Mechanical Engineering Principles

LINEAR AND ANGULAR MOTION 131

Table 11.1

s =arc length (m) r =radius of circle (m)

t =time (s) θ =angle (rad)

v =linear velocity (m/s) ω =angular velocity (rad/s)

v 1 =initial linear velocity (m/s) ω 1 =initial angular velocity (rad/s)

v 2 =final linear velocity (m/s) ω 2 =final angular velocity (rad/s)

a =linear acceleration (m/s^2 ) α =angular acceleration (rad/s^2 )

n =speed of revolution (rev/s)

Equation number Linear motion Angular motion

(11.1) s=rθm

(11.2) 2 πrad= 360 °

(11.3) and (11.4) v=

s

t

ω=

θ

t

rad/s

(11.5) ω= 2 πnrad/s

(11.6) v=ωrm/s^2

(11.8) and (11.10) v 2 =(v 1 +at)m/s ω 2 =(ω 1 +αt)rad/s

(11.11) a=rαm/s^2

(11.12) and (11.13) s=

(

v 1 +v 2

2

)

tθ=

(

ω 1 +ω 2

2

)

t

(11.14) and (11.16) s=v 1 t+^12 at^2 θ=ω 1 t+^12 αt^2

(11.15) and (11.17) v 22 =v 12 +2as ω^22 =ω^21 + 2 αθ

From equation (11.13), angle turned through,

θ=

(

ω 1 +ω 2

2

)

t

=

⎛

⎜

⎝

300 × 2 π

60

+

800 × 2 π

60

2

⎞

⎟

⎠(^10 )rad

However, there are 2π radians in 1 revolution,
hence, number of revolutions

=

⎛

⎜

⎝

300 × 2 π

60

+

800 × 2 π

60

2

⎞

⎟

⎠

(

10

2 π

)

=

1

2

(

1100

60

)

( 10 )=

1100

12

= 91 .67 revolutions

Problem 6. The shaft of an electric motor,

initially at rest, accelerates uniformly for

0.4 s at 15 rad/s^2. Determine the angle (in

radians) turned through by the shaft in this

time.

From equation (11.16),

θ=ω 1 t+^12 αt^2

Since the shaft is initially at rest,ω 1 =0and

θ=^12 αt^2 ;

the angular acceleration,α= 15 rad/s^2 and time

t= 0 .4 s. Hence,angle turned through,

θ=^12 × 15 × 0. 42

= 1 .2rad

Problem 7. A flywheel accelerates

uniformly at 2.05 rad/s^2 until it is rotating at

1500 rev/min. If it completes 5 revolutions

during the time it is accelerating, determine

its initial angular velocity in rad/s, correct to

4 significant figures.

Since the final angular velocity is 1500 rev/min,

ω 2 = 1500

rev

min

×

1 min

60 s

×

2 πrad

1 rev

= 50 πrad/s