rotation - 5.35

5.7 PRACTICE PROBLEM SOLUTIONS

1.

T

θ 1

Ks1( θ 1 – θ 2)

θ 2

JM1

JM2

Ks1( θ 1 – θ 2)

Ks2θ

2

Bθ·2

+

∑M1 = T – Ks1( θ 1 – θ 2) = JM1 θ··1

··

Ks1

–Ks1

T

θ 1 + θ

-------

-----------

-------

1 J

+ θ

2 J

M1

=

J

M1

M1

+

∑M2 = Ks1( θ 1 – θ 2) – Bθ·2 – Ks2θ 2 = JM2 θ··2

··

· B

Ks1 + Ks2

–Ks1

θ 2 + θ

-------

+ θ

2

----------------------

+

θ

-----------

= 0

2 J

M2

J

M2

1 J

M2

2.

x·1

=

v1

·

=

x

–Ks1

+ x

Ks1

+ g

x

-----------

-------

1

1

M

1

2

M

1

x·2

=

v2

·

–R2K

–R2K

R2K

–FR

1

= v

2

d1

+ x

2

s1

+ x

2

s1

+

v

------------

2

1

------------------

1

-----------------

2

--------------

J1

J1

J1

J1

rotation - 5.36

3.

T1

T2

M1

JM1

JM2

Kdx·1

gM1

T1

T2

if T1,T2,Kdx1 > 0

θ

–x

1

x1

T

=

K

( x

– x )

1 = -------

θ 2 = -----

1

2

R2

R1

s

1

··

+

∑Fy = T1 – gM1 = –M1x2

··

=

T1

x2

g – ------

M1

+

∑M1 = – T1R1 + T2R1 = –JM1 θ··2

··

=

T1R1 – T2R1

θ 2

-----------------------------

JM1

+

∑M2 = T2R2 – R2Kdx·1 = –JM2 θ··1

··

·

–R2Kd

T2R2

θ 1 + x1

---------------

-----------

J

=

J

M2

M2

6 equations, 6 unknowns

rotation - 5.37

4.

θ Ks1			T		Ks2( x2 – θ R)
a)
JM
RT		M1			M2
RKs2( x2 – θ R)
	F		M1g			M2g

··

if(T < 0) T=0

∑FM1 = T – M1g – F = –M1x1

if(T >= 0) Rθ

··

··

T = – M1x1 + M1g + F = M1Rθ + M1g + F

∑MJ = – RT – θ Ks1 + RKs2( x2 – θ R) = JMθ··

+ R2M )θ··

– R( M

g + F) – θ ( K

s1

+ R2Ks2) + ( RK

) x

2

= ( J

M

1

s2

1

··

Ks1 + R2Ks2

–RKs2

=

–R( M1g + F)

θ + θ

+ x

--------------------------

--------------------------------

-----------------------------

2

J

M

+ R2M

1

J

M

+ R2M

J

M

+ R2M

1

··

1

∑FM2 = Ks2( x2 – θ

R) – M2g = –M2x2

··

Ks2

–RKs2

-------

+ θ

---------------

= g

x2 + x2 M

2

M

2

= –x1

(1)

(2)

b) θ·

= ω

– R2K

g – RF

·

– K

s1

RK

s2

– RM

1

= θ

s2

+ x

+

ω

---------------------------------

2

--------------------------

--------------------------------

J

M

+ R2M

1

J

M

+ R2M

J

M

+ R2M

1

x·2 = v2

1

·

RKs2

–Ks2

v2 = θ

-----------

+ x2

-----------

+ g

M

2

M

2

rotation - 5.38

5.

