Hamilton's Principle Equations Work Shown

T= (1/2)m v'(t).v'(t) = (1/2)m (x'(t)^2 + y'(t)^2 + z'(t)^2)if v(t) = (x(t),y(t),z(t)))v'(t) = (x'(t),y'(t),z'(t))) so in cylindrical coordinatesx= rcos(theta) y= rsin(theta)z = k.thetav(t) = (r cos(theta),r sin(theta),k theta)v'(t) = (r' cos(theta) - r theta' sin(theta),r' sin(theta) + r theta' cos
  • #1
johnq2k7
64
0
Hamilton's Principle Equations! Work Shown Please Help!

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


PLEASE HELP A LOT OF WORK SHOWN NEED HELP HERE!
 
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  • #2


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

johnq2k7 said:
my work:

a.) T= 1/2 m(x*time-deriv.)^2 + 1/2(y*time-deriv)^2 + 1/2(k(theta)*time-deriv.)^2
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)
[tex] \frac{dx}{dt} = x' [/tex]

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

johnq2k7 said:
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)
those derivatves (if I'm reading them right) don't look correct, now you have assumed r is constant , but what is
[tex] \frac{d}{dt} cos(\theta(t)) = ? [/tex]

johnq2k7 said:
therefore, kinetic energy in cylin. co-ord.

T= (1/2)*(m)[(r*time-deriv)+(r^2*angletimederiv^2)+(ztimederiv^2)]
once again i can;t totally work out what you mean

johnq2k7 said:
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
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

johnq2k7 said:
d.) using langrangian equation:

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

need help proving that


PLEASE HELP A LOT OF WORK SHOWN NEED HELP HERE!
 
Last edited:
  • #3


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


PLEASE HELP I HOPE MY STEPS AND CLARIFICATION HELPS HERE.. NEED A LOT OF HELP
 
  • #4


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
johnq2k7 said:
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
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

johnq2k7 said:
b.) if x= rcos(theta) and y=rsin(theta)

equation for kinetic energy in terms of cylin. co-ord is:
i would include z(t) = k.theta(t) here and differentiate that as well
johnq2k7 said:
x= rcos(theta)
y= rsin(theta)
therefore, x'= r*theta'. cos(theta)
y'= r*theta'.sin(theta)
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
johnq2k7 said:
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
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)
johnq2k7 said:
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


PLEASE HELP I HOPE MY STEPS AND CLARIFICATION HELPS HERE.. NEED A LOT OF HELP
 
  • #5


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


please help work shown!
 

Related to Hamilton's Principle Equations Work Shown

1. What is Hamilton's principle?

Hamilton's principle is a fundamental law in classical mechanics that states that the motion of a physical system is determined by a single function called the "action". This principle is based on the conservation of energy and the principle of least action.

2. How is Hamilton's principle used in equations?

In equations, Hamilton's principle is used to derive the equations of motion for a system by minimizing the action function. This results in a set of differential equations known as the Hamilton's equations.

3. What is the significance of Hamilton's principle?

Hamilton's principle is significant because it provides a powerful tool for predicting the behavior of physical systems. It is used extensively in many fields of physics, including classical mechanics, quantum mechanics, and optics.

4. How is Hamilton's principle related to Newton's laws of motion?

Hamilton's principle is a more general and powerful form of Newton's laws of motion. It includes the principle of least action, which is not explicitly stated in Newton's laws, and can be used to derive the equations of motion from a single function rather than multiple laws.

5. Can Hamilton's principle be applied to all physical systems?

Yes, Hamilton's principle can be applied to all physical systems as long as they can be described by a Lagrangian function. This includes both classical and quantum systems, making it a very versatile and widely applicable principle in physics.

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