Mechanical Engineering Principles

MOTION IN A CIRCLE 187

However, 1 kg m/s^2 =1 N, hence,tension in the
string,T=15 N
From equation (16.13),

T=

mg

cosθ

from which,

cosθ=

mg

T

=

0 .3kg× 9 .81 m/s^2

15 N

= 0. 1962

Hence, the cone angle,θ =cos−^1 ( 0. 1962 )

= 78. 685 °

From equation (16.15),

sinθ=

r

L

,

from which,radius of turning circle,

r=Lsinθ=2m×sin 78. 685 °= 1 .961 m

Exercise 81 Further problems on the con-

ical pendulum

1. A conical pendulum rotates about a hor-
   izontal circle at 100 rpm. If the speed of
   rotation of the mass increases by 5%, how
   much does the mass of the pendulum rise?
   [8.36 mm]


2. If the speed of rotation of the mass of
   Problem 1 decreases by 5%, how much
   does the mass fall? [9.66 mm]


3. A conical pendulum rotates at a horizontal
   angular velocity of 2 rad/s. If the length of
   the string is 3 m and the pendulum mass
   is 0.25 kg, determine the tension in the
   string. Determine also the radius of the
   turning circle. [3 N, 1.728 m]

16.4 Motion in a vertical circle

This Problem is best solved by energy considerations.
Consider a particleProtating in a vertical circle of
radiusrabout a pointO, as shown in Figure 16.6.
Neglect losses due to friction.
LetT=tension in a mass-less string,
r=radius of turning circle,
m=mass of particle.

q

q

T

mg

P C

A

O

r

B

T

Figure 16.6 Motion in a vertical circle

Problem 8. Determine the minimum

tangential velocity atA, namely,vA,which

will just keep the string taut at the pointB

for the particle moving in the vertical circle

of Figure 16.6.

At the pointBthe

potential energy=mg× 2 r( 16. 18 )

and the kinetic energy =

mv^2 B

2

( 16. 19 )

At the pointA, the kinetic energy

KE=

mvA^2

2

( 16. 20 )

and the potential energy

PE= 0

As there are no energy losses,

KE atA=(KE+PE)atB

Hence, from equations (16.18) to (16.20):

mv^2 A

2

=mg× 2 r+

mv^2 B

2

or

vA^2

2

=

vB^2

2

+ 2 gr

from which, v^2 A=v^2 B+ 4 gr ( 16. 21 )