MEDIUM
11th CBSE
IMPORTANT
Earn 100

A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is ML23)

Important Points to Remember in Chapter -1 - Systems of Particles and Rotational Motion from NCERT PHYSICS PART 1 TEXTBOOK FOR CLASS XI Solutions

1. Centre of mass of a system of n discrete particles:

rcm= m1r1+m2r2++mnrnm1+m2+..+mn

2. Centre of mass of a continuous mass distribution:

xcm= xdmdm,ycm= ydmdm,zcm= zdmdm

3. Velocity of centre of mass of system:

(i) vcm= m1v1+m2v2++mnvnM

(ii) Momentum of a system given by, PSystem= Mvcm

4. Acceleration of centre of mass of system:

(i) acm= m1a1+m2a2++mnanM
(ii) The net force acting on a system is given by, Fext= Macm

5. Cross product:

(i) Cross product of two vectors A and B can be written as A×B= ABsinθ n^, where θ is the angle between the vectors, n^ is unit vector perpendicular to both A and B.

(ii) If A= Axi^+Ayj^ +Azk^ and B= Bxi^+Byj^ +Bzk^ then, A×B= i^j^k^AxAyAzBxByBz

6. Moment of Inertia:

(i) For a single particle, I= mr2 where m is the mass of the particle and r is the perpendicular distance of the particle from the axis about which moment of Inertia is to be calculated.

(ii) Moment of inertia of a system of particles: I= mr12+m2r22+...= I1+I2+I3+....

(iii) Moment of inertia of a continuous object: I= dmr2, where dm is the mass of a small element, r is the perpendicular distance of the element from the axis

7. Perpendicular Axis Theorem (Only applicable to plane lamina):

Iz= Ix+Iy (when object is in x-y plane).

8. Parallel Axis Theorem (Applicable to any type of object):

I= Icm+Md2, here Icm is the moment of inertia about an axis passing through centre of mass and parallel to required axis and d is the perpendicular distance between required axis and axis passing through centre of mass.

9. Moment of inertia of few structures:

(i) Solid sphere: 25MR2

(ii) Hollow sphere: 23MR2

(iii) Ring: MR2

(iv) Disc: 12MR2

(v) Hollow cylinder: MR2

(vi) Solid cylinder: 12MR2

(vii) Rod about centre:ML212

(viii) Rectangular plate: Ma2+b212

10. Moment of inertia (I) in terms of radius of gyration K:

I= MK2

11. Torque:

(i) τ= r×F

(ii) τ= rFsinθ= rF= Fr

12. Angular momentum L of a particle about a point:

(i) L= r×P

(ii) L= rP sinθ= r P= P r

13. Angular momentum of a rigid body:

(i) LH= IHω, where LH=angular momentum of object about axis HIH=Moment of Inertia of rigid object about axis Hω=angular velocity of the object.

(ii) LAB= Icmω+rcm×Mvcm

14. Conservation of angular momentum:

According to conservation of angular momentum, if τext= 0 about a point or axis of rotation, angular momentum of the particle or the system remains constant about that axis or point.

15. Relation between Torque and Angular Momentum:

τ= dLdt

16. Kinetic energy:

Total K.E. = 12Mvcm2+12Icmω2