EASY
11th Tamil Nadu Board
IMPORTANT
Earn 100

A disc of moment of inertia Ia is rotating in a horizontal plane about its symmetry axis with a constant angular speed ω. Another disc initially at rest of moment of inertia Ib is dropped coaxially on to the rotating disc. Then, both the discs rotate with the same constant angular speed. The loss of kinetic energy due to friction in this process is :-

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Important Points to Remember in Chapter -1 - Motion of System of Particles and Rigid Bodies from Tamil Nadu Board Physics Standard 11 Vol I 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

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

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

10. 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.

11. 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

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

I=MK2

13. Torque:

(i) τ=r×F

(ii) τ=rFsinθ=rF=Fr

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

(i) L=r×P

(ii) L=rP sinθ=r P=P r

15. 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

16. 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.

17. Relation between Torque and Angular Momentum:

 τ=dLdt

18. Kinetic energy:

Total K.E. =12Mvcm2+12Icmω2