0.27

0.475

0.5

0.73

315°

160°

45°

Steve Goddard

Mechanical Principles – Assignment 4

Dynamics of Rotating Systems

1. A turbine rotor has an out of balance mass equivalent to 0.05kg at a radius of 3mm. If the speed of the rotor is 18000 rev/min find the unbalanced force produced on the bearing of the rotor.

First of all I converted rpm into rad/s

18000 rpm = 1884.96 rad/s

Force =

= = 532.96 N

2. A system of rotating masses is as shown below

If the system rotates at 250 rev/min find the out-of-balance force on the shaft.

M (kg) r (m) Mr (Kgm) θA 2.5 0.2 0.5 0B 1.9 0.25 0.475 45C 1.5 0.18 0.27 160

Balancing 0.73

250 rpm=26 . 179rad

sec

Unbalanced Force = M×ω2r

Mr×ω2

0 .73×26 . 1792

¿500 .298 N

20° 45°

1.5kg

C B

A0.2m

0.25m0.18m

1.9kg

2.5kg

Steve Goddard3. Using the system shown above and with a speed of 400 rev/min find the radius at which a mass of 2.1kg must be attached in order to balance the system and the angular position of this mass

M (kg) r (m) Mr (Kgm) θA 2.5 0.2 0.5 0B 1.9 0.25 0.475 45C 1.5 0.18 0.27 160

Balancing 2.1 0.73

Mr=0 .732 .1 r=0. 73

r=0 .732. 1

=0 . 347

4. a circular plate rotates about the axis through its centre with four masses A,B,C and D located on its surface but off centre. Mass A is 2.5kg a radial distance of 350mm, mass B is 4kg at a radial distance of 300mm at right angles to mass A, mass C is 3.75kg at a radial distance of 200mm at 150° to the radius A and mass D is at 5kg at a radial distance 250mm at an angle of 240° to radius A.

Determine the balancing mass required and its angular position, if it is to be placed on the plate at a radial distance of 250mm.

M (kg) r (m) Mr (Kgm) θA 2.5 0.35 0.875 0B 4 0.3 1.2 90C 3.75 0.2 0.75 150D 5 0.25 1.25 240

Balancing 2.48 0.25 0.62 309

To work out M:

Mr=0 .62

M=0 . 62r

=0. 620. 25

=2. 48 Kg

Steve GoddardDiagram of vectors to accompany question 4.

5. Two rotors A and B are fixed to a shaft 0.3m apart. The out of balance mass of A is equivalent to 0.1kg at 100mm radius and B is 0.05kg at 80mm radius. The mass centres are 45° apart.Two balancing masses C and D are to be placed in planes C and D at 100mm radius. The planes C and D are equally spaced between A and B.

Steve GoddardFind the magnitude and angular positions of the masses C and D.

Firstly I drew out the question:

Plane M (Kg) R (m) Mr (Kgm) L (m) MrL (Kgm2)

Ө

A 0.1 0.1 0.01 -0.3 0.03 0°B 0.1 -0.2 170°C 0.1 -0.1 205°D 0.05 0.08 0.004 0 0.0008 45°

Then I make C the reference plane

Plane M (Kg) R (m) Mr (Kgm) L (m) MrL (Kgm2)

Ө

A 0.1 0.1 0.01 -0.1 0.001 0°B 0.174 0.1 0.174 -0.2 0 172°C 0.07 0.1 0.007 -0.1 0.0007 82°D 0.05 0.08 0.004 0.2 0.0008 45°

6. A rotating shaft length 2.75m is supported by bearings at each end and carries three cranks A, B and C mounted on it. Crank A has a mass of 3.25kg acting at a radial distance of 195mm and 2.45m from the left hand end

Steve Goddardbearing. B has a mass of 2.0kg at a radial distance of 250mm and 0.95m from the left hand end bearing.

Determine the forces acting on the two bearings when the shaft is rotating at 5rev/s and when the cranks relative to a datum plane are as given in the following diagram:

Firstly I drew out my problem:

Plane M (Kg) R (m) Mr (Kgm) L (m) MrL (Kgm2) ӨA 3.25 0.195 0.63325 2.45 1.5526 30°B 2.0 0.325 0.65 1.50 0.975 145°C 1.25 0.250 0.3125 0.95 0.2968 75°

Left 0 0Right 2.75

7. A flywheel has a moment of inertia of 22.5 x 103 kgm2 and is rotating at 500rpm. If a constant resisting torque of 700Nm is applied to the wheel, determine.

45°

30°

B

C

A

70°Datum

Steve Goddard7.1 The initial kinetic energy of the flywheel7.2 The time taken to bring it to rest

7.1

First of all I converted the rpm to rad/s

500rpm = 52.36 rad/s

The equation for Kenetic energy is I put my values in:

= 30.842 MJ

7.2

Firstly rearrange Newton’s Second Law to make α the subject.

T=Iα ∴ α=TI

So :700

22. 5×103=0 .031

Using the following equation:

ω2=ω1+α t

0=52.36+0 .031 t

−52 .36=0 .031 t

−1689 .032=t

Time taken to slow down = 1689.032 Seconds

Steve Goddard8. A flywheel is required to absorb 6kJ of kinetic energy while its speed increases from 40 rev/s to 42 rev/s. Determine the moment of inertia that the flywheel should have.

