A bead on a rotating helical wire

Proposed Solution by Alejandro Cárdenas-Avendaño Cárdenas-Avendaño (acardenasavendano.org)

A bead on a rotating helical wire A bead (mass m (mass  m)) slides frictionlessly on a wire which is wound in a helix about the vertical ( z ( z)) axis. The helix is described in cylindrical coordinates by r  =  R ϕ =  kz  k z + ϕ 0 , where k  where k  is  is the right-handed pitch of the helix. A motor is causing the helical wire to spin about the axis (picture a  ω t , where ω  zˆ. At t  barber’s pole) so that φ that φ0 =  ωt where ω is  is the constant rate of driving. Gravity is − g z. At  t  =  0 the bead is at rest  h . at the top of the wire, z wire,  z  =  h. a) Introduce coordinates and constraint equations to enable the calculation of all components of forces on the wire. Find the Lagrangian and equations of motion.

The helix is described in cylindrical coordinates by r = R φ = kz  + φ0 , where k  where k  is  is the the right handed pitch of the helix. Adding the motor causing the helical wire to spin we modify the last restriction as φ0  =  ωt  ω t, where ω where ω is  is the constant rate of driving. The latter conditions can be cast into two restriction functions as

= r−R = φ − kz − ω t,

C1 C2

(1) (2)

 φ  and z  z as and use r use r,,  φ and  as the coordinates to write the EOMs to get the components of forces on the wire. Nevertheless, it is important to note that, only z only  z is  is the generalized coordinate. Let us start by writing the energies to build the Lagrangian. The potential energy is V  =  mgz  m gz and the kinetic energy is T  =  =

1 m r˙ 2 + r2 φ˙ 2 + z˙ 2 2



so the Lagrangian is

1 m r˙ 2 + r2 φ˙ 2 + z˙ 2 2 which allows us to write the EOM for each variable as L  =

⇒

  

d ∂L ∂L − dt ∂r˙ ∂r d  ( mr˙ ) − mr  ˙φ2 dt mr¨ − mr  ˙φ2 =  λ 1 , d dt

⇒

   

∂L ∂L − ∂φ ∂ ˙φ d mr 2 φ˙ dt 2mr r˙  ˙φ + mr 2 φ¨  =  λ 2 , 1



−

 mgz, mgz ,

= λ1

∂C1 ∂r

= λ1

= λ2 = λ2

∂C2 ∂φ

Proposed Solution by Alejandro Cárdenas-Avendaño (acardenasavendano.org)

and d dt

⇒

  ∂L ∂ z˙

−

∂L ∂ z

= λ2

d  ( m z˙ ) + mg dt m z¨ + mg  = −k λ2 .

=

−

∂C2 ∂ z

k λ2

We can understand λ 1  as a normal force that the wire exerts outwards and λ 2  as the normal force that it exerts upwards. b) Solve for motion of the bead and find the forces of constraint. When does the bead hit the bottom, z = 0?

From the constraints we get that r˙  = r¨  =  0 and φ˙ φ¨

 z + ω = k ˙  z = k ¨

(3)

which allow us to write them as −

mR (k ˙  z + ω )2 mR 2 φ¨ φ¨ + g k 

From the last one we get m − k 

= λ1

(4)

= λ2

(5)

=

−

k  λ2 . m

 ¨φ + g  =  λ 2 . k 

 

Now, let us write this result in Eq. (5) to get m mR φ¨ = − k  2

 1  ¨φ + g R φ¨ + k  k  2

 

 ¨φ + g k 

 

= 0

R2 k 2 + 1  g φ¨ + = 0 k  k 2 φ¨ =

−

kg R2 k 2 + 1

subject to φ (0) =  kh and φ˙ (0) =  ω. Furthermore, let us define γ≡

kg R2 k 2 + 1

to write the non-homogeneous differential equation as φ¨  =

−

γ.

Therefore φ  =  kh + ωt − Using the constraint we also know that

2

γ 2 t . 2

(6)

Proposed Solution by Alejandro Cárdenas-Avendaño (acardenasavendano.org)

 z¨

=

−

γ k 

and finally  z  =  h −

γ 2 t . 2k 

(7)

Using this equation we can get one force from Eq. (4) as −

2 γ  t + ω k  2 −mR (−γt + ω )

  

mR k 

= λ1

−

= λ1

and therefore −

mR (ω − γt)2 =  λ 1 .

From Eq. (5) we get −

mR 2 γ  =  λ 2 .

(8)

Now, from Eq. (7) we can calculate when the beat hits the bottom, i.e., γ 2 t 2k 

0 = h− ±

t

= =

q  s 

4  γ 2k h

γ 2k 

±2

2k   h γ

where the solution to be physical must be the positive one, i.e., t  =  2

s 

2k   h. γ

3

.