 Impulse and MomentumAngularMomentum (contd.)

• This expression holds at any particular instant of time about O, which may be a fixed or moving point on or off the body.

• When a body rotates about a fixed point O on the body or body extended,

𝐻𝑂 = 𝐼𝑂𝜔

𝑀𝑂 = ሶ𝐻𝑂 𝑎𝑛𝑑 𝐻𝑂 1 +න𝑡1

𝑡2

𝑀𝑂𝑑𝑡 = 𝐻𝑂 2

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5.7 Impulse and MomentumInterconnectedRigid Bodies

𝐅= ሶ𝐆𝑎 + ሶ𝐆𝑏 +⋯

𝐌𝑂 = ( ሶ𝐇𝑂)𝑎+( ሶ𝐇𝑂)𝑏+⋯

න𝑡1

𝑡2

𝐅𝑑𝑡 = ∆𝐆 𝑠𝑦𝑠𝑡𝑒𝑚

න𝑡1

𝑡2

𝐌𝑂𝑑𝑡 = ∆𝐇𝑂 𝑠𝑦𝑠𝑡𝑒𝑚 49

5.7 Impulse and MomentumInterconnectedRigid Bodies (contd.)

• We note that the equal and opposite actions and reactions inthe connections are internal to the system and cancel oneanother so they are not involved in the force and momentsummations. Also, point O is one fixed reference point for theentire system.

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5.7 Impulse and MomentumConservation of Momentum

• Principles for the particles are applicable to either a singlerigid body or a system of interconnected rigid bodies. Thus, ifΣF=0 for a given interval of time, then

𝐆1 = 𝐆2

• Similarly, if the resultant moment about a given fixed point Oor about the mass center is zero during a particular interval oftime for a single rigid body or for a system of interconnectedrigid bodies, then

𝐻𝑂 1 = 𝐻𝑂 2 𝑜𝑟 𝐻𝐺 1 = 𝐻𝐺 2 51

5.7 Impulse and MomentumSample Problem (11)

• The grooved drums in the two systems shown are identical. Inboth cases, (a) and (b), the system is at rest at time t=0Determine the angular velocity of each grooved drum at timet=4s. Neglect friction at the pivot O.

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Solution

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Solution

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5.7 Impulse and MomentumSample Problem (12)

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Solution

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Solution

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5.7 Impulse and MomentumSample Problem (13)

The uniform rectangular block of dimensions shown is sliding to theleft on the horizontal surface with a velocity v1 when it strikes thesmall step at O. Assume negligible rebound at the step and computethe minimum value of v1 which will permit the block to pivot freelyabout O and just reach the standing position A with no velocity.Compute the percentage energy loss n for b=c.

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Solution

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Solution (Cntd.)

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5.7 Impulse and MomentumSample Problem (14)

The sheave E of the hoisting rig shown has a mass of 30 kg and a centroidalradius of gyration of 250 mm. The 40-kg load D which is carried by the sheavehas an initial downward velocity 𝑣1 = 1.2 m/s at the instant when a clockwisetorque is applied to the hoisting drum A to maintain essentially a constantforce F =380 N in the cable at B. Compute the angular velocity 𝜔2 of thesheave 5 seconds after the torque is applied to the drum and find the tensionT in the cable at O during the interval. Neglect all friction.

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Solution

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Solution (Cntd.)

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