Mechanical Engineering Principles

TORQUE 113

In general, this may be written as:

Total kinetic energy=

(∑

mr^2

)ω^2

2

=I

ω^2

2

whereI(=mr^2 )is called themoment of inertia
of the system about the axis of rotation and has units
of kg m^2.
The moment of inertia of a system is a measure
of the amount of work done to give the system an
angular velocity ofωrad/s, or the amount of work
that can be done by a system turning atωrad/s.
From Section 9.2, work done = Tθ, and if this
work is available to increase the kinetic energy of a
rotating body of moment of inertiaI, then:

Tθ=I

(

ω^22 −ω^21

2

)

whereω 1 andω 2 are the initial and final angular
velocities, i.e.

Tθ=I

(

ω 2 +ω 1

2

)

(ω 2 −ω 1 )

However,

(

ω 2 +ω 1

2

)

is the mean angular velocity, i.e.

θ

t

, wheretis the

time, and(ω 2 −ω 1 )is the change in angular velocity,
i.e.αt,whereαis the angular acceleration. Hence,

Tθ=I

(

θ

t

)

(αt)

from which, torqueT=Iα

whereIis the moment of inertia in kg m^2 ,αis the
angular acceleration in rad/s^2 andT is the torque
in N m.

Problem 10. A shaft system has a moment

of inertia of 37.5 kg m^2. Determine the

torque required to give it an angular

acceleration of 5.0 rad/s^2.

Torque,T = Iα, where moment of inertiaI =
37 .5kgm^2 and angular acceleration,α= 5 .0 rad/s^2.
Hence,torque,T=( 37. 5 )( 5. 0 )= 187 .5Nm

Problem 11. A shaft has a moment of

inertia of 31.4 kg m^2. What angular

acceleration of the shaft would be produced

by an accelerating torque of 495 N m?

Torque,T=Iα, from which, angular acceleration,

α=

T

I

, where torque,T=495 N m and moment

of inertiaI= 31 .4kgm^2.

Hence,angular acceleration,

α=

495

31. 4

= 15 .76 rad/s^2

Problem 12. A body of mass 100 g is

fastened to a wheel and rotates in a circular

path of 500 mm in diameter. Determine the

increase in kinetic energy of the body when

the speed of the wheel increases from 450

rev/min to 750 rev/min.

From above, kinetic energy=I

ω^2

2

Thus, increase in kinetic energy=I

(

ω^22 −ω 12

2

)

where moment of inertia,I=mr^2 ,

mass,m=100 g= 0 .1kg

and radius,

r=

500

2

=250 mm= 0 .25 m.

Initial angular velocity,

ω 1 =450 rev/min=

450 × 2 π

60

rad/s

= 47 .12 rad/s,

and final angular velocity,

ω 2 =750 rev/min=

750 × 2 π

60

rad/s

= 78 .54 rad/s.

Thus,increase in kinetic energy

=I

(

ω^22 −ω^21

2

)

=(mr^2 )

(

ω^22 −ω^21

2

)

=( 0. 1 )( 0. 252 )

(

78. 542 − 47. 122

2

)

= 12 .34 J