θ·

= ω

·

– K

s1

– r2K

–K

d1

K

s2

r

Fr

= θ

s2

+

ω

+ ω

+ x

----------

-----

--------------------------------

-----------

JM

JM

JM

JM

·

= v

x

·

Ks2r

–Kd2

– Ks2

– Ks3

v = θ

----------

+ v

-----------

+ x

--------------------------

M

M

M

6.

x·1 = v1

·

–Kd1

–Ks1

Kd1R2R4

Ks1R2R4

v1 = v1

-----------

-----------

+ ω 1

---------------------

+ θ 1

--------------------

+ g

M

1

+ x1

M

1

M

1

R

3

M

1

R

3

θ·1

= ω 1

–Kd1R22R42

–Ks1R22R42

ω

·

= v1

Kd1R2R4

+ x1

Ks1R2R4

1

+ θ

+

–F1R1

1

+ ω

2

1

2

---------------

---------------------

--------------------

------------------------

------------------------

J1R3

J1R3

J1R3

J1R3

J1

rotation - 5.39

7.

	F1R1		θ 1
		JM1
	JM1θ··1		τ
	θ	1N1 = θ 2N2

x·1 = v1

θ 2

R

K

·

·

2

d1

( x

1

– x )

3

JM2

JM2θ··2

τ

N2

-----

( x

– x )

N1

R

2

K

s1

1

3

θ 2 =

x3

-----

R2

Kd1( x·1 – x·3)

Ks1( x1 – x3)

		M3
··			M3g
M3x1

v·1

θ·2

ω·2

–K

d1

–K

s1

R

2

K

d1

R2K

s1

-----------

-----------

+ ω

--------------

--------------

– g

= v1 M

3

+ x1

M

3

2

M

3

+ θ 1

M

3

= ω 2

2

2

–N2

v1( R2Kd1) + x1( R2Ks1)

---------

+ω 2( –R2Kd1) +θ 1( –R2Ks1) + F N1

R1

= -----------------------------------------------------------------------------------------------------------------------------------------------------------------

N22

+ JM2

JM1 -----

N12

rotation - 5.40

8.

					Kd1x·1
	Ks1x1
								x1
µ			x1			M1
	kN		-------				K		( x		– x )
			x1
								s2		1	2
N = M1g cos ( θ						3)
						M1g sin ( θ 3)
						x·1 = v1
						v·1	=
						x·2 = v2
						v·2	=

9.

x·2 = v2

Ks2( x1 – x2)

··		x2
M2x2	JM2, R2
		θ 2
	M2
	Fc	F2
θ 2 =	x2	M2g sin ( θ 3)
	-----
	R2

·		R2R2	( – K				s2		– K		)								R2R2K						s2				R2R2( 2F				1	+ M	2	g)
	= x	1 2									s3		+ x						1 2								+			1 2
v																-----------------------------------------------------------												-----------------------------------------------------------
2	3 4-----------------------------------------------------------J R2 + J				R2			+ R2R2M							2
																4J		R2	+ J			R2		+ R2R2M				4J		R2 + J		R2		+ R2R2M
	x·3	1= 2v3		2		1				1 2 2							1	2		2			1		1 2 2				1	2	2		1	1 2			2
	·		–Kd3								– Ks2 – Ks3								Ks2
	v3	= v3 -----------M				3				+ x3 --------------------------			M	3			+ x2 -------M				3			+ g

rotation - 5.41

10.

state equations

·

= v1

x1

·

R1 + R2

v1

= F1 -----------------R

1

M

1

·

–Ks2

Ks2

F1

q = x2 -----------K

d2

+ x3 K--------

+ K--------

d1

d2

·

–Ks1

R1Ks1

R1 + R2

p = x1 -----------K

d1

+ x2 K--------------------------------

d1

( R

1

+ R )

+ F1 ------------------R

1

K

d1

2

output equations

R1 + R2

x2

-----------------

= ( x1 – p)

R1

x3

= – q + x2

11.

x·1 = v1
	·		–Ks1					+ θ		R2Ks1N1
v1 = x1			-----------						1		---------------------								+ g
				M		1						M	1	N	2
θ·1
		= ω 1																	–Kd1N12								–R22Ks1N12							F1R1N22
ω	·	R2Ks1N1N2												ω									+ θ								+
		= x		-----------------------------								+																				-----------------------------
																	-----------------------------									-----------------------------
	1	1 J			1	N2		+ J	2	N2					1	J		1	N2 + J	2	N2			1	J		1	N2 + J	2	N2		J	1	N2 + J	2	N2
							2				1								2			1						2		1				2			1

12.

