2452650 Kinematics Rigid Bodies

Kinematics of rigid bodies relations between time and the positions, velocities, and accelerations of the particles forming a rigid body.

(1)

Rectilinear translation (2)

Rotation about a fixed axis

Curvilinear translation (3)

Curvilinear translation rotation R.Ganesh Narayanan Plane motion

(4)

General Plane motion

(5) Motion about a fixed point

R.Ganesh Narayanan

(1) Translation Consider rigid body in translation: direction of any straight line inside the body is constant, all particles forming the body move in parallel lines. For any two particles in the body, rB = rA + rB Differentiating with respect to time, rB = rA + rB

A

= rA

rB/A = 0

vB = v A All particles have the same velocity. Differentiating with respect to time again, aB = a A All particles have the same acceleration. R.Ganesh Narayanan

A

vB = v A

aB = a A

When a rigid body is in translation, all the points of the body have the same velocity and same acceleration at any given instant In curvilinear translation, the velocity and acceleration change in direction and in magnitude at every instant; In rectilinear translation, velocity, acceleration direction are same during entire motion

R.Ganesh Narayanan

(2) Rotation About a Fixed Axis: Velocity & Acceleration Consider rotation of rigid body about a fixed axis AA’ The length ∆s of the arc described by P when the body rotates through ∆θ, ∆s = (BP )∆θ = (r sin φ )∆θ ∆θ ds = lim (r sin φ ) = rθ sin φ v= ∆t dt ∆t →0 V=

dr =ω ×r dt

Vector direction along the rotation axis AA’, angular velocity R.Ganesh Narayanan

• Differentiating to determine the acceleration, dv d = (ω × r ) dt dt dω dr = ×r +ω × dt dt dω = ×r +ω ×v dt

a=

Angular acceleration, α

a = α x r + ω x (ω x r) Tangential acceleration component

Radial acceleration component

R.Ganesh Narayanan

Equations Defining the Rotation of a Rigid Body About a Fixed Axis

ω=

dθ dt

or

dt =

dθ

ω

dω d ω d 2θ α = = 2 =ω dθ dt dt

v = dx/dt a = d2x/dt2 a = v (dv/dx)

Uniform Rotation, α = 0:

θ = θ

0

Equns. Can be used only when α = 0 & constant

+ ωt

Uniformly Accelerated Rotation, α = constant: ω = ω0 +αt

θ = θ 0 + ω 0t + ω

2

1αt2 2

x = x0 + vt

v = v0 + at x = x0 + v0t + ½ at2

= ω 02 + 2 α (θ − θ 0

)

R.Ganesh Narayanan

v2 = v02 + 2a (x-x0)

The rectangular block shown rotates about the diagonal OA with a constant angular velocity of 6.76 rad/s. Knowing that the rotation is counterclockwise as viewed from A, determine the velocity and acceleration of point B at the instant shown.

Angular velocity.

Velocity of point B.

vB = ω x RB/O = (i + 6.24j + 2.4 k) x (0.127i + 0.396j) = -0.95i + 0.305j – 0.396 k

Acceleration of point B.

m/s

aB =ω×vB = -3.2 i – 1.88j + 6.23k R.Ganesh Narayanan

m/s2

The assembly shown consists of two rods and a rectangular plate BCDE which are welded together. The assembly rotates about the axis AB with a constant angular velocity of 10 rad/s. Knowing that the rotation is counterclockwise as viewed from B, determine the velocity and acceleration of corner E.

Find rB/A, rE/B Find angular velocity vector; ω = ω (rB/A / lAB) 1) vE = ω x rE/B; 2) aE = ω x vE

R.Ganesh Narayanan

Ring B has an inner radius r2 and hangs from the horizontal shaft A as shown. Shaft A rotates with a constant angular velocity of 25 rad/s and no slipping occurs. Knowing that r1 = 1.27 cm, r2 = 6.35 cm, and r3 = 8.9 cm, determine (a) the angular velocity of ring B, (b) the acceleration of the points of shaft A and ring B which are in contact, (c) the magnitude of the acceleration of a point on the outside surface of ring B.

