2019-03-01_Physics_Times

becomes zero before tension vanishes, in this

case the particle oscillates about its lowest

position with angular amplitude 0     90

The angle made by the string with the vertical

when its velocity becomes zero is given by

2

cos 1

2

vA

gr

 .

(IV) Condition for leaving the circular path

without looping

If 2 gr v A 5 gr the particle is not able

to complete the vertical circle, it goes to

certain height and leaves the circular path

(90o<<180o) while leaving the circular path

T 0 but v 0

16. When car moves on a concave bridge

17. When car moves on a convex bridge

2

cos

mv

N mg

r

 

2

cos

mv

N mg

r

 

6. Centre of Mass, Linear Momentum & Collision

Consider a system of N point masses m 1 , m 2 , m 3 , ....
mn whose position vectors from origin O are given
by r r r1 2 3, , ,.......rn
   
respectively. Then the position
vector of the centre of mass C of the system is
given by.^

For continuous mass distribution the centre of

mass can be located by replacing summation sign

with an integral sign.

1 1 2 2

1 2

........

;

........

n n

CM

n

m r m r m r

r

m m m

  



  

  



m r m r1 1 2 2

or

2

1

2 1

m

r r

m m

 

 

   and

1. Centre of Mass 2. Position of C.M of two Particles

3. Centre of mass of a continuous mass

distribution

1

2

1 2

m

r r

m m

 

 

  

CM , CM , CM

xdm y dm z dm

x y z

dm dm dm

  

  

  