Mechanical Engineering Principles

186 MECHANICAL ENGINEERING PRINCIPLES

Equating equations (16.12) and (16.13) gives:

mω^2 r

sinθ

=

mg

cosθ

Rearranging gives:

mω^2 r

mg

=

sinθ

cosθ

i.e. tanθ=

ω^2 r

g

Hence, the cone angle,

θ=tan−^1

(

ω^2 r

g

)

( 16. 14 )

From Figure 16.5,

sinθ=

r

L

( 16. 15 )

Hence, from equation (16.12),

T=

mω^2 r

r

L

i.e.the tension in the string,

T=mω^2 L ( 16. 16 )

From equation (16.14),

ω^2 r

g

=tanθ

But, from Figure 16.5, tanθ=

r

h

Hence,

ω^2 r

g

=

r

h

and ω^2 =

g

h

Thus, angular velocity aboutC,

ω=

√

g

h

( 16. 17 )

Problem 6. A conical pendulum rotates

about a horizontal circle at 90 rpm. If the

speed of rotation of the mass increases by

10%, how much does the mass of the

pendulum rise (in mm)? Takegas 9.81 m/s^2.

Angular velocity,

ω=

2 πn

60

=

2 π× 90

60

= 9 .425 rad/s

From equation (16.17),

ω=

√

g

h

or ω^2 =

g

h

from which, height,

h=

g

ω^2

=

9. 81

9. 4252

= 0 .11044 m(see Figure 16.5)

When the speed of rotation rises by 10%,

n 2 = 90 × 1. 1 =99 rpm. Hence,

ω 2 =

2 πn 2

60

=

2 π× 99

60

= 10 .367 rad/s

From equation (16.17),

ω 2 =

√

g

h 2

or ω^22 =

g

h 2

Hence,

h 2 =

g

ω^22

=

9. 81

10. 3672

i.e. the new value of height,h 2 = 0 .09127 m.

Rise in height of the pendulum mass

=‘old’h−‘new’h

=h−h 2 = 0. 11044 − 0. 09127

= 0 .01917 m= 19 .17 mm

Problem 7. A conical pendulum rotates at a

horizontal angular velocity of 5 rad/s. If the

length of the string is 2 m and the pendulum

mass is 0.3 kg, determine the tension in the

string. Determine also the radius of the

turning circle. Takegas 9.81 m/s^2.

Angular velocity,ω=5 rad/s

From equation (16.16), tension in the string,

T=mω^2 L

= 0 .3kg×(5 rad/s)^2 ×2m

i.e. T=15 kg m/s^2