R.Ganesh Narayanan

R.Ganesh Narayanan

(3) General Plane Motion

• General plane motion is neither a translation nor a rotation. • General plane motion can be considered as the sum of a translation and rotation. • Displacement of particles A and B to A2 and B2 can be divided into two parts: - translation to A2 and B1′ - rotation of B1′ about A2 to B2

R.Ganesh Narayanan

Absolute and Relative Velocity in Plane Motion Absolute and Relative Velocity in Plane Motion

• Any plane motion can be replaced by a translation of an arbitrary reference point A and a simultaneous rotation about A. vB = v A + vB vB

A

A

= ω × rB

vB

A

v B = v A + ω × rB

A

A

R.Ganesh Narayanan

= rω

Point A as reference

• Assuming that the velocity vA of end A is known, wish to determine the velocity vB of end B and the angular velocity ω in terms of vA, l, and θ. • The direction of vB and vB/A are known. Complete the velocity diagram.

vA = = cos θ v B A lω vA v B = v A tan θ ω= l cos θ R.Ganesh Narayanan

vB = tan θ vA

vA

Point B as reference

• Selecting point B as the reference point and solving for the velocity vA of end A and the angular velocity ω leads to an equivalent velocity triangle. • vA/B has the same magnitude but opposite sense of vB/A. The sense of the relative velocity is dependent on the choice of reference point. • Angular velocity ω of the rod in its rotation about B is the same as its rotation about A. Angular velocity is not dependent on the choice of reference point.

R.Ganesh Narayanan

The crank AB has a constant clockwise angular velocity of 2000 rpm. For the crank position indicated, determine (a) the angular velocity of the connecting rod BD, and (b) the velocity of the piston P. r • The velocity vB is obtained from the crank rotation data.

ω AB =  2000 

rev  min  2π rad    = 209.4 rad s  min  60 s  rev 

vB = ( AB )ω AB = (3 in.)(209.4 rad s )

= 628.3 in/s

The velocity direction is as shown. • The direction of the absolute velocity vD is horizontal. The direction of the relative velocity vD B is perpendicular to BD. Compute the angle between the horizontal and the connecting rod from the law of sines. sin 40° sin β = 8 in. 3 in.

β = 13.95°

R.Ganesh Narayanan

vD B vD 628.3 in. s = = sin 53.95° sin 50° sin76.05°

vD = 523.4 in. s = 43.6 ft s vD vD

B

B

vP = vD = 43.6 ft s

= 495.9 in. s

= lω BD vD

495.9 in. s 8 in. l = 62.0 rad s

ω BD =

B

=

R.Ganesh Narayanan

ω BD = (62.0 rad s )k

Instantaneous Center of Rotation in Plane Motion • Plane motion of all particles in a slab can always be replaced by the translation of an arbitrary point A and a rotation about A with an angular velocity that is independent of the choice of A. • The same translational and rotational velocities at A are obtained by allowing the slab to rotate with the same angular velocity about the point C on a perpendicular to the velocity at A. • The velocity of all other particles in the slab are the same as originally defined since the angular velocity and translational velocity at A are equivalent. • As far as the velocities are concerned, the slab seems to rotate about the instantaneous center of rotation C. R.Ganesh Narayanan

How to obtain instantaneous center of rotation Fig. 1

Fig. 2

• If the velocity at two points A and B are known, the instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity vectors through A and B . (fig. 1) • If the velocity vectors at A and B are perpendicular to the line AB, the instantaneous center of rotation lies at the intersection of the line AB with the line joining the extremities of the velocity vectors at A and B. (fig. 2) • If the velocity vectors are parallel, the instantaneous center of rotation is at infinity and the angular velocity is zero. (fig. 1) • If the velocity magnitudes are equal, the instantaneous center of rotation is at infinity and the angular velocity is zero. (fig. 2)

R.Ganesh Narayanan

• The particle at the center of rotation has zero velocity. • The particle coinciding with the center of rotation changes with time and the acceleration of the particle at the instantaneous center of rotation is not zero. • The acceleration of the particles in the slab cannot be determined as if the slab were simply rotating about C.

• The trace of the locus of the center of rotation on the body is the body centrode and in space is the space centrode.

R.Ganesh Narayanan

Same problem

The crank AB has a constant clockwise angular velocity of 2000 rpm. For the crank position indicated, determine (a) the angular velocity of the connecting rod BD, and (b) the velocity of the piston P. vB = 628.3 in/s; β = 13.95° • The instantaneous center of rotation is at the intersection of the perpendiculars to the velocities through B and D.

γ B = 40° + β = 53.95° γ D = 90° − β = 76.05° BC CD 8 in. = = sin 76.05° sin 53.95° sin50° BC = 10.14 in. CD = 8.44 in. R.Ganesh Narayanan

• Determine the angular velocity about the center of rotation based on the velocity at B.

vB = (BC )ω BD

ω BD =

vB 628.3 in. s = BC 10.14 in.

