Mechanical Engineering Principles

188 MECHANICAL ENGINEERING PRINCIPLES

AtB,T= 0
Thus, weight=centrifugal force atB,

or mg=

mvB^2

r

from which, v^2 B=gr ( 16. 22 )

Substituting equation (16.22) into equation (16.21)
gives:

v^2 A=gr+ 4 gr= 5 gr

Hence,the minimum tangential velocity atA,

vA=

√

5 gr ( 16. 23 )

Problem 9. A mass of 0.1 kg is being

rotated in a vertical circle of radius 0.6 m. If

the mass is attached to a mass-less string and

the motion is such that the string is just taut

when the mass is at the top of the circle,

what is the tension in the string when it is

horizontal? Neglect losses and takegas

9.81 m/s^2.

At the top of the circle,

potential energy=PE = 2 mgr and KE=

mvT^2
2 ,
wherevT=velocity of mass at the top.
When the string ishorizontal,PE=mgr
and kinetic energy,

KE=

mv 12

2

,

wherev 1 =velocity of mass at this point.
From the conservation of energy,(PE+KE)at the
top=(PE+KE)when the string is horizontal

i.e. 2 mgr=mgr+

mv^21

2

−

mvT^2

2

but CF at top =

mvT^2

r

=mgorv^2 T=gr

or

v^21

2

= 2 gr−gr+

gr

2

=

3 gr

2

i.e. v^21 = 3 gr

and v 1 =

√

3 gr=

√

3 × 9. 81 × 0. 6

= 4 .202 m/s

Resolving forces horizontally,

Centrifugal force=T=tension in the string

Therefore,

T=

mv 12

r

=

0 .1kg×( 4. 202 )^2 m^2 /s^2

0 .6m

i.e.the tension in the string,T= 2 .943 N

Problem 10. What is the tension in the

string for Problem 9 when the mass is at the

bottom of the circle?

From equation (16.23), the velocity at the bottom of

the circle=v=

√

5 gr

i.e.v=

√

5 × 9. 81 × 0. 6 = 5 .4249 m/s.

Resolving forces vertically, T = tension in the

string=centrifugal force+the weight of the mass

i.e. T=

mv^2

r

+mg=m

(

v^2

r

+g

)

= 0. 1 ×

(

5. 42492

0. 6

+ 9. 81

)

= 0. 1 ×( 49. 05 + 9. 81 )= 0. 1 × 58 .86 N

i.e.tension in the string,T= 5 .886 N

Problem 11. If the mass of Problem 9 were

to rise, so that the string is at 45°to the

vertical axis and below the halfway mark,

what would be the tension in the string?

At 45°, PE=

mgr

2

andKE=

mv 22

2

wherev 2 =velocity of the mass at this stage.

From the conservation of energy,

PE+KE(at top)=(PE+KE)at this stage

Therefore, 2 mgr=

mgr

2

+

mv 22

2

−

mv^2 T

2

From problem 9, v^2 T=gr

hence

v^22

2

=

(

2 r−

r

2

+

r

2

)

g= 2 gr

from which, v^22 = 2 gr