Kinematics, Dynamics, and Vibrations

FE Review Session

Dr. David Herrin March 27, 2012

Example 1 A 10 g ball is released vertically from a height of 10 m. The ball strikes a horizontal surface and bounces back. The coefficient of restitution between the surface and the ball is 0.75. The height that the ball will reach after bouncing is most nearly

00

0202

1

mghv

mghmv

PEKE

=

=

=

mh

mghmv

vve

r

rr

r

6.521 2

0

=

=

=

Example 2 A 5 g mass is to be placed on a 50 cm diameter horizontal table that is rotating at 50 rpm. The mass must not slide away from its position. The coefficient of friction between the mass and the table is 0.2. What is the maximum distance that the mass can be placed from the axis of rotation?

5 g

ω

R

ω

R 5 g

mgR

RmNmaF r

072.2

2

==

−=−

=∑

ωµ

ωµ

Example 3 A truck of 4000 kg mass is traveling on a horizontal road at a speed of 95 km/h. At an instant of time its brakes are applied, locking the wheels. The dynamic coefficient of friction between the wheels and the road is 0.42. The stopping distance of the truck is most nearly (A)  55 m (B)  70 m (C)  85 m (D)  100 m

mN

mvx

xNmv

WKEKE

k

k

5.84121

21

2

2

21

==Δ

Δ=

+=

µ

µ

Example 4 A translating and rotating ring of mass 1 kg, angular speed of 500 rpm, and translational speed of 1 m/s is placed on a horizontal surface. The coefficient of friction between the ring and the surface is 0.35. For an outside radius of 3 cm, the time at which skidding stops and rolling begins is most nearly

ω

v0

a

N µN

mg

maNmaFx

=

=∑

µ mgNFy=

=∑ 0

αµ

α2mRNR

IM=

=∑

tatvv o

αωω −=

+=

0

Skidding Stops when ωRv =

Example 5 A stationary uniform rod of length 1 m is struck at its tip by a 3 kg rigid ball moving horizontally with velocity of 8 m/s as shown. The mass of the rod is 7 kg, and the coefficient of restitution between the rod and the ball is 0.75. The velocity of the ball after impact is most nearly

v

1 m

v

ωr

v’

ω’r Before After

r

r

r

vvvve

LmLmvmvL

−

−=

+=

''

'31' 2ω

smv /88.1'=

Example 6 The D’Alembert force is

(A)  Force of gravity in France. (B)  Force due to inertia. (C)  Resisting force due to static friction (D)  Atomic force discovered by D’Alembert.

Example 7 A 5 kg pendulum is swung on a 7 m long massless cord from rest at 5° from center. The time required for the pendulum to return to rest is most nearly

Lg

n =ωLgf

π21

=Hzf 1885.0=

fT 1=

Example 8 A 2 kg mass is suspended by a linear undamped spring with a spring constant 3.2 kN/m. The mass is given an initial velocity of 10 m/s from the equilibrium position.

sradmk

n /40==ω

Example 8.1 Calculate the natural frequency of the mass.

nf ωπ21

=

Example 8.2 How long does it take for the mass to complete one complete cycle.

sf

T 156.01==

Example 8.3 What is the maximum deflection of the spring from the equilibrium position?

( ) ( ) ( )tvtxtx nn

n ωω

ω sincos 00 +=

Example 9 A 115 kg motor turns at 1800 rpm, and it is mounted on a pad having a stiffness of 500 kN/m. Due to an unbalanced condition, a periodic force of 85 N is applied in a vertical direction, once per revolution. Neglecting damping and horizontal movement, the amplitude of vibration is

mmD

m

C

mEkF

srad

sradmk

pst

n

f

n

f

opst

f

n

024.0

14.0

21

1

47.1

/5.188

/66

22

==

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

−==

=

==

βδ

ω

ω

ω

ωβ

δ

ω

ω

Example 10 A uniform disk of 10 kg mass and 0.5 m diameter rolls without slipping on a flat horizontal surface, as shown. When its horizontal velocity is 50 km/h, the total kinetic energy of the disk is most nearly