ω BD = 62.0 rad s

• Calculate the velocity at D based on its rotation about the instantaneous center of rotation.

vD = (CD )ω BD = (8.44 in.)(62.0 rad s )

vP = vD = 523 in. s = 43.6 ft s

R.Ganesh Narayanan

Absolute and Relative Acceleration in Plane Motion

• Absolute acceleration of a particle of the slab, aB = a A + aB

A

r • Relative acceleration a B A associated with rotation about A includes tangential and normal components,

(a B A )t = α × rB A (a B A )n = −ω 2 rB A

(a B A )t = rα 2 (a BR.Ganesh ) = r ω A n Narayanan

aB = a A + aB

= a A + (a B

A

) + (aB A )t

A n

Rate of Change With Respect to a Rotating Frame • Frame OXYZ is fixed. • Frame Oxyz rotates about fixed axis OA with angular velocity Ω • Vector function Q(t ) varies in direction and magnitude. Rate of change of Q depends on frame of reference • With respect to the rotating Oxyz frame, • If Q were fixed within Oxyz then Q is Q = Qx i + Q y j + Qz k OXYZ equivalent to velocity of a point in a rigid body Q Oxyz = Q x i + Q y j + Qz k attached to Oxyz and Q x di/dt + Q y dj/dt + Q zdk/dt = Ω × Q • With respect to the fixed OXYZ frame,

()

()

(Q)

OXYZ

(Q)

= Q x i + Q y j + Qz k + Qx di/dt + Q y dj/dt + Qz dk/dt

Oxyz

= Qx i + Q y j + Qz k

Represent velocity of particle, ΩxQ

R.Ganesh Narayanan

• With respect to the fixed OXYZ frame,

(Q)

OXYZ

()

= Q Oxyz + Ω × Q

This relation is useful to find rate of change of Q w.r.t. fixed frame of reference OXYZ when Q is defined by its components along the rotating frame Oxyz

R.Ganesh Narayanan

Coriolis Acceleration • Frame OXY r is fixed and frame Oxy rotates with angular velocity Ω . • Position vector rP for the particle P is the same in both frames but the rate of change depends on the choice of frame. • The absolute velocity of the particle P is,

v P = (r )OXY = Ω × r + (r )Oxy

• The absolute acceleration of the particle P is,

ap = vp = Ωxr + Ωxr + d/dt [(r)Oxy]

(Q)

OXYZ

()

= Q Oxyz + Ω × Q

v P = (r )OXY = Ω × r + (r )Oxy

ap = Ω x r + Ω x (Ω x r) + 2Ω x (r)Oxy + (r)Oxy R.Ganesh Narayanan

ap = Ω x r + Ω x (Ω x r) + 2Ω x (r)Oxy + (r)Oxy a = α x r + ω x (ω x r)

Coriolis acceleration, ac Two vectors are normal to each other, 2ΩvOxy

R.Ganesh Narayanan

Disk D of the Geneva mechanism rotates with constant counterclockwise angular velocity ωD = 10 rad/s. At the instant when φ = 150o, determine (a) the angular velocity of disk S, and (b) the velocity of pin P relative to disk S. • Direction of velocity of P with respect to S is parallel to slot. From the law of cosines, r 2 = R2 + l 2 − 2Rl cos30° = 0.551R2

From the law of sine, sinβ sin 30° sin 30° = sin β = R r 0.742

r = 37.1 mm

β = 42.4°

• Magnitude and direction of absolute velocity of pin P are calculated from radius and angular velocity of disk D.

) = 500 mm vP = Rω D = (50 mm )(10 rad sR.Ganesh s Narayanan

• The absolute velocity of the point P may be written as

v P = (r )OXY = Ω × r + (r )Oxy v

P

= v

P ′

+ v

P

s

The interior angle of the vector triangle is γ = 90° − 42.4° − 30° = 17.6°

vP′ = vP sin γ = (500 mm s )sin 17.6° = 151.2 mm s = rω s

ωs =

151.2 mm s 37.1 mm

ωs =

4.08 rad s

vP s = vP cos γ = (500 m s ) cos17.6° vP s = (477 m s )

42.4°

Coriolis acceleration, ac = 2 ωs vp/s = 2 (4.08) (477) = 3890 mm/s2 R.Ganesh Narayanan