First convert the rev/s into rad/s:

40 rev/s = 2400 rpm2400 rpm = 251.32 rad/s

42rev/s = 2520 rpm2520 rpm = 263.89 rad/s

Change of energy =

12I (ω2

2−ω1

2)6×103=

12I (263 . 892−251. 322 )

6×103=12I (69637 . 932−63161 . 742 )

6×103=3238 .095 I

I=1 .852 Kgm2

9. Determine the constant torque that must be applied to a flywheel to bring it to rest in 20 revolutions from rotating at 10rev/s if it has a moment of inertia of 6kgm2.

10 rev/s = 62.83 rad/s

Therefore:

ω1=62 . 83 rad / s And

ω2=00 rad /s

Torque = Iα

So to work out α :

α=ω2−ω1

t

α=0−62 .832

=−31 . 415

Or a retardation of 31.415 r/s2

T=Iα=6×3 .1415=18 .849 Nm

ω1

ω2

If the fly wheel is rotating at 10/ rev/s and needs to stop in 20 revs this will take 2

Steve Goddard10. Determine the output torque from a gearbox, which gives a reduction of 4:1 with an efficiency of 98%, when the input is a power of 20kW and a shaft rotation of 50rev/s.

P=Tω

T Input=20×103

314 . 159=63 . 662 Nm

Assuming the gearbox at a 100% efficiency.

T iωi=T oωo

Gear Ratio=ωi

ωo

T output=T i

ωi

ωo

=63 .662×4=254 .648 Nm

Therefore in a gearbox with 98% efficiency:

T output=254 .648×0 .98=249 . 555 Nm

11. A gearbox has an input rotor with a moment of inertia of 3kgm2 and an output rotor with a moment of inertia of 24kgm2 and gives a reduction of 4:1 with an efficiency of 80%.

Determine the angular acceleration of the input shaft when there is a torque of 100Nm applied to the input shaft.

If T A=(I A+ I B

G2 )α A

Putting in my own values I get:

100=(3+2416 )αA

100=4 . 5 α A

α A=22 . 22 rad /s Assuming 100% Efficiency

With 80% efficiency:

22 .22×0 .80=17 . 776 rad /s2

12. A power press makes steel pressings. The energy required for each pressing is 5000J with each pressing taking 1 second. The time taken to

Steve Goddardremove the pressing and get the material ready for the next pressing is 4 seconds. If the press is to be operated by an electric motor which runs continuously with a flywheel used to supply the extra torque required for the pressing, determine:

12.1 The power of the motor that is required 12.2 The energy that has to be supplied by the flywheel12.3 The moment of inertia of the flywheel

If its speed is not to drop by more than 30 rev/min from a speed of 600 rev/min.

12.1

Energy required = 5000J, total time = 5 seconds.

Power=Energy RequiredTime

=50005

=1 kW

12.2

Total Energy = Power x Time

Therfore:

1000 W x 4 = 4 kJ

12.3

Firstly I converted all the rev/m into rad/s:

600 rpm = 62.832 rad/s570 rpm = 59.690 rad/s

Then using the equation:

δEnergy=12I (ω2

2−ω1

2)

4000 J=0 .5 I ( 62. 8322−59 .6902)=0 .5 I (384 . 964 )

I=20 .78 Kgm2

13. A flywheel and shaft have a moment of inertia of 720kgm2 and is rotating at 5 rev/s. Another flywheel and shaft, on the same axis, is

Steve Goddardsuddenly connected by means of a clutch. If this flywheel and shaft has a moment of inertia of 720kgm2 and is initially at rest, calculate:

13.1 the common speed of rotation after the two are connected13.2 the loss in kinetic energy

13.1

5 rev/s = 31.41 rad/s

Angular momentum before connection = Angular momentum after connection

(720×31.416 )+ (0 )=(720+720 )ω

22619 .52=1440ω

ω=15 .708 rad /s

13.2

Loss of Kinetic Energy = KE before – KE after

KELoss=12

((720×31. 4162)+( 720×02)− (1440×15 .7082))

¿12

=(710614 . 84−355307. 42 )

¿177 . 653KJ

14. In the diagram below (See assignment pack) the following distances apply:-

OA = 150mm OB = 75mm AC= 300mm CD = 250mm

Steve GoddardOB rotates at a uniform angular velocity of 20red/s clockwise.

14.1 Find the velocity of the piston D14.2 Find the angular velocity of the link CD

15. (See assignment pack) The mechanism consist of links AB = 100mm, BCD = 450mm, FC = 150mm, DE = 250mm and point C is 250mm from B.The slider at E is constrained to move in a vertical guide on the frame. CF rotates at a constant angular velocity of 120 rev/min in an anticlockwise direction.

For the position shown, find:

15.1 The velocity of E relative to the frame15.2 The acceleration of E relative to the frame

16. A bar mechanism is shown below (see assignment pack). OA is vertical. The link BC has an angular velocity of 8 red/s clockwise and an angular acceleration of 12 rad/s2 anticlockwise.

Find for the position shown,

16.1 The acceleration of A relative to the frame16.2 The angular acceleration of the link AB

Steve Goddard

Bibliography

Lecture Notes

HNC/HND Book – Tooley & Dingle