22

21

21

ωoo ImvKE +=

Example 11 A homogeneous disk of 5 cm radius and 10 kg mass rotates on an axle AB of length 0.5 m and rotates about a fixed point A. The disk is constrained to roll on a horizontal floor. Given an angular velocity of 30 rad/s in the x direction and -3 rad/s in the y direction, the kinetic energy of the disk is

222

21

21

21

zzyyxx IIIKE ωωω ++=

Example 12 The natural frequency of the system is designed to be ωn = 10 rad/s. The spring constant k2 is half of k1, and the mass is 1 kg. The mass associated with the other components may be assumed to be negligible. For the given natural frequency, the spring constant k1 is most nearly

0=+ xkxm eq

21

1211

kkkeq+=

mkeq

n =ω

Example 13 A two-bar linkage rotates about the pivot point O as shown. The length of members AB and OA are 2.0 m and 2.5 m, respectively. The angular velocity and acceleration of member OA is ωOA = 0.8 rad/s CCW and αOA = 0 rad/s2. The angular velocity of member AB is ωAB = 1.2 rad/s CW, and the acceleration of member AB is αAB = 3 rad/s2 CCW. When the bars are in the position shown, the magnitude of the acceleration of point B is most nearly

( )( )

ABABAB

ABABABABABAB

AOOAOAOAOAA

avrraa

rra

//

//

/

2

+×

+×+××+=

×+××=

ω

αωω

αωω

Gears – Quick Review

Spur Gears" Helical Gears"

Bevel Gears"

Types of Gears

Worm Gear"

Rack and Pinion"

Concepts Review

Terminology

Fundamental Law of Gearing

N1 N2 θ2

θ1 1

2

1

2

NN

rrN ==

2211 θ=θ rr

Gear Ratio Fixed Shafts

2

1

2

1

1

2

1

2

ωω

=θθ

===NN

rrN

Gear Compatibility USCS

rNp2

= Units???"

SI

Nrm 2

=

Idlers

N3

N2 θ1

θ3

3

1

2

1

3

2

1

2

2

3

1

3

NN

NN

NN

==ωω

ωω

=ωω

Gear Ratio Fixed Shafts

2

1

2

1

1

2

1

2

ωω

=θθ

===NN

rrN

θ2

N1

Double Reductions Gear Ratio

Fixed Shafts

⎟⎟⎠

⎞⎜⎜⎝

⎛−==

2

1123 NNnnn

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

4

334 NNnn

42

31

1

4

NNNN

nn

=

Double Reductions Gear Ratio

Fixed Shafts

thdriven tee ofproduct teethdriving ofproduct

input

output =nn

42

31

1

4

NNNN

nn

=

Example 1 - Speed Reducer

502 =N

minrev4032 == nn

Example 2 The compound gear train shown is attached to a motor that drives gear A at ω in clockwise as viewed from below. What is the expression for the angular velocity of gear H in terms of the number of teeth on the gears? What is the direction of rotation of gear H as viewed from below.

Gear Forces

Wt

Wr

t

r

WW

=φtan

mNT

dTWt

22==

SI Units: N, mm, kW

ωω mNH

dHWt

22==

•  Grashof Mechanism – One link can perform a full rotation relative to another link

•  Grashof Criterion

ba LLLL +<+ minmax

Classification of 4-Bar Linkages

min1 LL =

Crank-Rocker Mechanism

min0 LL =

Drag Link Mechanism

min2 LL =

Double-Rocker Mechanism

ba LLLL +=+ minmax

Also called Crossover-Position Mechanism Change-Point Mechanism

ba LLLL +>+ minmax

•  No link can rotate through 360° •  Double Rocker Mechanism of the 2nd Kind •  Triple-Rocker Mechanism

Non-Grashof Mechanism

Crank-Slider Mechanism

Quick Return Mechanism

Geneva Mechanism