2. Moment of Inertia Let the figure represent a rigid body which is rotating about a fixed axis, the angular velocity. With a suitable system of cylindrical polar coordinates, as follows; P - particle of mass dm, of the rigid body. Kinetic energy of P is.

3. Total kinetic energy of the body is where denotes the summation over the whole of the body. is called the moment of inertia of the body. Example 1 : A uniform rod of mass m and length 2a:

4. Normal Modes Consider the equation of motion of a given system. If it is of the form then the motion is periodic, By choosing generalized coordinates in a suitable way we can see the body vibrates or performs small oscillations. Such a combination of coordinates is called normal coordinates. So, if y is a normal coordinate. Then equation of motion is of the form Here is called normal frequency corresponding to the normal coordinate y. with period T

5. Mg x Zero potential energy level Lagrange’s equation becomes For small oscillations, above equation becomes Hence, period is Let 2a be the length of the rod, then Q#3

6. Now attach a particle of mass m to the free end. Mg x mg Now Lagrangian becomes Lagrange’s equation for this system is Hence if period is

7. Initially x Lagrangian of the system is Q#4

8. Lagrange’s equation becomes Using initial condition

9. When the particle comes to the level of the axis When the particle comes to the bottom

10. Tute : Q # 7 X Y Zero energy level

11. Lagrange’s equation is Consider the transformation Eq 1

12. by

13. Lagrangian of the system Q10

14. So Lagrange’s Equations are Second equation implies First equation becomes

15. Using initial conditions If Q just reach the hole then when

17. 6. A uniform rod of mass M is free to rotate about one end which is fixed and its period as a compound pendulum is T. A particle of mass m is attached to the free end of the rod and the period of the new compound pendulum is kT. Prove that (2 k 2 -3)m=(1-k 2 )M. Mg x Zero potential energy level Lagrange’s equation becomes Let 2a be the length of the rod, then Eg.

18. For small oscillations, above equation becomes Hence, period is 1 Now attach a particle of mass m to the free end. Mg x mg Lagrange’s equation for this system is

19. Hence if period kT is 2 2 1

20. E.g. consider double pendulum. Mg b mg a Kinetic energy of the system is

21. Potential energy of the system is

22. E.g. consider the following compound pendulum. mg O A B C G Here AB is a uniform rod of length 12a, OC is a string of length 5a and AC = 4a. v

23. Kinetic energy of the system is Potential energy of the system is

24. Lagrange’s equations are Eq 1 Eq 2 +.k Eq 1Eq 2 Find k such that

25. When k=1 the equation becomes is normal coordinate. When k= -2/5 is normal coordinate.