For area:

Jarea

=

R

2

R

2

( 2

π rdr) =

R

3

dr

= 2π

r4

R

π

R4

∫r

dA = ∫r

2π ∫r

----

= ---------

4

0

2

0

0

0

For mass:

ρ =

M

M

----

=

---------

A

π

R2

R

2

R

2

R

3

r4

R

π R4

MR2

Jmass

=

∫

r dM =

∫

r ( ρ

2π rdr)

= 2πρ

∫

r dr =

2πρ

4

= ρ

2

2

----

0

---------

= ----------

0

0

0

The mass moment can be found by multiplying the area moment by the area density.

rotation - 5.42

13.

a)

T

+ ∑M = T – Ksθ – Bθ· = JMθ··

JM

JMθ·· + Bθ· + Ksθ

·

Ksθ

= T

··

·

B

Ks

T

Bθ

θ

+ θ

-----

-----

-----

JM

+ θ JM

=

JM

b)

θ '

=

ω

·

=

T

Ks

–

B

ω

----- –

-----θ

-----ω

JM

JM

JM

Nm

1

Nms

·

10Nm

10--------

----------

rad

rad

ω

=

-----------------

–

-----------------

–

-----------------

ω

2

1Kgm

2

1Kgm

2 θ

1Kgm

2

·

Nm

10θ

s

1

ω

=

-------------

--------

--------

Kgm

2

10 – rad

– radω

Kgm

m

0

-----------

·

s

2

10θ

s

5s

ω

=

---------------------

--------

--------

Kgm

2

10 – rad

– radω

·

–2

10θ

s

ω

= s

--------

--------

ω

10 – rad – rad

rotation - 5.43

(c

homogeneous:

θ·· + ( 1s–1)θ· + ( 10s–2)θ

=

0

guess:

θ h = e

At

·

= Ae

At

θ

··

= A

2

At

θ h

h

e

A

2

At

–1

At

–2

At

= 0

e + (

1s

) Ae

+ ( 10s

) e

2

–1

–2

– 1s

–1

±

–1

)

2

–2

A

+

(

1s

10s

= 0

A =

( 1s

– 4( 1) ( 10s

)

) A +

------------------------------------------------------------------------------

2(

1)

A

=

– 1s

–1 ±

1s–2

– 40s–2

=

( – 0.5

± j3.123) s

–1

-------------------------------------------------------

2

( 1)

θ

h

= C

1

e–0.5s–1t cos ( 3.123s–1t + C )

particular:

2

θ '' + ( 1s–1)θ

' + ( 10s–2)θ

=

10s–2

guess:

θ p = A

θ·p = 0

θ··p

= 0

10s–2

+ ( 1s–1) (

+ ( 10s–2) ( A)

10s–2

( 0)

0)

=

A =

------------ = 1

θ

= 1

10s–2

p

Initial conditions:

θ ( t)

= C

1

e–0.5s–1 t cos ( 3.123s–1t + C )

+ 1

2

θ ( 0)

= C

1

e–0.5s–1 0 cos ( 3.123s–10 + C )

+ 1 = 0

2

C1 cos ( C2)

+ 1 = 0

θ '( t) = – 0.5s–1C1e–0.5s–1t cos ( 3.123s–1t + C )

– 3.123s–1C

1

e–0.5s–1 t sin ( 3.123s–1t + C )

2

2

θ '( 0)

= – 0.5s–1C1( 1) cos ( C

2

) – 3.123s–1C

1

( 1) sin ( C )

= 0

2

– 0.5 cos ( C2) – 3.123 sin ( C2)

= 0

sin ( C2)

=

–0.5

=

tan

( C2)

C2

=

–0.159

-------------------

------------

cos (

C2)

3.123

–1

C

cos ( –0.159)

+ 1 =

0

C1

=

–1.013

1

----------------------------- =

cos ( –0.159)

θ ( t)

= –1.013e–0.5s–1t cos ( 3.123s–1t – 0.159)

+ 1

14.

θ ( t) =	10	+ ( –1.283) e	–1.5t		27
	-----				---------
	9			cos	2	t – 0.524