Concepts based on location of CM

CONCEPTS BASED ON LOCATION OF CM

Concept (1) Addition of masses

Given a small disc of radius R, thickness t is put over a large disc of same material and thickness, radius 2R- Finding C.M. of Combined System of two disc.

Step:- Fixed your Origin

    

Here  is density and t is thickness so m α A

 

Area of small disc = p(R)² = pR²

 

           

Concept 2. Cutting of masses (Cavity)

In this case a small disc of radius R is cutting from a large disc of radius 2R. Now finding C.M. of remaining part of the disc

 

 So CM is located at left side form origin at distance R/3

Concep.3 Motion of CM

 

 

  

  

 

Concept 4. Shifting of C. M.

 

 

 

Example

Given if 2 kg block displace by 4 meter due right and 8 kg block displace by 2 meter due left then changing in position of C.M.

 

 

So C.M. will shift by 0.2 meter due left side

Example:

If 6 kg block moves towards down ward by 2 meter and if CM is to be Shift by +3 met in upward direction then amount and direction to which we have to shift 4 kg block.

D Y_cm = + 3 mt

 

 

30 = - 12 + 4 DY₂

Concept: 5 If F_ext = 0 then C.M. will not move

Example: Given a boat of mass 50 kg is stationary on a lake. Two Person A and B are initially at two ends of the boat. Now both start to move towards each other and reach the center of the boat. So what will be the distance and direction boat will move.

Solution

 

                       

                                

 

(x_cm)_f = 7.5 mt,

Boat displace

(x_cm)_f – (x_cm)_i

= 7.5  - 6.5

= 1 met

Boat will move due left by 1 meter

Example:  A trolley of mass 100 kg, length 10 mt along with a person of mass 50 kg is initially at rest. But now person moves towards other end of the trolley and reach to other end of the trolley.  During this process distance with respect to ground moved by person and trolley. Here we are assuming that trolley is on the smooth surface.

Solution

 

                               

 

                    

Trolly will move by (x_cm)_f  – (x_cm)_i

 6.6 – 3.3

       ≈ 3.3 met due left side

CONCEPT .6 ACCELERATION OF CM

Example

Finding a_cm = ?

8g - T = 8a

T - 6g = 6 a

2g = 14 a

 

 

                       

                         = -

                        

a_cm = - 0.2 m/s²

 

 

 

A uniform bar of length 12 L and mass 48 m is supported horizontally on two fixed smooth tables as shown in figure. A small moth (an insect) of mass 8m is sitting on end A of the rod and a spider (an insect) of mass 16 m is sitting on the other end. B. Both the insects moving towards each other along the rod with moving at speed 2v and then spider at half this speed (absolute). They met at appoint P on the rod and the spider eats the moth. After this the spider moves with a velocity  relative the end A. The spider takes negligible time in eating on the other . Also, let  where T is a having value 4s.

Question. Displacement of the rod by the time the insect meet the moth is

                                                    (b) L

                                                   (d) zero

Solution: p_i = p_f

   0 = (8m) (2v) – (16m) (v) + (48m) v’

Here v’ = absolute speed of rod

             = 0

   Displacement of rod = 0                                  Answer (d)

 

Question.. The point P is at

(a) The center of the rod

(b) The edge of the table supporting the end B

(c) The edge of the table supporting end A

(d) None of the above

Solution:

x_A + x_B = 12 L

   2vt + vt = 12 L

     vt = 4L

Þ   X_B = vt = 4L

Question. The speed of the rod after the spider eats up the moth and moves toward A is

                                                    (b) v

                                                     (d) 2v

Solution:

                                                                      

                                                                         Answer (c)

Question.. After starting from end B of the rod the spider reaches the end A at a time

(a) 40 s                                                (b) 30 s

(c) 80 s                                                (d) 10 s

Solution:

But   Lv=T=4s  

∴      t=80 s                                                              

                                                                          Answer (c)

Question. By what distance the center of mass of the rod shift during this time?

                                      

(c) L                                       

Solution: Till t₁, rod is stationary. For time t₂ rod is moving with absolute speed u (= v/6)

 

                                      Answer (a)