# Hamilton's Principle Equations! Work Shown

1. Oct 28, 2009

### johnq2k7

A particle of mass m moves under the influence of gravity alng the helix z=k(theta), and r=R, where R and k are constants and z is vertical.

a.) Using cartesian co-ordinates, write down the expressions for the kinetic energy of the system.

b.) Change to cylindrical co-ordinate system using x= rcos(theta) y= rsin(theta)

express your eq for the kinetic energy as a function of the new co-ordinates. Give also expressions for the potential energy and the Langrangian of the system (in cylindrical co-ordinates).

c.) How many degrees of freedom do you have for this system? Name them (or it)

d.) Calculate the equation(s) of motion using the Langrange equation for this system.

my work:

a.) T= 1/2 m(x*time-deriv.)^2 + 1/2(y*time-deriv)^2 + 1/2(k(theta)*time-deriv.)^2

b.) if x= rcos(theta) and y=rsin(theta)

equation for kinetic energy in terms of cylin. co-ord is:
x= rcos(theta)
y= rsin(theta)
therefore, x-time deriv.= r*angle-timederiv. cos(theta)
y-time deriv.= r*angle-timederiv.sin(theta)

therefore, kinetic energy in cylin. co-ord.

T= (1/2)*(m)[(r*time-deriv)+(r^2*angletimederiv^2)+(ztimederiv^2)]

and therefore potential energy equals:
U=mgz

therefore the langrangian of the system in cylin. co-ord equals:

since z=k(theta) and r=R

therefore langrangian equals (L)= (1/2)*(m)[(R^2/k^2)(ztimederiv^2)+(ztimederiv^2)] -mgz

c.) need some help here, i believe the degree of freedom is six:

since in three dimensions, the six DOFs of a rigid body are sometimes described using these nautical names:
Moving up and down (heaving);
Moving left and right (swaying);
Moving forward and backward (surging);
Tilting forward and backward (pitching);
Turning left and right (yawing);
Tilting side to side (rolling).

not sure if this is correct need help
d.) using langrangian equation:

z-2ndtime deriv= g/[(r^2/k^2)+1]

need help proving that

2. Oct 28, 2009

### lanedance

hi john - i (& others) will able to help more if things are written a little easier to understand, so i've given some tips below

what is x*time-deriv ???
do you mean dx/dt?, if so use x', it will make things clearer, or even better use tex, click below to see code (or the sum sign in the advanced tab)
$$\frac{dx}{dt} = x'$$

also theta isn't a cartesian coordinate, if theta is to hard to write & you don;t want to use tex, give it a single letter varieable ie. q = theta

those derivatves (if i'm reading them right) don't look correct, now you have assumed r is constant , but what is
$$\frac{d}{dt} cos(\theta(t)) = ?$$

once again i can;t totally work out what you mean

that sounds like all the DOF for a rigid mass. Here you have a particle follwing a constrained path. A better question is how many parameters does it take to characterise the particles motion, I think that is the number of DOF

Last edited: Oct 28, 2009
3. Oct 28, 2009

### johnq2k7

my work:

a.) T= 1/2 m(x')^2 + 1/2(y')^2 + 1/2(k(theta)*z')^2 , I believe i made a mistake here need help with the cartesian eqn for the kinetic energy here

b.) if x= rcos(theta) and y=rsin(theta)

equation for kinetic energy in terms of cylin. co-ord is:
x= rcos(theta)
y= rsin(theta)
therefore, x'= r*theta'. cos(theta)
y'= r*theta'.sin(theta)

therefore, kinetic energy in cylin. co-ord.

T= (1/2)*(m)[(r')+(r^2*theta'^2)+(z'^2)]

i believe i made a few mistakes here, need some help

and therefore potential energy equals:
U=mgz

therefore the langrangian of the system in cylin. co-ord equals:

since z=k(theta) and r=R which is a const.

therefore langrangian equals (L)= (1/2)*(m)[(R^2/k^2)(z'^2)+(z'^2)] -mgz

c.) need some help here, i believe the degree of freedom is 3:

since in three dimensions and it is a rigid mass, and you said u have to use the parameters to explain the particle of motion, since it's three dimensional and it contains components for radius which is a constant R, and z which theta times a const. k, and z is vertical

how do i describe the DOF's and how many are there?
d.) using langrangian equation:

z''= g/[(r^2/k^2)+1]

I need help getting to this process

4. Oct 28, 2009

### lanedance

one other thing may help, if a function is a variable of something write it so, so you know what you're differentiating against eg. x = x(t)

and the dashes here correspond only to time derivatives
ok so if
v(t) = (x(t),y(t),z(t)))

the time derivative is
v'(t) = (x'(t),y'(t),z'(t)))

the kinetic energy is something like
T= (1/2)m v'(t).v'(t) = (1/2)m (x'(t)^2 + y'(t)^2 + z'(t)^2)

the only other thing you could do here is relate the x, y & z's together form the know equations, but that is much easier in cylindrical coordinates, so I would wait until next step

i would include z(t) = k.theta(t) here and differentiate that as well
ok so you need to note r is constant here before differntiating (ie r'(t) = 0)

the derivtives are still not right, what is

(d/dt) cos(theta(t)) = ?

update below with you r new derivtives & show how you simplify
this is not a rigid mass, it is a particle in sapce so has at maximum only has 3 DOF

so at the end of the day - how many independent (and changing varibales are there)

5. Oct 29, 2009

### johnq2k7

for a.) i got v(t)= 1/2*m*[x'(t)^2+ y'(t)^2 + z'(t)^2]

for cartesian co-ord. expression for KE
is this correct?

b.) i got 1/2*m*[(theta'(t)^2*r^2)+ z'(t)^2]

U=mgz

since L= T-U
since z=k(theta) z'(t)= k(theta'(t)

therefore sub. L= 1/2*m[ ((z'(t)^2)/(k^2))*(r^2)+z'(t)^2]- mgz

is this correct?

c.) for degrees of freedom, since it's a particle of space it has 3 dimensions so degrees of
freedom, how do i specify, do i say it's x,y, and z?

d.) i got the equation of motion as: z''(t)= g/ ((r^2/k^2) + 1) is this correct