Mechanical Engineering Principles

114 MECHANICAL ENGINEERING PRINCIPLES

Problem 13. A system consists of three

small masses rotating at the same speed

about the same fixed axis. The masses and

their radii of rotation are: 15 g at 250 mm,

20 g at 180 mm and 30 g at 200 mm.

Determine (a) the moment of inertia of the

system about the given axis, and (b) the

kinetic energy in the system if the speed of

rotation is 1200 rev/min.

(a) Moment of inertia of the system,I=mr^2

i.e. I=[( 15 × 10 −^3 kg)( 0 .25 m)^2 ]

+[( 20 × 10 −^3 kg)( 0 .18 m)^2 ]

+[( 30 × 10 −^3 kg)( 0 .20 m)^2 ]

=( 9. 375 × 10 −^4 )+( 6. 48 × 10 −^4 )

+( 12 × 10 −^4 )

= 27. 855 × 10 −^4 kg m^2

= 2. 7855 × 10 −^3 kg m^2

(b) Kinetic energy = I

ω^2

2

, where moment of

inertia,I= 2. 7855 × 10 −^3 kg m^2 and angular

velocity,

ω= 2 πn= 2 π

(

1200

60

)

rad/s= 40 πrad/s

Hence,kinetic energy in the system

=( 2. 7855 × 10 −^3 )

( 40 π)^2

2

= 21 .99 J

Problem 14. A shaft with its rotating parts

has a moment of inertia of 20 kg m^2 .Itis

accelerated from rest by an accelerating

torque of 45 N m. Determine the speed of

the shaft in rev/min (a) after 15 s, and

(b) after the first 5 revolutions.

(a) Since torqueT =Iα, then angular accelera-

tion,α=

T

I

=

45

20

= 2 .25 rad/s^2.

The angular velocity of the shaft is initially

zero, i.e.ω 1 =0.

From chapter 11, page 129, the angular veloc-

ity after 15 s,

ω 2 =ω 1 +αt= 0 +( 2. 25 )( 15 )

= 33 .75 rad/s,

i.e.speed of shaft after 15 s

=( 33. 75 )

(

60

2 π

)

rev/min= 322 .3rev/min

(b) Work done=Tθ, where torqueT=45 N m

and angular displacementθ=5 revolutions=

5 × 2 π= 10 πrad.

Hence work done=( 45 )( 10 π)=1414 J.

This work done results in an increase in kinetic

energy, given byI

ω^2

2

, where moment of inertia

I=20 kg m^2 andω=angular velocity.

Hence, 1414=( 20 )

(

ω^2

2

)

from which,

ω=

√(

1414 × 2

20

)

= 11 .89 rad/s

i.e.speed of shaft after the first 5 revolutions

= 11. 89 ×

60

2 π

= 113 .5rev/min

Problem 15. The accelerating torque on a

turbine rotor is 250 N m.

(a) Determine the gain in kinetic energy of

the rotor while it turns through 100

revolutions (neglecting any frictional

and other resisting torques).

(b) If the moment of inertia of the rotor is

25 kg m^2 and the speed at the beginning

of the 100 revolutions is 450 rev/min,

determine its speed at the end.

(a) The kinetic energy gained is equal to the work

done by the accelerating torque of 250 N m

over 100 revolutions,

i.e.gain in kinetic energy = work done =

Tθ=( 250 )( 100 × 2 π)= 157 .08 kJ