Motion on Smooth Vertical Curves, Study notes of Mathematics

Download Motion on Smooth Vertical Curves and more Study notes Mathematics in PDF only on Docsity!

B.Sc. Mathematics – 2

nd

Semester

MTB 202 – Statics and Dynamics

by

Dr. Krishnendu Bhattacharyya

Department of Mathematics,

Institute of Science, Banaras Hindu University

Part – VI

Motion on Smooth Vertical Curves

Cycloid

Parametric equation: x  a t  sin t  and y  a (^)  1 cos t 

Parametric equation: x  a t  sin t  and y  a (^)  1 cos t 

I) Vertical Motion on a Smooth Circle

A. Downward motion along the inner surface:

Let the motion occur towards the bottom point from one end of a

horizontal diameter of a circle of radius a and under gravity only. The

other force on the particle will be the reaction of the surface in the

direction perpendicular to the tangent, i.e., along inward normal.

Let the particle of mass m be moving in the inner surface of the circle

OADB with the centre at C and let at time t P s  ,^  be the position of the

particle. Then OP = s and s  a .

Integrating (1) we get

2 2 cos 1

g c a

     

2

2

 a   2 ag cos  c

2 2

 v  s  2 ag cos  c 2

Now, at 0, , 0 2 0 2

t     v   c 

So,  

2 2

v^ ^2 ag^ cos^  or^ ,^ a ^ ^2 ag cos.

From (2),

2 cos   cos

R

g g m

  R  3 mg cos .

So, at the starting point the reaction is zero.

Now, if the time taken from

2

  to   0 is

1 t , then

1 0

0 2

sec 2

t (^) a dt d g

  

 





 

[ v  a    2 ag cos. For this case a    2 ag cos , as t

increases  decreases, i.e., the motion along  decreasing direction.]

Now, if the particle starts from a position very close to the vertex O , i.e.,

 is very small, then we can write sin   then the equation (1) becomes

g

a

B. Upward motion along the inner surface:

Let the particle be projected from the vertex O (^)  s ^ 0,^  ^0  with velocity

u. The equations of motion are same as previous,

i.e., ms^   mg sin^  ,

2

cos

v m R mg a

i.e., sin

g

a

    and

2

cos

v m R mg a

Hence, we have

2 v^ ^2 ag^ cos^   c.

Now, at

2

  0, v  u  c  u  2 ag.

So,  

2 2

v  u  2 ag 1 cos . Therefore

2

2 3 cos

u R m g g a

Let at    1 , u  0 and    2 , R  0.

Now

2

0 cos 1 1 2

u v ag

     and

2

2 1

0 cos 1 cos 3 2 3

u R ag

Thus for

2

u  2 ag ,cos  1  cos  2 or 1  2.

These all are true up to 2

 ^ 

So, v  0 before R  0 (up to 2

The particles come at rest and its contact with the surface at end A of

horizontal diameter also vanishes. From this position it comes down and

starts oscillating.

Now, when

2 u  2 ag. Also, let at the top the particle be at rest.

Then  

2 2

0  u  2 ag 1  cos   u  4 ag

Thus if the particles projected velocity is equal to 4 ag then it comes at

rest at top position.

Then R  m  2 g  3 g cos and it will be zero at

cos 131. 3

     .

It means for

2 u  4 ag the contact of the particle with the surface vanishes

before the top position.

Thus, the reaction become zero earlier to the linear velocity between

A  ^  and D   if

2 2 ag  u  4 ag

If the reaction at the top is zero, then

2 2 0 2 3 cos 5

u m g g u ag a

It means if the particle reaches the top then it’s projected velocity is equal

to 5 ag.

So, at

2 , 5 and 5. 3

 v  ag R  mg

 

2 , 4 and 4 2

 v  ag R  mg

 

, 3 and 2. 3

 v  ag R  mg

 

2

   , v  2 ag and R  mg.

Thus one may observed that the velocity and the reaction on the particle,

both decrease up to ^ . Beyond that both again increase and at the

lowest are

2 2 v  4 ag  2 ag cos 2   6 ag  u  4 3 cos 2 (^)  7

m R ag ag mg a

_Time of coming at position of no contact when projected with_*

2 u  4 ag

If  1 be angle at the point of no contact, then

1

2 3 cos 0

ag m g g a

1

cos 3

   . So,

1 1

cos 3

 ^ 

We have  

2

v  4 ag  2 ag 1 cos 

  ^ 

2 2 2 1 cos 4 cos 2

 a   ag    ag^ 

2 cos 2

 a   ag

 [ as t increases  decrease]

2

cos

g

a

