Mechanical Engineering Principles

TORQUE 111

Power P= 2 πnT(from above),from which,

torque, T=

P

2 πn

Nm

where power,P = 2 .50 kW=2500 W and speed,
n= 1000 /60 rev/s Thus,

torque, T=

P

2 πn

=

2500

2 π

(

1000

60

)

=

2500 × 60

2 π× 1000

= 23 .87 N m

Problem 6. An electric motor develops a

power of 5 hp and a torque of 12.5 N m.

Determine the speed of rotation of the motor

in rev/min.

Power, P= 2 πnT,from which,

speed n=

P

2 πT

rev/s

where power, P=5hp= 5 × 745. 7

= 3728 .5W

and torqueT= 12 .5Nm.

Hence, speedn=

3728. 5

2 π( 12. 5 )

= 47 .47 rev/s

The speed of rotation of the motor
= 47. 47 × 60 =2848 rev/min.

Problem 7. In a turning-tool test, the

tangential cutting force is 50 N. If the mean

diameter of the work-piece is 40 mm,

calculate (a) the work done per revolution of

the spindle, (b) the power required when the

spindle speed is 300 rev/min.

(a) Work done=Tθ,whereT= Fr

ForceF =50 N, radiusr=

40

2

=20 mm=

0 .02 m and angular displacement,θ=1 rev=

2 πrad.

Hence, work done per revolution of

spindle=Frθ=( 50 )( 0. 02 )( 2 π)= 6 .28 J

(b) Power,P= 2 πnT, where torque,T=Fr=

( 50 )( 0. 02 )=1 N m and speed,n=

300

60

=

5 rev/s.

Hence,power required,P= 2 π( 5 )( 1 )

= 31 .42 W.

Problem 8. A pulley is 600 mm in diameter

and the difference in tensions on the two

sides of the driving belt is 1.5 kN. If the

speed of the pulley is 500 rev/min, determine

(a) the torque developed, and (b) the work

done in 3 minutes.

(a) TorqueT =Fr, where forceF = 1 .5kN=

1500 N, and

radiusr=

600

2

=300 mm= 0 .3m.

Hence,torque developed=( 1500 )( 0. 3 )

=450 N m

(b) Work done=Tθ, where torqueT=450 N m

and angular displacement in 3 minutes

=( 3 × 500 )rev=( 3 × 500 × 2 π)rad.

Hence,work done=( 450 )( 3 × 500 × 2 π)=

4. 24 × 106 J= 4 .24 MJ

Problem 9. A motor connected to a shaft

develops a torque of 5 kN m. Determine the

number of revolutions made by the shaft if

the work done is 9 MJ.

Work done=Tθ, from which, angular displace-

ment,

θ=

work done

torque

Work done=9MJ= 9 × 106 J

and torque=5kNm=5000 N m.

Hence, angular displacement,

θ=

9 × 106

5000

=1800 rad.

2 πrad=1 rev, hence,

the number of revolutions made by the shaft

=

1800

2 π

= 286 .5revs