Introduction to Dynamics (N. Zabaras)

HW8 SolutionsMechanical Vibrations

Prof. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: nzabaras@gmail.com

URL: http://www.zabaras.com/

May 3, 2016

1

Introduction to Dynamics (N. Zabaras)

Problem 1

k

A cylinder of weight W is

suspended as shown.

Determine the period and natural

frequency of vibrations of the

cylinder.

SOLUTION:

• From the kinematics of the system,

relate the linear displacement and

acceleration to the rotation of the

cylinder.

• Based on a free-body-diagram

equation for the equivalence of the

external and effective forces, write the

equation of motion.

• Substitute the kinematic relations to

arrive at an equation involving only the

angular displacement and

acceleration.

2

Introduction to Dynamics (N. Zabaras)

Problem 1

SOLUTION:

• From the kinematics of the system, relate the linear

displacement and acceleration to the rotation of the cylinder.

rx rx 22

rra

ra

• Based on a free-body-diagram equation for the equivalence of

the external and effective forces, write the equation of motion.

: effAA MM IramrTWr 22

rkWkTT 2but21

02

• Substitute the kinematic relations to arrive at an equation

involving only the angular displacement and acceleration.

21 12 2

2 2

80

3

Wr W kr r m r r mr

k

m

m

kn

3

8

k

m

nn

8

32

2

m

kf nn

3

8

2

1

2

3

Introduction to Dynamics (N. Zabaras)

Problem 2

s13.1

lb20

n

W

s93.1n

The disk and gear undergo torsional

vibration with the periods shown.

Assume that the moment exerted by the

wire is proportional to the twist angle.

Determine a) the wire torsional spring

constant, b) the centroidal moment of

inertia of the gear, and c) the maximum

angular velocity of the gear if rotated

through 90o and released.

SOLUTION:

• Using the free-body-diagram equation

for the equivalence of the external

and effective moments, write the

equation of motion for the disk/gear

and wire.

• With the natural frequency and

moment of inertia for the disk known,

calculate the torsional spring

constant.

• With natural frequency and spring

constant known, calculate the

moment of inertia for the gear.

• Apply the relations for simple

harmonic motion to calculate the

maximum gear velocity.

4

Introduction to Dynamics (N. Zabaras)

Problem 2

s13.1

lb20

n

W

s93.1n

SOLUTION:

• Using the free-body-diagram equation for the

equivalence of the external and effective moments,

write the equation of motion for the disk/gear and wire.

: effOO MM

0

I

K

IK

K

I

I

K

nnn

2

2

• With the natural frequency and moment of inertia for

the disk known, calculate the torsional spring

constant.2

22

21 sftlb 138.0

12

8

2.32

20

2

1

mrI

K

138.0213.1 radftlb27.4 K

5

Introduction to Dynamics (N. Zabaras)

Problem 2

s13.1

lb20

n

W

s93.1n

radftlb27.4 K

K

I

I

K

nnn

2

2

• With natural frequency and spring constant known,

calculate the moment of inertia for the gear.

27.4293.1

I 2sftlb 403.0 I

• Apply the relations for simple harmonic motion to

calculate the maximum gear velocity.

nmmnnmnm tt sinsin

rad 571.190 m

s 93.1

2rad 571.1

2

nmm

srad11.5m

6

Introduction to Dynamics (N. Zabaras)

Problem 3

7

The 10-kg rectangular plate shown.is

suspended at its center from a rod

having a torsional stiffness k=1.5 N · m

/ rad. Determine the natural period of

vibration of the plate when it is given a

small angular displacement in the

plane of the plate.

The torsional restoring moment created by the rod is

M = k . This moment acts in the direction opposite to

the angular displacement . The angular acceleration

acts in the direction of positive .

Equation of Motion.

0 0

0

O n

kM I k I

I

2 2 2 2 2 2

0

1 110 0.2 0.3 0.1083 .

12 12I m a b kg m kg m

02 0.10832 2 1.69

1.5n

Is

k

Introduction to Dynamics (N. Zabaras)

Problem 4

8

The bent rod shown has a negligible

mass and supports a 5-kg collar at its

end. If the rod is in the equilibrium

position shown. determine the natural

period of vibration for the system.

Since the spring is subjected to an initial

compression of xst for equilibrium, then

when the displacement x > xst the spring

exert a force of Fs=k (x -xst) on the rod.

5ay must act upward, in accordance with

positive

Introduction to Dynamics (N. Zabaras)

Problem 4

9

Equation of Motion.

:

(0.1 ) (0.1 ) 49.05 (0.2 ) (5 ) (0.2 )

B k B

st y

M

kx m kx m N m kg a m

M

-k xst(0.1m), represents the moment created by the

spring force which is necessary to hold the collar in

equilibrium, i.e., at x = 0. This moment is equal and

opposite to 49.05 N(0.2 m) created by the weight

of the collar. Thus

(0.1 ) (5 ) (0.2 )ykx m kg a m

Introduction to Dynamics (N. Zabaras)

Problem 4

10

Kinematics. Since is small,

x = (0.1 m) and y = (0.2 m) .

Therefore:

2

(0.1 ) (5 ) (0.2 ) 5(0.2 )0.2

20 0 20 4.47 /

2 / 2 / 4.47 1.40

y

n n

n

kx m kg a m

rad s

s

Introduction to Dynamics (N. Zabaras)

Problem 5

11

A 10-lb block is suspended from a cord that passes

over a 15-lb disk. The spring has a stiffness k = 200

lb/ft. Determine the natural period of vibration for

the system.

:O k OM M

.. ..

0.75 0.75bs a s

200 / (0.75 ) 10sF lb ft ft lb

2

368 0

368 19.18 /

2 / 2 /19.18 0.328

n n

n

rad s

s

..2

2 2

1 15 1010 (0.75 ) (0.75 ) (0.75 ) (0.75 )

2 32.2 / 32.2 /s b

lb lblb ft F ft ft a ft

ft s ft s

Introduction to Dynamics (N. Zabaras)

Problem 6

12

The thin hoop show is supported by the peg at O .

Determine the natural period of oscillation for small

amplitudes of swing. The hoop has a mass m.

2 2 2 2 2 2

0

2 2

1 1( )

2 2

( cos )

cos

nT I mr mr mr

V mg r

T V mr mgr

Take time derivative:

2 2 sin 0

2 sin 0

sin 0 ( )2

20 2 / 2

2 2n n

mr mgr

mr r g

gsmall

r

g g r

r r g

Introduction to Dynamics (N. Zabaras)

Problem 7

13

A 10-kg block is suspended from a cord that passes

over a 5-kg disk. The spring has a stiffness k = 200

N/m. Determine the natural period of vibration for

the system.

2

2 2 2 2

0

2 2

2 2

1 1 1 1 1(10 ) (0.15 ) ( 5 ) (0.15 ) 0.1406

2 2 2 2 2

1 1( ) (200 / )( 0.15 ) 98.1 (0.15 )

2 2

0.1406 100( 0.15 ) 14.715

b b d

st st

st

T m v I kg m kg m

V k s s Ws N m s N m

T V s

Take time derivative:

0.28125 200( 0.15 )0.15 14.715 0

98.1/ 200 0.4905

16 0 16 4 / 2 / 4 1.57

st

st

n

s

s m

rad s s

0.15m

0.15m

0.15

98.1N

0.15s

Datum

sts s

200 /k N m

200 /k N m