25.4. EXERCISES 473

5. ↑ Now show from Newton’s second law ( mass times acceleration equals force ) thatfor F the force,

ddt

(∂T∂ ẋk

)− ∂T

∂xk = m ÿ· ∂y

∂xk = F · ∂y∂xk . (∗∗∗)

6. ↑ In the example of the simple pendulum above,

y =

(l sinθ

l− l cosθ

)= l sinθ i+ (l− l cosθ)j.

Use ∗∗∗ to find a differential equation which describes the vibrations of the pendu-lum in terms of θ . First write the kinetic energy and then consider the force actingon the mass which is −mgj.

7. Of course, the idea is to write equations of motion in terms of the variables xk, insteadof the rectangular variables yk. Suppose y = y (x) and x is a function of t. Letting Gdenote the metric tensor, show that the kinetic energy is of the form 1

2 mẋT Gx wherem is a point mass with m its mass.

8. The pendulum problem is fairly easy to do without the formalism developed. Nowconsider the case where x = (ρ,θ ,φ) , spherical coordinates, and write differentialequations for ρ,θ , and φ to describe the motion of an object in terms of these coor-dinates given a force, F.

9. Suppose the pendulum is not assumed to vibrate in a plane. Let it be suspended atthe origin and let φ be the angle between the negative z axis and the positive x axiswhile θ is the angle between the projection of the position vector onto the xy planeand the positive x axis in the usual way. Thus

x = ρ sinφ cosθ ,y = ρ sinφ sinθ ,z =−ρ cosφ

10. If there are many masses, mα ,α = 1, · · · ,R, the kinetic energy is the sum of thekinetic energies of the individual masses. Thus,

T ≡ 12

R

∑α=1

mα |ẏα |2 .

Generalize the above problems to show that, assuming

yα = yα (x,t) ,

ddt

(∂T∂ ẋk

)− ∂T

∂xk =R

∑α=1

F α ·∂yα

∂xk

where F α is the force acting on mα .

11. Discuss the equivalence of these formulae with Newton’s second law, force equalsmass times acceleration. What is gained from the above so called Lagrangian for-malism?