Mechanical Engineering Principles

MOTION IN A CIRCLE 185

Problem 4. What angle of banking of the

rails is required for Problem 3 above, for the

outer rail to have a zero value of thrust?

Assume the speed of the locomotive is

40 km/h.

From Problem 2, angle of banking,

θ=tan−^1

(

v^2

rg

)

v=40 km/h=

40

3. 6

= 11 .11 m/s

Hence, θ=tan−^1

(

11. 112 m^2 /s^2

700 m× 9 .81 m/s^2

)

=tan−^1 ( 0. 01798 )

i.e.angle of banking,θ= 1. 03 °

Exercise 80 Further problems on motion

in a circle

Where needed, takeg= 9 .81 m/s^2

1. A locomotive travels around a curve of
   500 m radius. If the horizontal thrust on
   the outer rail is 501 of the locomotive
   weight, determine the speed of the loco-
   motive. The surface that the rails are on
   may be assumed to be horizontal and the
   horizontal force on the inner rail may be
   assumed to be zero. [35.64 km/h]


2. If the horizontal thrust on the outer rail
   of Problem 1 is 1001 of the locomotive’s
   weight, determine its speed. [25.2 km/h]


3. What angle of banking of the rails of
   Problem 1 is required for the outer rail
   to have a zero value of outward thrust?
   Assume the speed of the locomotive is
   15 km/h. [0.203°]


4. What angle of banking of the rails is
   required for Problem 3, if the speed of
   the locomotive is 30 km/h? [0.811°]

16.3 Conical pendulum

If a massm were rotated at a constant angular
velocityω, in a horizontal circle of radiusr,by

a mass-less taut string of lengthL, its motion will

be in the form of a cone, as shown in Figure 16.5.

r

C

O

CF

P

T

T

mg

w

q

h

L

Figure 16.5 Conical pendulum

Let r=radius of horizontal turning circle,

L=length of string,

h=OC,

ω=constant angular velocity aboutC,

m=mass of particleP,

T=tension in string, and

θ=cone angle

Problem 5. Determine an expression for the

cone angleθand the tension in the stringT,

for the conical pendulum of Figure 16.5.

Determine also an expression forω.

Resolving forces horizontally gives:

CF=Tsinθ

i.e. mω^2 r=Tsinθ

from which, T=

mω^2 r

sinθ

( 16. 12 )

Resolving forces vertically gives:

Tcosθ=mg

from which, T=

mg

cosθ

( 16. 13 )