EASY
11th Tamil Nadu Board
IMPORTANT
Earn 100

What is radius of gyration?

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